Author Archives: Priyanka

Exercise Ex. 13

Question 1

Draw a line segment AB = 5.6 cm and draw its perpendicular bisector. Measure the length of each part.Solution 1

Steps of construction:

1. Draw line segment AB = 5.6 cm
2. With A as centre and radius more than half of AB, draw two arcs, one on each side of AB.
3. With B as centre and the same radius as in step 2, draw arcs cutting the arcs drawn in the previous step at P and Q respectively.
4. Join PQ to intersect AB at M.

Thus, PQ is the required perpendicular bisector of AB.

AM = BM = 2.8 cmQuestion 2

Draw an angle of 80° with the help of a protractor and bisect it. Measure each part of the bisected angle.Solution 2

Steps of construction:

1. Draw ray OB.

2. With the help of protractor construct an angle AOB of measure 80°.

3. With centre O and convenient radius draw an arc cutting sides OA and OB at Q and P respectively.

4. With centre Q and radius more than half of PQ, draw an arc.

5. With centre P and the same radius, as in the previous step, draw another arc intersecting the arc drawn in the previous step at R.

6. Join OR and produce it to form ray OX.

Then, OX is the required bisector of ∠AOB.

∠AOX = ∠BOX = 40° Question 3

_{Construct an angle of  using ruler and compasses and bisect it.}Solution 3

Step of Construction:

(i) Draw a line segment OA.

(ii) With O as centre and any suitable radius draw an arc, cutting OA at B.

(iii) With B as centre and the same radius cut the previously drawn arc at C.

(iv) With C as centre and the same radius cut the arc at D.

(v) With C as centre and the radius more than half CD draw an arc.

(vi) With D as centre and the same radius draw another arc which cuts the previous arc at E.

(vii) Join ENow, AOE =90⁰

(viii) Now with B as centre and radius more than half of CB draw an arc.

(iv) With C as centre and same radius draw an arc which cuts the previousat F.

(x) Join OF.

_(xi) F is the bisector of right AOE.

Question 4(i)

Construct each of the following angles, using ruler and compasses:

75° Solution 4(i)

Steps of construction:

1. Draw a line segment PQ.
2. With centre P and any radius, draw an arc which intersects PQ at R.
3. With centre R and same radius, draw an arc which intersects previous arc at S.
4. With centre S and same radius, draw an arc which intersects arc in step 2 at T.
5. With centres T and S and radius more than half of TS, draw arcs intersecting each other at U.
6. Join PU which intersects arc in step 2 at V.
7. With centres V and S and radius more than half of VS, draw arcs intersecting each other at W.
8. Join PW.

∠WPQ = 75° 

Question 4(ii)

Construct each of the following angles, using ruler and compasses:

37.5° Solution 4(ii)

Steps of construction:

1. Draw a line segment PQ.
2. With centre P and any radius, draw an arc which intersects PQ at R.
3. With centre R and same radius, draw an arc which intersects previous arc at S.
4. With centre S and same radius, draw an arc which intersects arc in step 2 at T.
5. With centres T and S and radius more than half of TS, draw arcs intersecting each other at U.
6. Join PU which intersects arc in step 2 at V.
7. With centres V and S and radius more than half of VS, draw arcs intersecting each other at W.
8. Join PW. ∠WPQ = 75°
9. Bisect ∠WPQ.

Then, ∠ZPQ = 37.5° 

Question 4(iii)

Construct each of the following angles, using ruler and compasses:

135° Solution 4(iii)

Steps of construction:

1. Draw a line segment AB and produce BA to point C.
2. With centre A and any radius, draw an arc which intersects AC at D and AB at E.
3. With centres D and E and radius more than half of DE, draw two arcs which intersect each other at F.
4. Join FA which intersects the arc in step 2 at G.
5. With centres G and D and radius more than half of GD, draw two arcs which intersect each other at H.
6. Join HA.

Then, ∠HAB = 135° 

Question 4(iv)

Construct each of the following angles, using ruler and compasses:

105° Solution 4(iv)

Steps of construction:

1. Draw a line segment PQ.
2. With centre P and any radius, draw an arc which intersects PQ at R.
3. With centre R and same radius, draw an arc which intersects previous arc at S.
4. With centre S and same radius, draw an arc which intersects arc in step 2 at T.
5. With centres T and S and radius more than half of TS, draw arcs intersecting each other at U.
6. Join PU which intersects arc in step 2 at V.
7. Now taking T and V as centres, draw arcs with radius more than half the length TV. 
8. Let these arcs intersect each other at W. 
9. Join PW, which is the required ray making 105°with the given ray PQ. 

 Then, ∠WPQ = 105° 

Question 4(v)

Construct each of the following angles, using ruler and compasses:

22.5° Solution 4(v)

Steps of construction:

1. Draw a line segment AB.
2. With centre A and any radius, draw an arc which intersects AB at C.
3. With centre C and same radius, draw an arc which intersects previous arc at D.
4. With centre D and same radius, draw an arc which intersects arc in step 2 at E.
5. With centres E and D and radius more than half of ED, draw arcs intersecting each other at F.
6. Join AF which intersects arc in step 2 at G.
7. Now taking G and C as centres, draw arcs with radius more than half the length GC. 
8. Let these arcs intersect each other at H. 
9. Join AH which intersect the arc n step 2 at I. 
10. With centres I and C and radius more than half of IC, draw arcs intersecting each other at J. 
11. Join AJ. 

 Then, ∠JAB = 22.5° 

Question 5

Construct a Δ ABC in which BC = 5 cm, AB = 3.8 cm and AC = 2.6 cm. Bisect the largest angle of this triangle.Solution 5

Steps of construction:

1. Draw line segment AC = 2.6 cm.
2. With A as centre and radius 3.8 cm, draw an arc.
3. With C as centre and radius 5 cm, draw arc to intersect the previous arc at B.
4. Join AB and BC.

Thus, ΔABC is the required triangle.

Largest side = BC = 5 cm

⇒ Largest angle = ∠A

Steps of construction:

1. With A as centre and any radius, draw an arc, which intersect AB at P and AC at Q.
2. With P as centre and radius more than half of PQ, draw an arc.
3. With Q as centre and the same radius, draw an arac to intersect the previous arc at R.
4. Join AR and extend it.

Thus, ∠A is bisected by ray AR.

Question 6

Construct a ΔABC in which BC = 4.8 cm, ∠B = 45° and ∠C = 75°. Measure ∠A.Solution 6

Steps of construction:

1. Draw a line segment BC = 4.8 cm.
2. With centre B and any radius, draw an arc which intersects BC at P.
3. With centre P and same radius, draw an arc which intersects previous arc at Q.
4. With centre Q and same radius, draw an arc which intersects arc in step 2 at R.
5. With centres R and Q and radius more than half of RQ, draw arcs intersecting each other at S.
6. Join BS which intersects arc in step 2 at G. Then, ∠SBC = 90°
7. With centre P and radius more than half of PG, draw an arc.
8. With centre G and same radius, draw an arc which intersects previous arc at X.
9. Join B and extend it. Then ∠B = 45°
10. Construct ∠TCB = 90° following the steps given above.  
11. With centres M and H and radius more than half of MH, draw arcs intersecting each other at Y.
12. Join CY and extend it. Then, ∠C = 75°
13. Extended BX and CY intersect at A.

Thus, ΔABC is the required triangle.

m∠A = 60° 

Question 7

_{Construct an equilateral trianagle each of whose sides measures 5cm.}Solution 7

Step of construction:

(i) Draw a line segment BC=5cm.

(ii) With B as centre and radius equal to BC draw an arc. Question 8

Construct an equilateral triangle each of whose altitudes measures 5.4 cm. Measure each of its sides.Solution 8

Steps of construction:

1. Draw a line XY.
2. Mark any point P on it.
3. From P, draw PQ ⊥ XY.
4. From P, set off PA = 5.4 cm, cutting PQ at A.
5. Construct ∠PAB = 30° and ∠PAC = 30°, meeting XY at B and C respectively.

Then, ABC is the required equilateral triangle.

Question 9

Construct a right-angled triangle whose hypotenuse measures 5 cm and the length of one whose sides containing the right angle measures 4.5 cm.Solution 9

Steps of construction:

1. Draw a line segment BC = 5 cm.
2. Find the midpoint O of BC.
3. With O as centre and radius OB, draw a semicircle on BC.
4. With B as centre and radius equal to 4.5 cm, draw an arc, cutting the semicircle at A.
5. Join AB and AC.

Then, ΔABC is the required triangle.

Question 10

Construct a ΔABC in which BC = 4.5 cm, ∠B = 45° and AB + AC = 8 cm. Justify your construction.Solution 10

Steps of construction:

1. Draw BC = 4.5 cm
2. Draw ∠CBX = 45° 
3. From ray BX, cut-off line segment BD equal to AB + AC, i.e. 8 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD meeting BD at A.
6. Join CA to obtain the required triangle ABC.

Justification:

Clearly, A lies on the perpendicular bisector of CD.

∴ AC = AD

Now, BD = 8 cm

⇒ BA + AD = 8 cm

⇒ AB + AC = 8 cm

Hence, ΔABC is the required triangle.Question 11

Construct a ΔABC in which AB = 5.8 cm, ∠B = 60° and BC + CA = 8.4 cm. Justify your construction.Solution 11

Steps of construction:

1. Draw AB = 5.8 cm
2. Draw ∠ABX = 60° 
3. From ray BX, cut off line segment BD = BC + CA = 8.4 cm.
4. Join AD.
5. Draw the perpendicular bisector of AD meeting BD at C.
6. Join AC to obtain the required triangle ABC.

Justification:

Clearly, C lies on the perpendicular bisector of AD.

∴ CA = CD

Now, BD = 8.4 cm

⇒ BC + CD = 8.4 cm

⇒ BC + CA = 8.4 cm

Hence, ΔABC is the required triangle.Question 12

Construct a ΔABC in which BC = 6 cm, ∠B = 30° and AB – AC = 3.5 cm. Justify your construction.Solution 12

Steps of construction:

1. Draw base BC = 6 cm
2. Construct ∠CBX = 30° 
3. From ray BX, cut off line segment BD = 3.5 cm (= AB – AC)
4. Join CD.
5. Draw the perpendicular bisector of CD which cuts BX at A.
6. Join CA to obtain the required triangle ABC.

Justification:

Since A lies on the perpendicular bisector of CD.

∴ AD = AC

Now, BD = 3.5 cm

⇒ AB – AD = 3.5 cm

⇒ AB – AC = 3.5 cm

Hence, ΔABC is the required triangle.Question 13

Construct a ΔABC in which base AB = 5 cm, ∠A = 30° and AC – BC = 2.5 cm. Justify your construction.Solution 13

Steps of construction:

1. Draw base AB = 5 cm
2. Construct ∠BAX = 30° 
3. From ray AX, cut off line segment AD = 2.5 cm (= AC – BC)
4. Join BD.
5. Draw the perpendicular bisector of BD which cuts AX at C.
6. Join BC to obtain the required triangle ABC.

Justification:

Since C lies on the perpendicular bisector of BD.

∴ CD = BC

Now, AD = 2.5 cm

⇒ AC – CD = 2.5 cm

⇒ AC – BC = 2.5 cm

Hence, ΔABC is the required triangle. Question 14

_{Construct a  whose perimeter is 12 cm and the lengths of whose sides are in the ratio 3:2:4.}Solution 14

Question 15

Construct a triangle whose perimeter is 10.4 cm and the base angles are 45° and 120°.Solution 15

Steps of Construction:

1. Draw a line segment PQ = 10.4 cm.
2. Construct a 45° angle and bisect it to get ∠NPQ. 
3. Construct a 120° angle and bisect it to get ∠MQP. 
4. Let the rays PN and QM intersect at A.
5. Construct the perpendicular bisectors of PA and QA, to intersect PQ at B and C respectively.
6. Join AB and AC.

So, ΔABC is the required triangle.

Question 16

Construct a ΔABC whose perimeter is 11.6 cm and the base angles are 45° and 60°.Solution 16

Steps of Construction:

1. Draw a line segment PQ = 11.6 cm.
2. Construct a 45° angle and bisect it to get ∠NPQ. 
3. Construct a 60° angle and bisect it to get ∠MQP. 
4. Let the rays PN and QM intersect at A.
5. Construct the perpendicular bisectors of PA and QA, to intersect PQ at B and C respectively.
6. Join AB and AC.

So, ΔABC is the required triangle.

Question 17(i)

In each of the following cases, given reasons to show that the construction of ΔABC is not possible:

AB = 6 cm, ∠A = 40° and (BC + AC) = 5.8 cm.Solution 17(i)

Given,

AB = 6 cm, ∠A = 40° and (BC + AC) = 5.8 cm

We know that the sum of any two sides of a triangle is greater than the third side.

Here, we find that BC + AC < AB

Hence, construction of triangle ABC with given measurements is not possible. Question 17(ii)

In each of the following cases, given reasons to show that the construction of ΔABC is not possible:

AB = 7 cm, ∠A = 50° and (BC – AC) = 8 cm.Solution 17(ii)

Given,

AB = 7 cm, ∠A = 50° and (BC – AC) = 8 cm

We know that the sum of any two sides of a triangle is greater than the third side.

That is,

AB + AC > BC

⇒ AB + AC – AC > BC – AC

⇒ AB > BC – AC

Here, we find that AB < BC – AC

Hence, construction of triangle ABC with given measurements is not possible. Question 17(iii)

In each of the following cases, given reasons to show that the construction of ΔABC is not possible:

BC = 5 cm, ∠B = 80°, ∠C = 50° and ∠A = 60°.Solution 17(iii)

Given,

BC = 5 cm, ∠B = 80°, ∠C = 50° and ∠A = 60° 

We know that the sum of the measures of three angles of a triangle is 180°.

Here, we find that

∠A + ∠B + ∠C = 60° + 80° + 50° = 190° > 180° 

Hence, construction of triangle ABC with given measurements is not possible. Question 17(iv)

In each of the following cases, given reasons to show that the construction of ΔABC is not possible:

AB = 4 cm, BC = 3 cm and AC = 7 cm.Solution 17(iv)

Given,

AB = 4 cm, BC = 3 cm and AC = 7 cm

We know that the sum of any two sides of a triangle is greater than the third side.

Here, we find that AB + BC = AC

Hence, construction of triangle ABC with given measurements is not possible.Question 18

Construct an angle of 67.5° by using the ruler and compasses.Solution 18

Steps for Construction:

1. Draw a line XY.

2. Take a point A on XY.

3. With A as the centre, draw a semi-circle, cutting XY at P and Q.

4. Construct ∠YAC = 90°.

5. Draw the bisector AB of ∠XAC. Then ∠YAB = 135°.

6. Draw the bisector AM of ∠YAB. Then ∠YAM = 67.5°. 

Question 19

Construct a square of side 4 cm.Solution 19

Steps of Construction:

1. Draw a line segment PQ = 4 cm.

2. Construct ∠QPX = 90° and ∠PQY = 90°.

3. Cut an arc PS = 4 cm and QR = 4 cm. Join SR.

So, PQRS is the required square.

Question 20

Construct a right triangle whose one side is 3.5 cm and the sum of the other side and the hypotenuse is 5.5 cm.Solution 20

Steps of construction:

1. Draw BC = 3.5 cm
2. Draw ∠CBX = 90° 
3. From ray BX, cut off line segment BD = AB + AC = 5.5 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD meeting BD at A.
6. Join AC to obtain the required triangle ABC.

Question 21

Construct a ΔABC in which ∠B = 45°, ∠C = 60° and the perpendicular from the vertex A to base BC is 4.5 cm.Solution 21

Steps of construction:

1. Draw any line XY.
2. Take any point P on XY and draw PQ ⊥ XY.
3. Along PQ, set off PA = 4.5 cm.
4. Through A, draw LM ∥ XY.
5. Construct ∠LAB = 45° and ∠MAC = 60°, meeting XY at B and C respectively.

Then, ΔABC is the required triangle. 

Exercise MCQ

Question 1

In a ∆ABC it is given that base = 12 cm and height = 5 cm. Its area is

(a) 60 cm²

(b) 30 cm²

(d) 45 cm²Solution 1

Question 2

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is

(a) 96 cm²

(b) 120 cm²

(c) 144 cm²

(d) 160 cm²Solution 2

Question 3

Each side of an equilateral triangle measures 8 cm. The area of the triangle is

Solution 3

Question 4

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is

Solution 4

Question 5

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of triangle is

Solution 5

Question 6

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is

Solution 6

Question 7

Each side of an equilateral triangle is 10 cm long. The height of the triangle is

Solution 7

Question 8

The height of an equilateral triangle is 6 cm. Its area is

Solution 8

Question 9

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is

(a) 480 m²

(b) 320m²

(c) 384 m²

(d) 360m²Solution 9

Question 10

The sides of a triangle are in the ratio 5:12:13 and its perimeter is 150 cm. The area of the triangle is

(a) 375 cm²

(b) 750 cm²

(c) 250 cm²

(d) 500 cm²Solution 10

Question 11

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is

(a) 24 cm

(b) 18 cm

(c) 30 cm

(d) 12 cmSolution 11

Question 12

The base of an isosceles triangle is 16 cm and its area is 48 cm². The perimeter of the triangle is

(a) 41 cm

(b) 36 cm

(c) 48 cm

(d) 324 cmSolution 12

Question 13

Solution 13

Question 14

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is

(a)156 cm²

(b)78 cm²

(c) 60 cm²

(d) 120 cm²Solution 14

Question 15

The base of a right triangle is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is

(a) 168 cm²

(b) 252 cm²

(c) 336 cm²

(d) 504 cm²Solution 15

Question 16

Solution 16

Exercise Ex. 14

Question 1

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.Solution 1

Question 2

The base of the triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.Solution 2

Question 3

Find the area of triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.Solution 3

Question 4

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.Solution 4

Question 5

Find the area of a triangular field whose sides are 91m, 98 m, 105m in length. Find the height corresponding to the longest side.Solution 5

Question 6

The sides of a triangle are in the ratio 5: 12:13 and its perimeter is 150 m. Find the area of the triangle.Solution 6

Question 7

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at Rs. 5 per m².Solution 7

It is given that the sides a, b, c of the triangle are in the ratio 25 : 17 : 12,

i.e. a : b : c = 25 : 17 : 12

⇒ a = 25x, b = 17x and c = 12x

Given, perimeter = 540 m

⇒ 25x + 17x + 12x = 540

⇒ 54x = 540

⇒ x = 10

So, the sides of the triangle are

a = 25x = 25(10) = 250 m

b = 17x = 17(10) = 170 m

c = 12x = 12(10) = 120 m

Cost of ploughing the field = Rs. 5/m²

⇒ Cost of ploughing 9000 m² = Rs. (5 × 9000) = Rs. 45,000Question 8

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.Solution 8

Question 9

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measure is 20 cm.Solution 9

Question 10

The base of the isosceles triangle measures 80 cm and its area is 360cm².Find the perimeter of the triangle.Solution 10

Question 11

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

HINT Ratio of sides = 3 : 3 : 2.

* Back answer incorrectSolution 11

It is given that the ratio of equal side to its base is 3 : 2.

⇒ Ratio of sides of isosceles triangle = 3 : 3 : 2

i.e. a : b : c = 3 : 3 : 2

⇒ a = 3x, b = 3x and c = 2x

Given, perimeter = 32 cm

⇒ 3x + 3x + 2x = 32

⇒ 8x = 32

⇒ x = 4

So, the sides of the triangle are

a = 3x = 3(4) = 12 cm

b = 3x = 3(4) = 12 cm

c = 2x = 2(4) = 8 cm

Question 12

The perimeter of a triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.Solution 12

Let the three sides of a triangle be a, b and c respectively such that c is the smallest side.

Then, we have

a = c + 4

And, b = 2c – 6

Given, perimeter = 50 cm

⇒ a + b + c = 50

⇒ (c + 4) + (2c – 6) + c = 50

⇒ 4c – 2 = 50

⇒ 4c = 52

⇒ c = 13

So, the sides of the triangle are

a = c + 4 = 13 + 4 = 17 cm

b = 2c – 6 = 2(13) – 6 = 20 cm

c = 13 cm

Question 13

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m, 15 m. The advertisements yield an earning of Rs.2000 per m² a year. A company hired one of its walls for 6 months. How much rent did it pay?Solution 13

Three sides of a wall are 13 m, 14 m and 15 m respectively.

i.e.

a = 13 m, b = 14 m and c = 15 m

Rent for a year = Rs. 2000/m²

⇒ Rent for 6 months = Rs. 1000/m²

Thus, total rent paid for 6 months = Rs. (1000 × 84) = Rs. 84,000 Question 14

The perimeter of the isosceles triangle is 42 cm and its base is times with each of the equal sides. Find (i) the length of the equal side of the triangle (ii) the area of the triangle (iii)the height of the triangle. (Given, )Solution 14

Question 15

If the area of an equilateral triangle is cm², find its perimeter.Solution 15

Question 16

If the area of an equilateral triangle is cm², find its height.Solution 16

Question 17

Each side of the equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take =1.732.Solution 17

(i) Area of an equilateral triangle=

Where a is the side of the equilateral triangle

Question 18

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take =1.732.Solution 18

Question 19

The base of the right -angled triangle measures 48 cm and its hypotenuse measures 50 cm; find the area of the triangle.Solution 19

Question 20

Find the area of the shaded region in the figure given below.

Solution 20

In right triangle ADB, by Pythagoras theorem,

AB² = AD² + BD² = 12² + 16² = 144 + 256 = 400

⇒ AB = 20 cm

For ΔABC, 

Thus, area of shaded region

= Area of ΔABC – Area of ΔABD

= (480 – 96) cm²

= 384 cm²Question 21

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, )Solution 21

Let ABCD be the given quadrilateral such that ∠ABC = 90° and AB = 6 cm, BC = 8 cm, CD = 12 cm and AD = 14 cm.

In ΔABC, by Pythagoras theorem,

AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100

⇒ AC = 10 cm

In ΔACD, AC = 10 cm, CD = 12 cm and AD = 14 cm

Let a = 10 cm, b = 12 cm and c = 14 cm

Thus, area of quadrilateral ABCD

= A(ΔABC) + A(ΔACD)

= (24 + 58.8) cm²

= 82.8 cm²Question 22

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 CM and ∠ABD = 90°.

Solution 22

In ΔABD, by Pythagoras theorem,

AB² = AD² – BD² = 17² – 15² = 289 – 225 = 64

⇒ AB = 8 cm

∴ Perimeter of quadrilateral ABCD = AB + BC + CD + AD

= 8 + 12 + 9 + 17

= 46 cm

In ΔBCD, BC = 12 cm, CD = 9 cm and BD = 15 cm

Let a = 12 cm, b = 9 cm and c = 15 cm

Thus, area of quadrilateral ABCD

= A(ΔABD) + A(ΔBCD)

= (60 + 54) cm²

= 114 cm² Question 23

Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, ∠BAC = 90°, AC = 20 cm, CD = 42 cm and AD = 34 cm.

Solution 23

In ΔBAC, by Pythagoras theorem,

BC² = AC² + AB² = 20² + 21² = 400 + 441 = 841

⇒ BC = 29 cm

∴ Perimeter of quadrilateral ABCD = AB + BC + CD + AD

= 21 + 29 + 42 + 34

= 126 cm

In ΔACD, AC = 20 cm, CD = 42 cm and AD = 34 cm

Let a = 20 cm, b = 42 cm and c = 34 cm

Thus, area of quadrilateral ABCD

= A(ΔABC) + A(ΔACD)

= (210 + 336) cm²

= 546 cm² Question 24

Find the area of the quadrilateral ABCD in which BCD is an equilateral triangle, each of whose side is 26 cm, AD=24 cm and.Also, find the perimeter of the quadrilateral [Given =1.73]

Solution 24

Perimeter of quad. ABCD = AB + BC + CD + DA = 10 + 26 + 26 + 24 = 86 cmQuestion 25

Find the area of a parallelogram ABCD in which AB= 28 cm , BC=26 cm and diagonal AC=30 cm.

Solution 25

Question 26

Find the area parallelogram ABCD in which AB=14 cm, BC=10 cm and AC= 16 cm. [Given =1.73]

Solution 26

Question 27

In the given figure ABCD is a quadrilateral in which diagonal BD=64 cm, AL  BD and CM BD such that AL= 16.8 cm and CM=13.2 cm. Calculate the area of quadrilateral ABCD.

Solution 27

Question 28

The area of a trapezium is 475 cm² and its height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.Solution 28

Let the smaller parallel side of trapezium = x cm

Then, larger parallel side = (x + 4) cm

Thus, the lengths of two parallel sides are 23 cm and 27 cm respectively.Question 29

In the given figure, a ΔABC has been given in which AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE of the same area as that of ΔABC is constructed. Find the height DL of the parallelogram.

Solution 29

In ΔABC, AB = 7.5 cm, BC = 7 cm and AC = 6.5 cm

Let a = 7.5 cm, b = 7 cm and c = 6.5 cm

Question 30

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs.5 to plough 1 m² of the field, find the total cost of ploughing the field.Solution 30

Construction: Draw BT ⊥ CD

In ΔBTC, by Pythagoras theorem,

BT² = BC² – CT² = 100² – 60² = 10000 – 3600 = 6400

⇒ BT = 80 m

⇒ AD = BT = 80 m

Cost of ploughing 1 m² field = Rs. 5

⇒ Cost of ploughing 4800 m² field = Rs. (5 × 4800) = Rs. 24,000Question 31

A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3-m-wide space should be left in the front and back each and 2 m wide space on each of the sides. Find the largest area where house can be constructed.Solution 31

Length of rectangular plot = 40 m

Width of rectangular plot = 15 m

Keeping 3 m wide space in the front and back,

length of rectangular plot = 40 – 3 – 3 = 34 m

Keeping 2 m wide space on both the sides,

width of rectangular plot = 15 – 2 – 2 = 11 m

Thus, largest area where house can be constructed

= 34 m × 11 m

= 374 m²Question 32

A rhombus -shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs.5 per cm². Find the cost of painting.Solution 32

Let ABCD be the rhombus-shaped sheet.

Perimeter = 40 cm

⇒ 4 × Side = 40 cm

⇒ Side = 10 cm

⇒ AB = BC = CD = AD = 10 cm

Let diagonal AC = 12 cm

Since diagonals of a rhombus bisect each other at right angles,

AO = OC = 6 cm

In right ΔAOD, by Pythagoras theorem,

OD² = AD² – AO² = 10² – 6² = 100 – 36 = 64

⇒ OD = 8 cm

⇒ BD = 2 × OD = 2 × 8 = 16 cm

Now,

Cost of painting = Rs. 5/cm²

∴ Cost of painting rhombus on both sides = Rs. 5 × (96 + 96)

= Rs. 5 × 192

= Rs. 960Question 33

The difference between the semiperimeter and sides of a ΔABC are 8 cm, 7 cm, and 5 cm respectively. Find the area of the triangle.Solution 33

Let the sides of a triangle be a, b, c respectively and ‘s’ be its semi-perimeter.

Then, we have

s – a = 8 cm

s – b = 7 cm

s – c = 5 cm

Now, (s – a) + (s – b) + (s – c) = 8 + 7 + 5

⇒ 3s – (a + b + c) = 20

⇒ 3s – 2s = 20

⇒ s = 20

Thus, we have

a = s – 8 = 20 – 8 = 12 cm

b = s – 7 = 20 – 7 = 13 cm

c = s – 5 = 20 – 5 = 15 cm

Question 34

A floral design on a floor is made up of 16 tiles , each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm.

Solution 34

Question 35

An umbrella is made by stitching by 12 triangular pieces of cloth, each measuring ((50 cmx20 cm x50 cm).Find the area of the cloth used in it.

Solution 35

Question 36

In the given figure, ABCD is a square with diagonal 44 cm. How much paper of each shade is needed to make a kite given in the figure?

Solution 36

 In ΔAEF, AE = 20 cm, EF = 14 cm and AF = 20 cm

Let a = 20 cm, b = 14 cm and c = 20 cm

Question 37

A rectangular lawn, 75 m by 60 m, has two roads, each road 4 m wide, running through the middle of the lawn, one parallel to length and the other parallel to breadth, as shown in the figure. Find the cost of gravelling the roads at Rs.50 per m².

Solution 37

For road ABCD, i.e. for rectangle ABCD,

Length = 75 m

Breadth = 4 m

Area of road ABCD = Length × Breadth = 75 m × 4m = 300 m²

For road PQRS, i.e. for rectangle PQRS,

Length = 60 m

Breadth = 4 m

Area of road PQRS = Length × Breadth = 60 m × 4 m = 240 m²

For road EFGH, i.e. for square EFGH,

Side = 4 m

Area of road EFGH = (Side)² = (4)² = 16 m²

Total area of road for gravelling

= Area of road ABCD + Area of road PQRS – Area of road EFGH

= 300 + 240 – 16

= 524 m²

Cost of gravelling the road = Rs. 50 per m²

∴ Cost of gravelling 524 m² road = Rs. (50 × 524) = Rs. 26,200Question 38

The shape of the cross section of canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 m², find the depth of the canal.Solution 38

Area of cross section = Area of trapezium = 640 m²

Length of top + Length of bottom

= sum of parallel sides

= 10 m + 6 m

= 16 m

Thus, the depth of the canal is 80 m.Question 39

Find the area of a trapezium whose parallel sides are 11 m and 25 m long, and the nonparallel sides are 15 m and 13 m long.Solution 39

From C, draw CE ∥ DA.

Clearly, ADCE is a parallelogram having AD ∥ EC and AE ∥ DC such that AD = 13 m and D = 11 m.

AE = DC = 11 m and EC = AD = 13 m

⇒ BE = AB – AE = 25 – 11 = 14 m

Thus, in ΔBCE, we have

BC = 15 m, CE = 13 m and BE = 14 m

Let a = 15 m, b = 13 m and c = 14 m

Question 40

The difference between the lengths of the parallel sides of a trapezium is 8 cm, the perpendicular distance between these sides is 24 cm and the area of the trapezium is 312 cm². Find the length of each of the parallel sides.Solution 40

Let the smaller parallel side = x cm

Then, longer parallel side = (x + 8) cm

Height = 24 cm

Area of trapezium = 312 cm²

Thus, the lengths of parallel sides are 9 cm and 17 cm respectively.Question 41

A parallelogram and a rhombus are equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one of the sides of the parallelogram measures 66 m, find its corresponding altitude.Solution 41

Area of parallelogram = Area of rhombus

Question 42

A parallelogram and a square have the same area. If the sides of the square measures 40 m and altitude of the parallelogram measures 25 m, find the length of the corresponding base of the parallelogram.Solution 42

Area of parallelogram = Area of square

Question 43

Find the area of a rhombus one side of which measures 20 cm and one of whose diagonals is 24 cm.Solution 43

Let ABCD be a rhombus and let diagonals AC and BD intersect each other at point O.

We know that diagonals of a rhombus bisect each other at right angles.

Thus, in right-angled ΔAOD, by Pythagoras theorem,

OD² = AD² – OA² = 20² – 12² = 400 – 144 = 256

⇒ OD = 16 cm

⇒ BD = 2(OD) = 2(16) = 32 cm

Question 44

The area of a rhombus is 480 cm², and one of its diagonals measures 48 cm. Find (i) the length of the other diagonal, (ii) the length of each of its sides, and (iii) its perimeter.Solution 44

(i) Area of a rhombus = 480 cm²

(ii) Let diagonal AC = 48 cm and diagonal BD = 20 cm

We know that diagonals of a rhombus bisect each other at right angles.

Thus, in right-angled ΔAOD, by Pythagoras theorem,

AD² = OA² + OD² = 24² + 10² = 576 + 100 = 676

⇒ AD = 26 cm

⇒ AD = BC = CD = AD = 26 cm

Thus, the length of each side of rhombus is 26 cm.

(iii) Perimeter of a rhombus = 4 × side = 4 × 26 = 104 cm

Exercise MCQ

Question 1

The equation of the x-axis is

(a) x = 0

(b) y = 0

(c) x = y

(d) x + y = 0Solution 1

Correct option: (b)

The equation of the x-axis is y = 0.Question 2

The equation of the y-axis is

(a) x = 0

(b) y = 0

(c) x = y

(d) x + y = 0Solution 2

Correct option: (a)

The equation of the y-axis is x = 0. Question 3

The point of the form (a,a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y = x

(d) the line x + y = 0Solution 3

Question 4

The point of the form (a,-a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y-x=0

(d) the line x + y = 0Solution 4

Question 5

The linear equation 3x – 5y = 15 has

(a) a unique solution

(b) two solutions

(c) infinitely many solutions

(d) no solutionSolution 5

Question 6

The equation 2x + 5y = 7 has a unique solution, if x and y are

(a) natural numbers

(b) rational numbers

(c) positive real numbers

(d) real numbersSolution 6

Correct option: (a)

The equation 2x + 5y = 7 has a unique solution, if x and y are natural numbers.

If we take x = 1 and y = 1, the given equation is satisfied. Question 7

The graph of y = 5 is a line

(a) making an intercept 5 on the x-axis

(b) making an intercept 5 on the y-axis

(c) parallel to the x-axis at a distance of 5 units from the origin

(d) parallel to the y-axis at a distance of 5 units from the originSolution 7

Correct option: (c)

The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin. Question 8

The graph of x = 4 is a line

(a) making an intercept 4 on the x-axis

(b) making an intercept 4 on the y-axis

(c) parallel to the x-axis at a distance of 4 units from the origin

(d) parallel to the y-axis at a distance of 4 units from the originSolution 8

Correct option: (d)

The graph of x = 4 is a line parallel to the y-axis at a distance of 4 units from the origin. Question 9

The graph of x + 3 = 0 is a line

(a) making an intercept -3 on the x-axis

(b) making an intercept -3 on the y-axis

(c) parallel to the y-axis at a distance of 3 units to the left of y-axis

(d) parallel to the x-axis at a distance of 3 units below the x-axisSolution 9

Correct option: (c)

The graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis. Question 10

The graph of y + 2 = 0 is a line

(a) making an intercept -2 on the x-axis

(b) making an intercept -2 on the y-axis

(c) parallel to the x-axis at a distance of 2 units below the x-axis

(d) parallel to the y-axis at a distance of 2 units to the left of y-axisSolution 10

Correct option: (c)

The graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis. Question 11

The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point

(a) (2, 0)

(b) (3, 0)

(c) (0, 2)

(d) (0, 3)Solution 11

Correct option: (c)

When a graph meets the y-axis, the x coordinate is zero.

Thus, substituting x = 0 in the given equation, we get

2(0) + 3y = 6

⇒ 3y = 6

⇒ y = 2

Hence, the required point is (0, 2).Question 12

The graph of the linear equation 2x + 5y = 10 meets the x-axis at the point

(a) (0, 2)

(b) (2, 0)

(c) (5, 0)

(d) (0, 5)Solution 12

Correct option: (c)

When a graph meets the x-axis, the y coordinate is zero.

Thus, substituting y = 0 in the given equation, we get

2x + 5(0) = 10

⇒ 2x = 10

⇒ x = 5

Hence, the required point is (5, 0). Question 13

The graph of the line x = 3 passes through the point

(a) (0,3)

(b) (2,3)

(c) (3,2)

(d) None of theseSolution 13

Question 14

The graph of the line y = 3 passes though the point

(a) (3, 0)

(b) (3, 2)

(c) (2, 3)

(d) none of theseSolution 14

Correct option: (c)

Since, the y coordinate is 3, the graph of the line y = 3 passes through the point (2, 3).Question 15

The graph of the line y = -3 does not pass through the point

(a) (2,-3)

(b) (3,-3)

(c) (0,-3)

(d) (-3,2)Solution 15

Question 16

The graph of the linear equation x-y=0 passes through the point

Solution 16

Question 17

If each of (-2,2), (0,0) and (2,-2) is a solution of a linear equation in x and y, then the equation is

(a) x-y=0

(b) x+y=0

(c) -x+2y=0

(d) x – 2y=0Solution 17

Question 18

How many linear equations can be satisfied by x = 2 and y = 3?

(a) only one

(b) only two

(c) only three

(d) Infinitely manySolution 18

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2 and y = 3. Question 19

A linear equation in two variable x and y is of the form ax+by+c=0, where

(a) a≠0, b≠0

(b) a≠0, b=0

(c) a=0, b≠0

(d) a= 0, c=0Solution 19

Question 20

If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is

(a) 6

(b) 5

(c) 2

(d) 4Solution 20

Correct option: (d)

Since, (2, 0) is a solution of the linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given equation, we have

2(2) + 3(0) = k

⇒ 4 + 0 = k

⇒ k = 4 Question 21

Any point on x-axis is of the form:

(a) (x,y), where x ≠0 and y ≠0

(b) (0,y), where y ≠0

(c) (x,0), where x ≠0

(d) (y,y), where y ≠0Solution 21

Question 22

Any point on y-axis is of the form

(a) (x,0), where x ≠ 0

(b) (0,y), where y ≠ 0

(c) (x,x), where x ≠ 0

(d) None of theseSolution 22

Question 23

x = 5, y = 2 is a solution of the linear equation

(a) x + 2y = 7

(b) 5x + 2y = 7

(c) x + y = 7

(d) 5x + y = 7Solution 23

Correct option: (c)

Substituting x = 5 and y = 2 in L.H.S. of equation x + y = 7, we get

L.H.S. = 5 + 2 = 7 = R.H.S.

Hence, x = 5 and y = 2 is a solution of the linear equation x + y = 7. Question 24

If the point (3, 4) lies on the graph of 3y = ax + 7 then the value of a is

(a) 

(b) 

(c) 

(d) Solution 24

Correct option: (b)

Since the point (3, 4) lies on the graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we get

3(4) = a(3) + 7

⇒ 12 = 3a + 7

⇒ 3a = 5

Exercise Ex. 4B

Question 1(vii)

Draw the graph of each of the following equation.

y + 5 = 0 Solution 1(vii)

y + 5 = 0

⇒ y = -5, which is a line parallel to the X-axis, at a distance of 5 units from it, below the X-axis.

Question 1(viii)

Draw the graph of each of the following equation.

y = 4Solution 1(viii)

y = 4 is a line parallel to the X-axis, at a distance of 4 units from it, above the X-axis.

Question 1(i)

Draw the graph of each of the following equation.

x = 4Solution 1(i)

x = 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to its right.

Question 1(ii)

Draw the graph of each of the following equation.

x + 4 = 0Solution 1(ii)

x + 4 = 0

⇒ x = -4, which is a line parallel to the Y-axis, at a distance of 4 units from it, to its left.

Question 1(iii)

Draw the graph of each of the following equation.

y = 3Solution 1(iii)

y = 3 is a line parallel to the X-axis, at a distance of 3 units from it, above the X-axis.

Question 1(iv)

Draw the graph of each of the following equation.

y = -3Solution 1(iv)

y = -3 is a line parallel to the X-axis, at a distance of 3 units from it, below the X-axis.

Question 1(v)

Draw the graph of each of the following equation.

x = -2Solution 1(v)

x = -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to its left.

Question 1(vi)

Draw the graph of each of the following equation.

x = 5Solution 1(vi)

x = 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to its right.

Question 2(i)

Draw the graph of the equation y = 3x.

From your graph, find the value of y when x = 2.Solution 2(i)

y = 3x

When x = 1, then y = 3(1) = 3

When x = -1, then y = 3(-1) = -3

Thus, we have the following table:

x	1	-1
y	3	-3

Now, plot the points A(1, 3) and B(-1, -3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of y = 3x.

Reading the graph

Given: x = 2. Take a point M on the X-axis such that OM = 2.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 6

Thus, when x = 2, then y = 6.Question 2(ii)

Draw the graph of the equation y = 3x. From your graph, find the value of y when x = -2.Solution 2(ii)

The given equation is y = 3x.

Putting x = 1, y = 3  1 = 3

Putting x = 2, y = 3  2 = 6

Thus, we have the following table:

x	1	2
y	3	6

Plot points (1,3) and (2,6) on a graph paper and join them to get the required graph.

Take a point P on the left of y-axis such that the distance of point P from the y-axis is 2 units.

Draw PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to x-axis meeting y-axis at N.

So, y = ON = -6.Question 3(ii)

Draw the graph of the equation x + 2y – 3 = 0.

From your graph, find the value of y when x = -5Solution 3(ii)

x + 2y – 3 = 0

⇒ 2y = 3 – x

When x = -1, then  

When x = 1, then 

Thus, we have the following table:

x	-1	1
y	2	1

Now, plot the points A(-1, 2) and B(1, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + 2y – 3 = 0.

Reading the graph

Given: x = -5. Take a point M on the X-axis such that OM = -5.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 4

Thus, when x = -5, then y = 4. Question 3(i)

Draw the graph of the equation x + 2y – 3 = 0. From your graph, find the value of y when x = 5.Solution 3(i)

The given equation is,

x + 2y – 3 = 0

x = 3 – 2y

Putting y = 1,x = 3 – (2 1) = 1

Putting y = 0,x = 3 – (2 0) = 3

Thus, we have the following table:

x	1	3
y	1	0

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P.

Through P, draw PM parallel to x-axis cutting y-axis at M.

So, y = OM = -1.Question 4

Draw the graph of the equation 2x – 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x when y = 3.Solution 4

The given equation is, 2x – 3y = 5

Now, if x = 4, then

And, if x = -2, then

Thus, we have the following table:

x	4	-2
y	1	-3

Plot points (4,1) and (-2,-3) on a graph paper and join them to get the required graph.

(i) When x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to its right cutting the line at Q and through Q draw a line parallel to x-axis cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

(ii) When y = 3, draw a line parallel to x-axis at a distance of 3 units from x-axis and above it, cutting the line at point P. Through P, draw a line parallel to y-axis meeting x-axis at a point which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.Question 5

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.Solution 5

The given equation is 2x + y = 6

 y = 6 – 2x

Now, if x = 1, then y = 6 – 2  1 = 4

And, if x = 2, then y = 6 – 2  2 = 2

Thus, we have the following table:

x	1	2
y	4	2

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).Question 6

Draw the graph of the equation 3x + 2y = 6. Find the coordinates of the point, where the graph cuts the y-axis.Solution 6

The given equation is 3x + 2y = 6

 2y = 6 – 3x

Now, if x = 2, then

And, if x = 4, then

Thus, we have the following table:

x	2	4
y	0	-3

Plot points (2, 0) and (4,-3) on a graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6 cuts the y-axis at a point P which is 3 units above the x-axis.

So, co-ordinates of P are (0,3).Question 7

Draw the graphs of the equations 3x – 2y = 4 and x + y – 3 = 0. On the same graph paper, find the coordinates of the point where the two graph lines intersect.Solution 7

Graph of the equation 3x – 2y = 4

⇒ 2y = 3x – 4

When x = 2, then  

When x = -2, then 

Thus, we have the following table:

x	2	-2
y	1	-5

Now, plot the points A(2, 1) and B(-2, -5) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 3x – 2y = 4.

Graph of the equation x + y – 3 = 0

⇒ y = 3 – x

When x = 1, then y = 3 – 1 = 2 

When x = -1, then y = 3 – (-1) = 4

Thus, we have the following table:

x	1	-1
y	2	4

Now, plot the points C(1, 2) and D(-1, 4) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x + y – 3 = 0.

The two graph lines intersect at point A(2, 1). Question 8(i)

Draw the graph of the line 4x + 3y = 24.

Write the coordinates of the points where this line intersects the x-axis and the y-axis.Solution 8(i)

4x + 3y = 24

⇒ 3y = 24 – 4x

When x = 0, then  

When x = 3, then 

Thus, we have the following table:

x	0	3
y	8	4

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

Reading the graph

The graph of line 4x + 3y = 24 intersects the X-axis at point C(6, 0) and the Y-axis at point A(0, 8). Question 8(ii)

Draw the graph of the line 4x + 3y = 24.

Use this graph to find the area of the triangle formed by the graph line and the coordinate axes.Solution 8(ii)

4x + 3y = 24

⇒ 3y = 24 – 4x

When x = 0, then  

When x = 3, then 

Thus, we have the following table:

x	0	3
y	8	4

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

Reading the graph

Required area = Area of ΔAOC

Question 9

Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these two lines and the x-axis. Find the area of the shaded region.Solution 9

Graph of the equation 2x + y = 6

⇒ y = 6 – 2x

When x = 1, then y = 6 – 2(1) = 6 – 2 = 4 

When x = 2, then y = 6 – 2(2) = 6 – 4 = 2

Thus, we have the following table:

x	1	2
y	4	2

Now, plot the points A(1, 4) and B(2, 2) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 2x + y = 6.

Graph of the equation 2x – y + 2 = 0

⇒ y = 2x + 2

When x = -1, then y = 2(-1) + 2 = -2 + 2 = 0 

When x = 2, then y = 2(2) + 2 = 4 + 2 = 6

Thus, we have the following table:

x	-1	2
y	0	6

Now, plot the points C(-1, 0) and D(2, 6) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x – y + 2 = 0.

The two graph lines intersect at point A(1, 4).

The area enclosed by the lines and X-axis is shown in the graph.

Draw AM perpendicular from A on X-axis.

PM = y-coordinate of point A(1, 4) = 4

And, CP = 4

Area of shaded region = Area of ΔACP

Question 10

Draw the graphs of the lines x – y = 1 and 2x + y = 8. Shade the area formed by these two lines and the y-axis. Also, find this area.Solution 10

Graph of the equation x – y = 1

⇒ y = x – 1

When x = 1, then y = 1 – 1 = 0 

When x = 2, then y = 2 – 1 = 1

Thus, we have the following table:

x	1	2
y	0	1

Now, plot the points A(1, 0) and B(2, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x – y = 1.

Graph of the equation 2x + y = 8

⇒ y = 8 – 2x

When x = 2, then y = 8 – 2(2) = 8 – 4 = 4 

When x = 3, then y = 8 – 2(3) = 8 – 6 = 2 

Thus, we have the following table:

x	2	3
y	4	2

Now, plot the points C(2, 4) and D(3, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x + y = 8.

The two graph lines intersect at point D(3, 2).

The area enclosed by the lines and Y-axis is shown in the graph.

Draw DM perpendicular from D on Y-axis.

DM = x-coordinate of point D(3, 2) = 3

And, EF = 9

Area of shaded region = Area of ΔDEF

Question 11

Draw the graph for each of the equations x + y = 6 and x – y = 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect.

*Back answer incorrect.Solution 11

Graph of the equation x + y = 6

⇒ y = 6 – x

When x = 2, then y = 6 – 2 = 4 

When x = 3, then y = 6 – 3 = 3

Thus, we have the following table:

x	2	3
y	4	3

Now, plot the points A(2, 4) and B(3, 3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 6.

Graph of the equation x – y = 2

⇒ y = x – 2

When x = 3, then y = 3 – 2 = 1 

When x = 4, then y = 4 – 2 = 2 

Thus, we have the following table:

x	3	4
y	1	2

Now, plot the points C(3, 1) and D(4, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x – y = 2.

The two graph lines intersect at point D(4, 2).Question 12

Two students A and B contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation to satisfy the above data and draw its graph.Solution 12

Let the amount contributed by students A and B be Rs. x and Rs. y respectively.

Total contribution = 100

⇒ x + y = 100

⇒ y = 100 – x

When x = 25, then y = 100 – 25 = 75

When x = 50, then y = 100 – 50 = 50

Thus, we have the following table:

x	25	50
y	75	50

Now, plot the points A(25, 75) and B(50, 50) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 100.

Exercise Ex. 4A

Question 1(i)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3x + 5y = 7.5 Solution 1(i)

We have,

3x + 5y = 7.5

⇒ 3x + 5y – 7.5 = 0

⇒ 6x + 10y – 15 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = 10 and c = -15 Question 1(ii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(ii)

On comparing this equation with ax + by + c = 0, we obtain

a = 10, b = -1 and c = 30 Question 1(iii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3y – 2x = 6Solution 1(iii)

We have,

3y – 2x = 6

⇒ -2x + 3y – 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = -2, b = 3 and c = -6 Question 1(iv)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

4x = 5ySolution 1(iv)

We have,

4x = 5y

⇒ 4x – 5y = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 4, b = -5 and c = 0 Question 1(v)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(v)

⇒ 6x – 5y = 30

⇒ 6x – 5y – 30 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = -5 and c = -30 Question 1(vi)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 1(vi)

On comparing this equation with ax + by + c = 0, we obtain

a = , b =  and c = -5 Question 2(i)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

x = 6Solution 2(i)

We have,

x = 6

⇒ x – 6 = 0

⇒ 1x + 0y – 6 = 0

⇒ x + 0y – 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 0 and c = -6 Question 2(ii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

3x – y = x – 1Solution 2(ii)

We have,

3x – y = x – 1

⇒ 3x – x – y + 1 = 0

⇒ 2x – y + 1 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = -1 and c = 1 Question 2(iii)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

2x + 9 = 0Solution 2(iii)

We have,

2x + 9 = 0

⇒ 2x + 0y + 9 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = 0 and c = 9 Question 2(iv)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

4y = 7Solution 2(iv)

We have,

4y = 7

⇒ 0x + 4y – 7 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 0, b = 4 and c = -7 Question 2(v)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

x + y = 4Solution 2(v)

We have,

x + y = 4

⇒ x + y – 4 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 1 and c = -4 Question 2(vi)

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case

Solution 2(vi)

We have,

⇒ 3x – 8y – 1 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 3, b = -8 and c = -1 Question 3(i)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(4, 0)Solution 3(i)

Given equation is 5x – 4y = 20

Substituting x = 4 and y = 0 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(4) – 4(0)

= 20 – 0

= 20

= R.H.S.

Hence, (4, 0) is the solution of the given equation.Question 3(ii)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(0, 5)Solution 3(ii)

Given equation is 5x – 4y = 20

Substituting x = 0 and y = 5 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(0) – 4(5)

= 0 – 20

= -20

≠ R.H.S.

Hence, (0, 5) is not the solution of the given equation. Question 3(iii)

Check which of the following are the solutions of the equation 5x – 4y = 20.

Solution 3(iii)

Given equation is 5x – 4y = 20

Substituting x = -2 and y =  in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(-2) – 4

= -10 – 10

= -20

≠ R.H.S.

Hence,  is not the solution of the given equation. Question 3(iv)

Check which of the following are the solutions of the equation 5x – 4y = 20.

(0, -5)Solution 3(iv)

Given equation is 5x – 4y = 20

Substituting x = 0 and y = -5 in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(0) – 4(-5)

= 0 + 20

= 20

= R.H.S.

Hence, (0, -5) is the solution of the given equation. Question 3(v)

Check which of the following are the solutions of the equation 5x – 4y = 20.

Solution 3(v)

Given equation is 5x – 4y = 20

Substituting x = 2 and y =  in L.H.S. of given equation, we get

L.H.S. = 5x – 4y

= 5(2) – 4

= 10 + 10

= 20

= R.H.S.

Hence,  is the solution of the given equation. Question 4(a)

Find five different solutions of each of the following equations:

2x – 3y = 6Solution 4(a)

Given equation is 2x – 3y = 6

Substituting x = 0 in the given equation, we get

2(0) – 3y = 6

⇒ 0 – 3y = 6

⇒ 3y = -6

⇒ y = -2

So, (0, -2) is the solution of the given equation.

Substituting y = 0 in the given equation, we get

2x – 3(0) = 6

⇒ 2x – 0 = 6

⇒ 2x = 6

⇒ x = 3

So, (3, 0) is the solution of the given equation.

Substituting x = 6 in the given equation, we get

2(6) – 3y = 6

⇒ 12 – 3y = 6

⇒ 3y = 6

⇒ y = 2

So, (6, 2) is the solution of the given equation.

Substituting y = 4 in the given equation, we get

2x – 3(4) = 6

⇒ 2x – 12 = 6

⇒ 2x = 18

⇒ x = 9

So, (9, 4) is the solution of the given equation.

Substituting x = -3 in the given equation, we get

2(-3) – 3y = 6

⇒ -6 – 3y = 6

⇒ 3y = -12

⇒ y = -4

So, (-3, -4) is the solution of the given equation.Question 4(b)

Find five different solutions of each of the following equations:

Solution 4(b)

Given equation is  

Substituting x = 0 in (i), we get

4(0) + 3y = 30

⇒ 3y = 30

⇒ y = 10

So, (0, 10) is the solution of the given equation.

Substituting x = 3 in (i), we get

4(3) + 3y = 30

⇒ 12 + 3y = 30

⇒ 3y = 18

⇒ y = 6

So, (3, 6) is the solution of the given equation.

Substituting x = -3 in (i), we get

4(-3) + 3y = 30

⇒ -12 + 3y = 30

⇒ 3y = 42

⇒ y = 14

So, (-3, 14) is the solution of the given equation.

Substituting y = 2 in (i), we get

4x + 3(2) = 30

⇒ 4x + 6 = 30

⇒ 4x = 24

⇒ x = 6

So, (6, 2) is the solution of the given equation.

Substituting y = -2 in (i), we get

4x + 3(-2) = 30

⇒ 4x – 6 = 30

⇒ 4x = 36

⇒ x = 9

So, (9, -2) is the solution of the given equation.Question 4(c)

Find five different solutions of each of the following equations:

3y = 4xSolution 4(c)

Given equation is 3y = 4x

Substituting x = 3 in the given equation, we get

3y = 4(3)

⇒ 3y = 12

⇒ y = 4

So, (3, 4) is the solution of the given equation.

Substituting x = -3 in the given equation, we get

3y = 4(-3)

⇒ 3y = -12

⇒ y = -4

So, (-3, -4) is the solution of the given equation.

Substituting x = 9 in the given equation, we get

3y = 4(9)

⇒ 3y = 36

⇒ y = 12

So, (9, 12) is the solution of the given equation.

Substituting y = 8 in the given equation, we get

3(8) = 4x

⇒ 4x = 24

⇒ x = 6

So, (6, 8) is the solution of the given equation.

Substituting y = -8 in the given equation, we get

3(-8) = 4x

⇒ 4x = -24

⇒ x = -6

So, (-6, -8) is the solution of the given equation.Question 5

If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.Solution 5

Since x = 3 and y = 4 is a solution of the equation 5x – 3y = k, substituting x = 3 and y = 4 in equation 5x – 3y = k, we get

5(3) – 3(4) = k

⇒ 15 – 12 = k

⇒ k = 3 Question 6

If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.Solution 6

Since x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, substituting these values in equation, we get

4(3k + 2) – 3(2k – 1) + 1 = 0

⇒ 12k + 8 – 6k + 3 + 1 = 0

⇒ 6k + 12 = 0

⇒ 6k = -12

⇒ k = -2 Question 7

The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be Rs. x and that of a ballpoint to be Rs. y).Solution 7

Let the cost of one pencil be Rs. x and that of one ballpoint be Rs. y.

Then,

Cost of 5 pencils = Rs. 5x

Cost of 2 ballpoints = Rs. 2y

According to given statement, we have

5x = 2y

⇒ 5x – 2y = 0

Exercise MCQ

Question 1

Solution 1

Question 2

In a ΔABC, if ∠A – ∠B = 42° and ∠B – ∠C = 21° then ∠B = ?

(a) 32° 

(b) 63° 

(c) 53° 

(d) 95° Solution 2

Correct option: (c)

∠A – ∠B = 42° 

⇒ ∠A = ∠B + 42° 

∠B – ∠C = 21° 

⇒ ∠C = ∠B – 21° 

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ ∠B + 42° + ∠B + ∠B – 21° = 180° 

⇒ 3∠B = 159

⇒ ∠B = 53° Question 3

In a ΔABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?

(a) 160° 

(b) 60° 

(c) 80° 

(d) 30° Solution 3

Correct option: (b)

∠ACD = ∠B + ∠A (Exterior angle property)

⇒ 110° = 50° + ∠A

⇒ ∠A = 60° Question 4

Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?

1. 50°
2. 55°
3. 65°
4. 75°

Solution 4

Correct option: (d)

Question 5

In the given figure, sides CB and BA of ΔABC have been produced to D and E respectively such that ∠ABD = 110° and ∠CAE = 135°. Than ∠ACB =?

1. 65°
2. 45°
3. 55°
4. 35°

Solution 5

Correct option: (a)

Question 6

The sides BC, CA and AB of ΔABC have been produced to D,E and F respectively.  ∠BAE + ∠CBF + ∠ACD =?

1. 240°
2. 300°
3. 320°
4. 360°

Solution 6

Question 7

In the given figure, EAD ⊥ BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also, in ΔBAC, ∠BAC = x° and ∠ABC = (x + 10)°. Then, the value of x is

(a) 20

(b) 25

(c) 30

(d) 35Solution 7

Correct option: (b)

∠EAF = ∠CAD (vertically opposite angles)

⇒ ∠CAD = 30° 

In ΔABD, by angle sum property

∠A + ∠B + ∠D = 180° 

⇒ (x + 30)° + (x + 10)° + 90° = 180° 

⇒ 2x + 130° = 180° 

⇒ 2x = 50° 

⇒ x = 25° Question 8

In the given figure, two rays BD and CE intersect at a point A. The side BC of ΔABC have been produced on both sides to points F and G respectively. If ∠ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?

(a) x + y – 180

(b) x + y + 180

(c) 180 – (x + y)

(d) x + y + 360° Solution 8

Correct option: (a)

∠ABF + ∠ABC = 180° (linear pair)

⇒ x + ∠ABC = 180° 

⇒ ∠ABC = 180° – x

∠ACG + ∠ACB = 180° (linear pair)

⇒ y + ∠ACB = 180° 

⇒ ∠ACB = 180° – y

In ΔABC, by angle sum property

∠ABC + ∠ACB + ∠BAC = 180° 

⇒ (180° – x) + (180° – y) + ∠BAC = 180° 

⇒ ∠BAC – x – y + 180° = 0

⇒ ∠BAC = x + y – 180° 

Now, ∠EAD = ∠BAC (vertically opposite angles)

⇒ z = x + y – 180°  Question 9

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively. such that ∠OAE = x° and ∠ DBF = y°.

If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?

(a) 190° 

(b) 230° 

(c) 210° 

(d) 270° Solution 9

Correct option: (b)

In ΔOAC, by angle sum property

∠OCA + ∠COA + ∠CAO = 180° 

⇒ 80° + 40° + ∠CAO = 180° 

⇒ ∠CAO = 60° 

∠CAO + ∠OAE = 180° (linear pair)

⇒ 60° + x = 180° 

⇒ x = 120° 

∠COA = ∠BOD (vertically opposite angles)

⇒ ∠BOD = 40° 

In ΔOBD, by angle sum property

∠OBD + ∠BOD + ∠ODB = 180° 

⇒ ∠OBD + 40° + 70° = 180° 

⇒ ∠OBD = 70° 

∠OBD + ∠DBF = 180° (linear pair)

⇒ 70° + y = 180° 

⇒ y = 110° 

∴ x + y = 120° + 110° = 230° Question 10

In a ΔABC it is given that ∠A:∠B:∠C = 3:2:1 and ∠ACD = 90^o. If it is produced to E, Then ∠ECD =?

1. 60°
2. 50°
3. 40°
4. 25°

Solution 10

Question 11

In the given figure , BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC= ?

1. 130°
2. 100°
3. 115°
4. 120°

Solution 11

Question 12

In the given figure, side BC of ΔABC has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠ACD = 7y°. Then, the value of x is

(a) 60

(b) 50

(c) 45

(d) 35Solution 12

Correct option: (a)

∠ACB + ∠ACD = 180° (linear pair)

⇒ 5y + 7y = 180° 

⇒ 12y = 180° 

⇒ y = 15° 

Now, ∠ACD = ∠ABC + ∠BAC (Exterior angle property)

⇒ 7y = x + 3y

⇒ 7(15°) = x + 3(15°)

⇒ 105° = x + 45° 

⇒ x = 60° 

Exercise Ex. 8

Question 1

In ABC, if B = 76^o and C = 48^o, find A.Solution 1

Since, sum of the angles of a triangle is 180^o

A + B + C = 180^o

 A + 76^o + 48^o = 180^o

 A = 180^o – 124^o = 56^o

 A = 56^oQuestion 2

The angles of a triangle are in the ratio 2:3:4. Find the angles.Solution 2

Let the measures of the angles of a triangle are (2x)^o, (3x)^o and (4x)^o.

Then, 2x + 3x + 4x = 180         [sum of the angles of a triangle is 180^o ]     

 9x = 180

 The measures of the required angles are:

2x = (2  20)^o = 40^o

3x = (3  20)^o = 60^o

4x = (4  20)^o = 80^oQuestion 3

In ABC, if 3A = 4B = 6C, calculate A, B and C.Solution 3

Let 3A = 4B = 6C = x (say)

Then, 3A = x

 A = 

4B = x

and 6C = x

 C = 

As A + B + C = 180^o

 A = 

B = 

C = Question 4

In ABC, if A + B = 108^o and B + C = 130^o, find A, B and C.Solution 4

A + B = 108^o [Given]

But as A, B and C are the angles of a triangle,

A + B + C = 180^o

 108^o + C = 180^o

 C = 180^o – 108^o = 72^o

Also, B + C = 130^o [Given]

 B + 72^o = 130^o

 B = 130^o – 72^o = 58^o

Now as, A + B = 108^o

 A + 58^o = 108^o

 A = 108^o – 58^o = 50^o

 A = 50^o, B = 58^o and C = 72^o.Question 5

In ABC, A + B = 125^o and A + C = 113^o. Find A, B and C.Solution 5

Since. A , B and C are the angles of a triangle .

So, A + B + C = 180^o

Now, A + B = 125^o [Given]

 125^o + C = 180^o

 C = 180^o – 125^o = 55^o

Also, A + C = 113^o [Given]

 A + 55^o = 113^o

 A = 113^o – 55^o = 58^o

Now as A + B = 125^o

 58^o + B = 125^o

 B = 125^o – 58^o = 67^o

 A = 58^o, B = 67^o and C = 55^o.Question 6

In PQR, if P – Q = 42^o and Q – R = 21^o, find P, Q and R.Solution 6

Since, P, Q and R are the angles of a triangle.

So,P + Q + R = 180^o(i)

Now,P – Q = 42^o[Given]

P = 42^o + Q(ii)

andQ – R = 21^o[Given]

R = Q – 21^o(iii)

Substituting the value of P and R from (ii) and (iii) in (i), we get,

42^o + Q + Q + Q – 21^o = 180^o

3Q + 21^o = 180^o

3Q = 180^o – 21^o = 159^o

Q = 

P = 42^o + Q

= 42^o + 53^o = 95^o

R = Q – 21^o

= 53^o – 21^o = 32^o

P = 95^o, Q = 53^o and R = 32^o.Question 7

The sum of two angles of a triangle is 116^o and their difference is 24^o. Find the measure of each angle of the triangle.Solution 7

Given that the sum of the angles A and B of a ABC is 116^o, i.e., A + B = 116^o.

Since, A + B + C = 180^o

So, 116^o + C = 180^o

 C = 180^o – 116^o = 64^o

Also, it is given that:

A – B = 24^o

 A = 24^o + B

Putting, A = 24^o + B in A + B = 116^o, we get,

24^o + B + B = 116^o

 2B + 24^o = 116^o

 2B = 116^o – 24^o = 92^o

 B = 

Therefore, A = 24^o + 46^o = 70^o

 A = 70^o, B = 46^o and C = 64^o.Question 8

Two angles of a triangle are equal and the third angle is greater than each one of them by 18^o. Find the angles.Solution 8

Let the two equal angles, A and B, of the triangle be x^o each.

We know,

A + B + C = 180^o

x^o + x^o + C = 180^o

2x^o + C = 180^o(i)

Also, it is given that,

C = x^o + 18^o(ii)

Substituting C from (ii) in (i), we get,

2x^o + x^o + 18^o = 180^o

3x^o = 180^o – 18^o = 162^o

x = 

Thus, the required angles of the triangle are 54^o, 54^o and x^o + 18^o = 54^o + 18^o = 72^o.Question 9

Of the three angles of triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.Solution 9

Let C be the smallest angle of ABC.

Then, A = 2C and B = 3C

Also, A + B + C = 180^o

 2C + 3C + C = 180^o

 6C = 180^o

 C = 30^o

So, A = 2C = 2  30^o = 60^o

B = 3C = 3  30^o = 90^o

 The required angles of the triangle are 60^o, 90^o, 30^o.Question 10

In a right-angled triangle, one of the acute angles measures 53^o. Find the measure of each angle of the triangle.Solution 10

Let ABC be a right angled triangle and C = 90^o

Since, A + B + C = 180^o

 A + B = 180^o – C = 180^o – 90^o = 90^o

Suppose A = 53^o

Then, 53^o + B = 90^o

 B = 90^o – 53^o = 37^o

 The required angles are 53^o, 37^o and 90^o.Question 11

In a right-angled triangle, one of the acute angles measures 53^o. Find the measure of each angle of the triangle.Solution 11

Let ABC be a right angled triangle and C = 90^o

Since, A + B + C = 180^o

 A + B = 180^o – C = 180^o – 90^o = 90^o

Suppose A = 53^o

Then, 53^o + B = 90^o

 B = 90^o – 53^o = 37^o

 The required angles are 53^o, 37^o and 90^o.Question 12

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.Solution 12

Let ABC be a triangle.

So, A < B + C

Adding A to both sides of the inequality,

2 A < A + B + C

2 A < 180^o [Since A + B + C = 180^o]

Similarly, B <A + C

B < 90^o

and C < A + B

C < 90^o

ABC is an acute angled triangle.Question 13

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.Solution 13

Let ABC be a triangle and B > A + C

Since, A + B + C = 180^o

 A + C = 180^o – B

_{Therefore, we get,}

B > 180^o – B

Adding B on both sides of the inequality, we get,

B + B > 180^o – B + B

 2B > 180^o

 B > 

i.e., B > 90^o which means B is an obtuse angle.

 ABC is an obtuse angled triangle.Question 14

In the given figure, side BC of ABC is produced to D. If ACD = 128^o and ABC = 43^o, find BAC and ACB.

Solution 14

Since ACB and ACD form a linear pair.

So, ACB + ACD = 180^o

 ACB + 128^o = 180^o

 ACB = 180^o – 128 = 52^o

Also, ABC + ACB + BAC = 180^o

 43^o + 52^o + BAC = 180^o

 95^o + BAC = 180^o

 BAC = 180^o – 95^o = 85^o

 ACB = 52^o and BAC = 85^o.Question 15

In the given figure, the side BC of ABC has been produced on the right-hand side from B to D and  on the right-hand side from C and E. If ABD = 106^o and ACE= 118^o, find the measure of each angle of the triangle.

Solution 15

As DBA and ABC form a linear pair.

So,DBA + ABC = 180^o

106^o + ABC = 180^o

ABC = 180^o – 106^o = 74^o

Also, ACB and ACE form a linear pair.

So,ACB + ACE = 180^o

ACB + 118^o = 180^o

ACB = 180^o – 118^o = 62^o

In ABC, we have,

ABC + ACB + BAC = 180^o

74^o + 62^o + BAC = 180^o

136^o + BAC = 180^o

BAC = 180^o – 136^o = 44^o

In triangle ABC, A = 44^o, B = 74^o and C = 62^oQuestion 16

Calculate the value of x in each of the following figures.

(i) 

(ii) 

(iii) 

Given: AB || CD

(vi) Solution 16

(i) EAB + BAC = 180^o [Linear pair angles]

110^o + BAC = 180^o

 BAC = 180^o – 110^o = 70^o

Again, BCA + ACD = 180^o [Linear pair angles]

 BCA + 120^o = 180^o

 BCA = 180^o – 120^o = 60^o

Now, in ABC,

ABC + BAC + ACB = 180^o

x^o + 70^o + 60^o = 180^o

 x + 130^o = 180^o

 x = 180^o – 130^o = 50^o

 x = 50

(ii)

In ABC,

A + B + C = 180^o

 30^o + 40^o + C = 180^o

 70^o + C = 180^o

 C = 180^o – 70^o = 110^o

Now BCA + ACD = 180^o [Linear pair]

 110^o + ACD = 180^o

 ACD = 180^o – 110^o = 70^o

In ECD,

ECD + CDE + CED = 180^o

 70^o + 50^o + CED = 180^o

 120^o + CED = 180^o

 CED = 180^o – 120^o = 60^o

Since AED and CED from a linear pair

So, AED + CED = 180^o

 x^o + 60^o = 180^o

 x^o = 180^o – 60^o = 120^o

 x = 120

(iii)

EAF = BAC [Vertically opposite angles]

 BAC = 60^o

In ABC, exterior ACD is equal to the sum of two opposite interior angles.

So, ACD = BAC + ABC

 115^o = 60^o + x^o

 x^o = 115^o – 60^o = 55^o

 x = 55

(iv)

Since AB || CD and AD is a transversal.

So, BAD = ADC

 ADC = 60^o

In ECD, we have,

E + C + D = 180^o

 x^o + 45^o + 60^o = 180^o

 x^o + 105^o = 180^o

 x^o = 180^o – 105^o = 75^o

 x = 75

(v)

In AEF,

Exterior BED = EAF + EFA

 100^o = 40^o + EFA

 EFA = 100^o – 40^o = 60^o

Also, CFD = EFA [Vertically Opposite angles]

 CFD = 60^o

Now in FCD,

Exterior BCF = CFD + CDF

 90^o = 60^o + x^o

 x^o = 90^o – 60^o = 30^o

 x = 30

(vi)

In ABE, we have,

A + B + E = 180^o

 75^o + 65^o + E = 180^o

 140^o + E = 180^o

 E = 180^o – 140^o = 40^o

Now, CED = AEB [Vertically opposite angles]

 CED = 40^o

Now, in CED, we have,

C + E + D = 180^o

 110^o + 40^o + x^o = 180^o

 150^o + x^o = 180^o

 x^o = 180^o – 150^o = 30^o

 x = 30Question 17

In the figure given alongside, AB ∥ CD, EF ∥ BC, ∠BAC = 60° and ∠DHF = 50°. Find ∠GCH and ∠AGH.

Solution 17

AB ∥ CD and AC is the transversal.

⇒ ∠BAC = ∠ACD = 60° (alternate angles)

i.e. ∠BAC = ∠GCH = 60° 

Now, ∠DHF = ∠CHG = 50° (vertically opposite angles)

In ΔGCH, by angle sum property,

∠GCH + ∠CHG + ∠CGH = 180° 

⇒ 60° + 50° + ∠CGH = 180° 

⇒ ∠CGH = 70° 

Now, ∠CGH + ∠AGH = 180° (linear pair)

⇒ 70° + ∠AGH = 180° 

⇒ ∠AGH = 110° Question 18

Calculate the value of x in the given figure.

Solution 18

Produce CD to cut AB at E.

Now, in BDE, we have,

Exterior CDB = CEB + DBE

 x^o = CEB + 45^o     …..(i)

In  AEC, we have,

Exterior CEB = CAB + ACE

= 55^o + 30^o = 85^o

Putting CEB = 85^o in (i), we get,

x^o = 85^o + 45^o = 130^o

 x = 130Question 19

In the given figure, AD divides BAC in the ratio 1: 3 and AD = DB. Determine the value of x.

Solution 19

The angle BAC is divided by AD in the ratio 1 : 3.

Let BAD and DAC be y and 3y, respectively.

As BAE is a straight line,

BAC + CAE = 180^o        [linear pair]

 BAD + DAC +  CAE = 180^o

 y + 3y + 108^o = 180^o

 4y = 180^o – 108^o = 72^o

Now, in ABC,

ABC + BCA + BAC = 180^o

y + x + 4y = 180^o

[Since, ABC = BAD (given AD = DB) and BAC = y + 3y = 4y]

 5y + x = 180

 5  18 + x = 180

 90 + x = 180

 x = 180 – 90 = 90Question 20

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so fomed is equal to four right angles.

Solution 20

Given : A ABC in which BC, CA and AB are produced to D, E and F respectively.

To prove : Exterior DCA + Exterior BAE + Exterior FBD = 360^o

Proof : Exterior DCA = A + B(i)

Exterior FAE = B + C(ii)

Exterior FBD = A + C(iii)

Adding (i), (ii) and (iii), we get,

Ext. DCA + Ext. FAE + Ext. FBD

= A + B + B + C + A + C

= 2A +2B + 2C

= 2 (A + B + C)

= 2 180^o

[Since, in triangle the sum of all three angle is 180^o]

= 360^o

Hence, proved.Question 21

In the given figure, show that

A + B + C + D + E + F = 360^o.

Solution 21

In ACE, we have,

A + C + E = 180^o (i)

In BDF, we have,

B + D + F = 180^o (ii)

Adding both sides of (i) and (ii), we get,

A + C+E + B + D + F = 180^o + 180^o

A + B + C + D + E + F = 360^o.Question 22

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, find ∠MAN.

Solution 22

In ΔABC, by angle sum property,

∠A + ∠B + ∠C = 180° 

⇒ ∠A + 70° + 20° = 180° 

⇒ ∠A = 90° 

In ΔABM, by angle sum property,

∠BAM + ∠ABM + ∠AMB = 180° 

⇒ ∠BAM + 70° + 90° = 180° 

⇒ ∠BAM = 20° 

Since AN is the bisector of ∠A,

Now, ∠MAN + ∠BAM = ∠BAN

⇒ ∠MAN + 20° = 45° 

⇒ ∠MAN = 25° Question 23

In the given figure, BAD ∥ EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.

Solution 23

BAD ∥ EF and EC is the transversal.

⇒ ∠AEF = ∠CAD (corresponding angles)

⇒ ∠CAD = 55° 

Now, ∠CAD + ∠CAB = 180° (linear pair)

⇒ 55° + ∠CAB = 180° 

⇒ ∠CAB = 125° 

In ΔABC, by angle sum property,

∠ABC + ∠CAB + ∠ACB = 180° 

⇒ ∠ABC + 125° + 25° = 180° 

⇒ ∠ABC = 30° Question 24

In the given figure, ABC is a triangle in which A : B : C = 3 : 2 : 1 and AC  CD. Find the measure of 

Solution 24

In the given ABC, we have,

A : B : C = 3 : 2 : 1

Let A = 3x, B = 2x, C = x. Then,

A + B + C = 180^o

 3x + 2x + x = 180^o

 6x = 180^o

 x = 30^o

 A = 3x = 3  30^o = 90^o

B = 2x = 2  30^o = 60^o

and, C = x = 30^o

Now, in ABC, we have,

Ext ACE = A + B = 90^o + 60^o = 150^o

 ACD + ECD = 150^o

 ECD = 150^o – ACD 

 ECD = 150^o – 90^o    [since , ]

  ECD= 60^o

Question 25

In the given figure, AB ∥ DE and BD ∥ FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.

Solution 25

∠FGH + ∠FGE = 180° (linear pair)

⇒ 120° + y = 180° 

⇒ y = 60° 

AB ∥ DF and BD is the transversal.

⇒ ∠ABC = ∠CDE (alternate angles)

⇒ ∠CDE = 50° 

BD ∥ FG and DF is the transversal.

⇒ ∠EFG = ∠CDE (alternate angles)

⇒ ∠EFG = 50° 

In ΔEFG, by angle sum property,

∠FEG + ∠FGE + ∠EFG = 180° 

⇒ x + y + 50° = 180° 

⇒ x + 60° + 50° = 180° 

⇒ x = 70° Question 26

In the given figure, AB ∥ CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, Find the value of x.

Solution 26

AB ∥ CD and EF is the transversal.

⇒ ∠AEF = ∠EFD (alternate angles)

⇒ ∠AEF = ∠EFG + ∠DFG

⇒ 65° = ∠EFG + 30° 

⇒ ∠EFG = 35° 

In ΔGEF, by angle sum property,

∠GEF + ∠EGF + ∠EFG = 180° 

⇒ x + 90° + 35° = 180° 

⇒ x = 55° Question 27

In the given figure, AB ∥ CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.

Solution 27

AB ∥ CD and AE is the transversal.

⇒ ∠BAE = ∠DOE (corresponding angles)

⇒ ∠DOE = 65° 

Now, ∠DOE + ∠COE = 180° (linear pair)

⇒ 65° + ∠COE = 180° 

⇒ ∠COE = 115° 

In ΔOCE, by angle sum property,

∠OEC + ∠ECO + ∠COE = 180° 

⇒ 20° + ∠ECO + 115° = 180° 

⇒ ∠ECO = 45° Question 28

In the given figure, AB ∥ CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.

Solution 28

AB ∥ CD and EF is the transversal.

⇒ ∠EGB = ∠GHD (corresponding angles)

⇒ ∠GHD = 35° 

Now, ∠GHD = ∠QHP (vertically opposite angles)

⇒ ∠QHP = 35° 

In DQHP, by angle sum property,

∠PQH + ∠QHP + ∠QPH = 180° 

⇒ ∠PQH + 35° + 90° = 180° 

⇒ ∠PQH = 55° Question 29

In the given figure, AB ∥ CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.

Solution 29

AB ∥ CD and GE is the transversal.

⇒ ∠EGF + ∠GED = 180° (interior angles are supplementary)

⇒ ∠EGF + 130° = 180° 

⇒ ∠EGF = 50° 

Exercise MCQ

Question 1

In ancient India, the shapes of altars used for household rituals were

(a) squares and rectangles

(b) squares and circles

(c) triangles and rectangles

(d) trapeziums and pyramidsSolution 1

Correct option: (b)

Squares and circular altars were used for household rituals.

Whereas altars having shapes as combinations of rectangles, triangles and trapeziums were used for public worship.Question 2

In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for

(a) household rituals

(b) public rituals

(c) both (a) and (b)

(d) none of (a), (b) and (c)Solution 2

Correct option: (b)

In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for public rituals. Question 3

The number of interwoven isosceles triangles in Sriyantra is

(a) five

(b) seven

(c) nine

(d) elevenSolution 3

Correct option: (c)

The Sriyantra consists of nine interwoven isosceles triangles.Question 4

In Indus Valley Civilization (about 300 BC) the bricks used for construction work were having dimensions in the ratio

(a) 5:3:2

(b) 4:2:1

(c) 4:3:2

(d) 6:4:2Solution 4

Correct option: (b)

In Indus Valley Civilization (about 300 BC) the bricks used for construction work were having dimensions in the ratio is 4:2:1.Question 5

Into how many chapters was the famous treatise, ‘The Elements’ divided by Euclid?

(a) 13

(b) 12

(c) 11

(d) 9Solution 5

Correct option: (a)

The famous treatise ‘The Elements’ was divided into 13 chapters by Euclid.Question 6

Euclid belongs to the country

(a) India

(b) Greece

(c) Japan

(d) EgyptSolution 6

Correct option: (b)

Euclid belongs to the country, Greece.Question 7

Thales belongs to the country

(a) India

(b) Egypt

(c) Greece

(d) BabyloniaSolution 7

Correct option: (c)

Thales belongs to the country, Greece.Question 8

Pythagoras was a student of

(a) Euclid

(b) Thales

(c) Archimedas

(d) BhaskaraSolution 8

Correct option: (b)

Pythagoras was a student of Thales.Question 9

Which of the following needs a proof?

(a) axiom

(b) postulate

(c) definition

(d) theoremSolution 9

Correct option: (d)

A statement that requires a proof is called a theorem.Question 10

‘Lines are parallel if they do not intersect’ is started in the form of

(a) a definition

(b) an axiom

(c) a postulate

(d) a theoremSolution 10

Correct option: (a)

‘Lines are parallel if they do not intersect’ is started in the form of a definition.Question 11

Euclid stated that ‘All right angles are equal to each other’ in the form of

(a) a definition

(b) an axiom

(c) a postulate

(d) a proofSolution 11

Correct option: (c)

Euclid stated that ‘All right angles are equal to each other’ in the form of a postulate.

This is Euclid’s Postulate 4.

Note: The answer in the book is option (a). But if you have a look at the Euclid’s postulate, the answer is a postulate.Question 12

A pyramid is a solid figure, whose base is

(a) only a triangle

(b) only a square

(c) only a rectangle

(d) any polygonSolution 12

Correct option: (d)

A pyramid is a solid figure, whose base is any polygon.Question 13

The side faces of a pyramid are

(a) triangles

(b) squares

(c) trapeziums

(d) polygonsSolution 13

Correct option: (a)

The side faces of a pyramid are triangles.Question 14

The number of dimensions of a solid are

(a) 1

(b) 2

(c) 3

(d) 5Solution 14

Correct option: (c)

A solid has 3 dimensions.Question 15

The number of dimensions of a surface is

(a) 1

(b) 2

(c) 3

(d) 0Solution 15

Correct option: (b)

A surface has 2 dimensions.Question 16

How many dimensions dose a point have

(a) 0 dimension

(b) 1 dimension

(c) 2 dimension

(d) 3 dimensionSolution 16

Correct option: (a)

A point is an exact location. A fine dot represents a point. So, a point has 0 dimensions.Question 17

Boundaries of solids are

(a) lines

(b) curves

(c) surfaces

(d) none of theseSolution 17

Correct option: (c)

Boundaries of solids are surfaces.Question 18

Boundaries of surfaces are

(a) lines

(b) curves

(c) polygons

(d) none of theseSolution 18

Correct option: (b)

Boundaries of surfaces are curves.Question 19

The number of planes passing through three non-collinear points is

(a) 4

(b) 3

(c) 2

(d) 1Solution 19

Correct option: (d)

The number of planes passing through three non-collinear points is 1.Question 20

Axioms are assumed

(a) definitions

(b) theorems

(c) universal truths specific to geometry

(d) universal truths in all branches of mathematicsSolution 20

Correct option: (d)

Axioms are assumed as universal truths in all branches of mathematics because they are taken for granted, without proof.Question 21

Which of the following is a true statement?

(a) The floor and a wall of a room are parallel planes

(b) The ceiling and a wall of a room are parallel planes.

(c) The floor and the ceiling of a room are the parallel planes.

(d) Two adjacent walls of a room are the parallel planes.Solution 21

Correct option: (c)

Two lines are said to be parallel, if they have no point in common.

Options (a), (b) and (d) have a common point, hence they are not parallel.

In option (c), the floor and the ceiling of a room are parallel to each other is a true statement.Question 22

Which of the following is true statement?

(a) Only a unique line can be drawn to pass through a given point

(b) Infinitely many lines can be drawn to pass through two given points

(c) If two circles are equal, then their radii are equal

(d)A line has a definite length.Solution 22

Correct option: (c)

In option (a), infinite number of line can be drawn to pass through a given point. So, it is not a true statement.

In option (b), only one line can be drawn to pass through two given points. So, it is not a true statement.

In option (c),

‘If two circles are equal, then their radii are equal’ is the true statement.

In option (d), A line has no end points. A line has an indefinite length. So, it is not a true statement.Question 23

Which of the following is a false statement?

(a) An infinite number of lines can be drawn to pass through a given point.

(b) A unique line can be drawn to pass through two given points.

(c) 

(d)A ray has one end point.Solution 23

Correct option: (c)

Option (a) is true, since we can pass an infinite number of lines through a given point.

Option (b) is true, since a unique line can be drawn to pass through two given points.

Consider option (c).

As shown in the above diagram, a ray has only one end-point. So, option (d) is true.

Hence, the only false statement is option (c).Question 24

A point C is called the midpoint of a line segment , if

(a) C is an interior point of AB

(b) AC = CB

(c) C is an interior point of AB such that =

(d) AC + CB = ABSolution 24

Correct option: (c)

A point C is called the midpoint of a line segment , if C is an interior point of AB such that =.

Question 25

A point C is said to lie between the points A and B if

(a) AC = CB

(b) AC + CB = AB

(c) points A, C and B are collinear

(d) options (b) and (c)

* Options modifiedSolution 25

Correct option: (d)

Observe the above figure. Clearly, C lies between A and B if AC + CB = AB.

That means, points A, B, C are collinear.Question 26

Euclid’s which axiom illustrates the statement that when x + y = 15, then x + y + z = 15 + z?

(a) first

(b) second

(c) third

(d) fourthSolution 26

Correct option: (b)

Euclid’s second axiom states that ‘If equals are added to equals, the wholes are equal’.

Hence, when x + y = 15, then x + y + z = 15 + z. Question 27

A is of the same age as B and C is of the same age as B. Euclid’s which axiom illustrates the relative ages of A and C?

(a) First axiom

(b) Second axiom

(c) Third axiom

(d) Fourth axiomSolution 27

Correct option: (a)

Euclid’s first axiom states that ‘Things which are equal to the same thing are equal to one another’.

That is,

A’s age = B’s age and C’s age = B’ age

⇒ A’s age = C’s age 

Exercise Ex. 6

Question 1

What is the difference between a theorem and an axiom?Solution 1

A theorem is a statement that requires a proof. Whereas, a basic fact which is taken for granted, without proof, is called an axiom.

Example of Theorem: Pythagoras Theorem

Example of axiom: A unique line can be drawn through any two points.
Question 2

Define the following terms:

(i) Line segment (ii) Ray (iii) Intersecting lines (iv) Parallel lines (v) Half-line (vi) Concurrent lines (vii) Collinear points (viii) PlaneSolution 2

(i) Line segment: The straight path between two points is called a line segment.

(ii) Ray: A line segment when extended indefinitely in one direction is called a ray.

(iii) Intersecting Lines: Two lines meeting at a common point are called intersecting lines, i.e., they have a common point.

(iv) Parallel Lines: Two lines in a plane are said to be parallel, if they have no common point, i.e., they do not meet at all.

(v) Half-line: A ray without its initial point is called a half-line.

(vi) Concurrent lines: Three or more lines are said to be concurrent, if they intersect at the same point.

(vii) Collinear points: Three or more than three points are said to be collinear, if they lie on the same line.

(viii) Plane: A plane is a surface such that every point of the line joining any two points on it, lies on it.Question 3

In the adjoining figure, name:

(i) Six points

(ii) Five line segments

(iii) Four rays

(iv) Four lines

(v) Four collinear points

Solution 3

(i) Six points: A,B,C,D,E,F

(ii) Five line segments: 

(iii) Four rays: 

(iv) Four lines: 

(vi) Four collinear points: M,E,G,BQuestion 4

In the adjoining figure, name:

(i) Two pairs of intersecting lines and their corresponding points of intersection

(ii) Three concurrent lines and their points of intersection

(iii) Three rays

(iv) Two line segments

Solution 4

(i)  and their corresponding point of intersection is R.

 and their corresponding point of intersection is P.

(ii)  and their point of intersection is R.

(iii) Three rays are:

.

(iv) Two line segments are:

.Question 5

From the given figure, name the following:

(i) Three lines

(ii) One rectilinear figure

(iii) Four concurrent pointsSolution 5

(i) Three lines: Line AB, Line PQ and Line RS

(ii) One rectilinear figure: EFGC

(iii) Four concurrent points: Points A, E, F and BQuestion 6

(i) How many lines can be drawn to pass through a given point?

(ii) How many lines can be drawn to pass through two given points?

(iii) In how many points can the two lines at the most intersect?

(iv) If A, B, C are three collinear points, name all the line segments determined by them.Solution 6

(i) An infinite number of lines can be drawn to pass through a given point.

(ii) One and only one line can pass through two given points.

(iii) Two given lines can at the most intersect at one and only one point.

(iv) Question 7

Which of the following statements are true?

(i) A line segments has no definite length.

(ii) A ray has no end point.

(iii) A line has a definite length.

(iv) A line is the same as line .

(v) A ray is the same as ray .

(vi) Two distinct points always determine a unique line.

(vii) Three lines are concurrent if they have a common point.

(viii) Two distinct lines cannot have more than one point in common.

(ix) Two intersecting lines cannot be both parallel to the same line.

(x) Open half-line OA is the same thing as ray 

(xi) Two lines may intersect in two points.

(xii) Two lines l and m are parallel only when they have no point in common.Solution 7

(i) False

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

(vii) True

(viii) True

(ix) True

(x) True

(xi) False

(xii) TrueQuestion 8

In the given figure, L and M are mid-points of AB and BC respectively.

(i) If AB = BC, prove that AL = MC.

(ii) If BL = BM, prove that AB = BC.Solution 8

(ii) BL = BM

 ⇒ 2BL = 2BM

 ⇒ AB = BC

Exercise MCQ

Question 1

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

(a) an isosceles triangle

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangleSolution 1

Correct option: (d)

In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.Question 2

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is

(a) 70° 

(b) 55° 

(c) 35° 

(d) Solution 2

Correct option: (b)

Let each interior opposite angle be x.

Then, x + x = 110° (Exterior angle property of a triangle)

⇒ 2x = 110° 

⇒ x = 55° Question 3

The angles of a triangle are in the ratio 3:5:7 The triangle is

1. Acute angled
2. Obtuse angled
3. Right angled
4. an isosceles triangle

Solution 3

Question 4

If one of the angles of triangle is 130° then the angle between the bisector of the other two angles can be

(a) 50° 

(b) 65° 

(c) 90° 

(d) 155Solution 4

Correct option: (d)

Let ∠A = 130° 

In ΔABC, by angle sum property,

∠B + ∠C + ∠A = 180° 

⇒ ∠B + ∠C + 130° = 180° 

⇒ ∠B + ∠C = 50° 

Question 5

In the given figure, AOB is a straight line. The value of x is

(a) 12

(b) 15

(c) 20

(d) 25Solution 5

Correct option: (b)

AOB is a straight line.

⇒ ∠AOB = 180° 

⇒ 60° + 5x° + 3x° = 180° 

⇒ 60° + 8x° = 180° 

⇒ 8x° = 120° 

⇒ x = 15° Question 6

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is

(a) 120° 

(b) 100° 

(c) 80° 

(d) 60° Solution 6

Correct option: (c)

By angle sum property,

2x + 3x + 4x = 180° 

⇒ 9x = 180° 

⇒ x = 20° 

Hence, largest angle = 4x = 4(20°) = 80° Question 7

In the given figure, ∠OAB = 110° and ∠BCD = 130° then ∠ABC is equal to

(a) 40° 

(b) 50° 

(c) 60° 

(d) 70° Solution 7

Correct option: (c)

Through B draw YBZ ∥ OA ∥ CD.

Now, OA ∥ YB and AB is the transversal.

⇒ ∠OAB + ∠YBA = 180° (interior angles are supplementary)

⇒ 110° + ∠YBA = 180° 

⇒ ∠YBA = 70° 

Also, CD ∥ BZ and BC is the transversal.

⇒ ∠DCB + ∠CBZ = 180° (interior angles are supplementary)

⇒ 130° + ∠CBZ = 180° 

⇒ ∠CBZ = 50° 

Now, ∠YBZ = 180° (straight angle)

⇒ ∠YBA + ∠ABC + ∠CBZ = 180° 

⇒ 70° + x + 50° = 180° 

⇒ x = 60° 

⇒ ∠ABC = 60° Question 8

If two angles are complements of each other, then each angle is

1. An acute angle
2. An obtuse angle
3. A right angle
4. A reflex angle

Solution 8

Correct option: (a)

Two angles are said to be complementary, if the sum of their measures is 90°.

Clearly, the measures of each of the angles have to be less than 90°.

Hence, each angle is an acute angle.Question 9

An angle which measures more than 180° but less than 360°, is called

1. An acute angle
2. An obtuse angle
3. A straight angle
4. A reflex angle

Solution 9

Correct option: (d)

An angle which measures more than 180^o but less than 360^o is called a reflex angle.Question 10

The measure of an angle is five times its complement. The angle measures

1. 25°
2. 35°
3. 65°
4. 75°

Solution 10

Question 11

Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures

1. 72°^o
2. 54°
3. 63°
4. 36°

Solution 11

Question 12

In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC =?

Solution 12

Question 13

In the given figure, AOB is a straight line. If ∠AOC = (3x + 10) ° and ∠BOC = (4x – 26) °, then ∠BOC =?

1. 96°
2. 86°
3. 76°
4. 106°

Solution 13

Question 14

In the given figure, AOB is a straight line. If ∠AOC = (3x – 10) °, ∠COD = 50° and ∠BOD = (x +20) °, then ∠AOC =?

1. 40°
2. 60°
3. 80°
4. 50°

Solution 14

Question 15

Which of the following statements is false?

1. Through a given point, only one straight line can be drawn
2. Through two given points, it is possible to draw one and only one straight line.
3. Two straight lines can intersect only at one point
4. A line segment can be produced to any desired length.

Solution 15

Correct option: (a)

Option (a) is false, since through a given point we can draw an infinite number of straight lines.Question 16

An angle is one-fifth of its supplement. The measure of the angle is

1. 15°
2. 30°
3. 75°
4. 150°

Solution 16

Question 17

In the adjoining figure, AOB is straight line. If x:y:z = 4:5:6, then y = ?

1. 60°
2. 80°
3. 48°
4. 72°

Solution 17

Question 18

In the given figure, straight lines AB and CD intersect at O. If ∠AOC =ϕ, ∠BOC = θ and θ = 3 ϕ, then ϕ =?

1. 30°
2. 40°
3. 45°
4. 60°

Solution 18

Question 19

In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD =?

1. 65°
2. 115°
3. 110°
4. 125°

Solution 19

Question 20

In the given figure AB is a mirror, PQ is the incident ray and and QR is the reflected ray. If ∠PQR = 108°, then ∠ AQP =?

1. 72°
2. 18°
3. 36°
4. 54°

Solution 20

Question 21

In the given figure, AB ∥ CD, If ∠BAO = 60° and ∠OCD = 110° then ∠AOC = ?

(a) 70° 

(b) 60° 

(c) 50° 

(d) 40° Solution 21

Correct option: (c)

Let ∠AOC = x° 

Draw YOZ ∥ CD ∥ AB.

Now, YO ∥ AB and OA is the transversal.

⇒ ∠YOA = ∠OAB = 60° (alternate angles)

Again, OZ ∥ CD and OC is the transversal.

⇒ ∠COZ + ∠OCD = 180° (interior angles)

⇒ ∠COZ + 110° = 180° 

⇒ ∠COZ = 70° 

Now, ∠YOZ = 180° (straight angle)

⇒ ∠YOA + ∠AOC + ∠COZ = 180° 

⇒ 60° + x + 70° = 180° 

⇒ x = 50° 

⇒ ∠AOC = 50° Question 22

In the given figure, AB ‖ CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD =?

1. 130°
2. 150°
3. 80°
4. 100°

Solution 22

Question 23

In the given figure, AB ‖ CD. If ‖CAB = 80^o and ∠EFC= 25°, then ∠CEF =?

1. 65°
2. 55°
3. 45°
4. 75°

Solution 23

Question 24

In the given figure, AB ‖ CD, CD ‖ EF and y:z = 3:7, then x = ?

1. 108°
2. 126°
3. 162°
4. 63°

Solution 24

Question 25

In the given figure, AB ‖ CD. If ∠APQ = 70° and ∠RPD = 120°, then ∠QPR =?

1. 50°
2. 60°
3. 40°
4. 35°

Solution 25

Question 26

In the given figure AB ‖ CD. If ∠EAB = 50° and ∠ECD=60°, then ∠AEB =?

1. 50°
2. 60°
3. 70°
4. 50°

Solution 26

Question 27

In the given figure, ∠OAB = 75°, ∠OBA=55° and ∠OCD = 100°. Then ∠ODC=?

1. 20°
2. 25°
3. 30°
4. 35°

Solution 27

Question 28

In the adjoining figure y =?

1. 36°
2. 54°
3. 63°
4. 72°

Solution 28

Exercise Ex. 7A

Question 1

Define the following terms:

(i) Angle (ii) Interior of an angle (iii) Obtuse angle (iv) Reflex angle (v) Complementary angles (vi) Supplementary anglesSolution 1

(i) Angle: Two rays having a common end point form an angle.

(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.

(iii) Obtuse angle: An angle whose measure is more than 90^o but less than 180^o, is called an obtuse angle.

(iv) Reflex angle: An angle whose measure is more than 180^o but less than 360^o is called a reflex angle.

(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90^o.

(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180^o.
Question 2(ii)

Find the complement of each of the following angles.

16^oSolution 2(ii)

Complement of 16^o = 90 – 16^o = 74^oQuestion 2(iv)

Find the complement of each of the following angles.

46^o 30Solution 2(iv)

Complement of 46^o 30′ = 90^o – 46^o 30′ = 43^o 30’Question 2(i)

Find the complement of each of the following angle:

55° Solution 2(i)

Complement of 55° = 90° – 55° = 35°  Question 2(iii)

Find the complement of each of the following angle:

90° Solution 2(iii)

Complement of 90° = 90° – 90° = 0° Question 3(iv)

Find the supplement of each of the following angles.

75^o 36’Solution 3(iv)

Supplement of 75^o 36′ = 180^o – 75^o 36′ = 104^o 24’Question 3(i)

Find the supplement of each of the following angle:

42° Solution 3(i)

Supplement of 42° = 180° – 42° = 138° Question 3(ii)

Find the supplement of each of the following angle:

90° Solution 3(ii)

Supplement of 90° = 180° – 90° = 90° Question 3(iii)

Find the supplement of each of the following angle:

124° Solution 3(iii)

Supplement of 124° = 180° – 124° = 56° Question 4

Find the measure of an angle which is

(i) equal to its complement, (ii) equal to its supplement.Solution 4

(i) Let the required angle be x^o

Then, its complement = 90^o – x^o

 The measure of an angle which is equal to its complement is 45^o.

(ii) Let the required angle be x^o

Then, its supplement = 180^o – x^o

 The measure of an angle which is equal to its supplement is 90^o.Question 5

Find the measure of an angle which is 36^o more than its complement.Solution 5

Let the required angle be x^o

Then its complement is 90^o – x^o

 The measure of an angle which is 36^o more than its complement is 63^o.Question 6

Find the measure of an angle which is 30° less than its supplement.Solution 6

Let the measure of the required angle = x° 

Then, measure of its supplement = (180 – x)° 

It is given that

x° = (180 – x)° – 30° 

⇒ x° = 180° – x° – 30° 

⇒ 2x° = 150° 

⇒ x° = 75° 

Hence, the measure of the required angle is 75°. Question 7

Find the angle which is four times its complement.Solution 7

Let the required angle be x^o

Then, its complement = 90^o – x^o

 The required angle is 72^o.Question 8

Find the angle which is five times its supplement.Solution 8

Let the required angle be x^o

Then, its supplement is 180^o – x^o

 The required angle is 150^o.Question 9

Find the angle whose supplement is four times its complement.Solution 9

Let the required angle be x^o

Then, its complement is 90^o – x^o and its supplement is 180^o – x^o_{That is we have,}

 The required angle is 60^o.
Question 10

Find the angle whose complement is one-third of its supplement.Solution 10

Let the required angle be x^o

Then, its complement is 90^o – x^o and its supplement is 180^o – x^o

 The required angle is 45^o.Question 11

Two complementary angles are in the ratio 4: 5. Find the angles.Solution 11

Let the two required angles be x^o and 90^o – x^o.

Then 

 5x = 4(90 – x)

 5x = 360 – 4x

 5x + 4x = 360

 9x = 360

Thus, the required angles are 40^o and 90^o – x^o = 90 ^o – 40^o = 50^o.
Question 12

Find the value of x for which the angles (2x – 5)° and (x – 10)° are the complementary angles.Solution 12

(2x – 5)° and (x – 10)° are complementary angles.

∴ (2x – 5)° + (x – 10)° = 90° 

⇒ 2x – 5° + x – 10° = 90° 

⇒ 3x – 15° = 90° 

⇒ 3x = 105° 

⇒ x = 35° 

Exercise Ex. 7B

Question 1

In the given figure, AOB is a straight line. Find the value of x.

Solution 1

Since BOC and COA form a linear pair of angles, we have

BOC + COA = 180^o

 x^o + 62^o = 180^o

 x = 180 – 62

 x = 118^oQuestion 2

In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC and ∠BOD.

Solution 2

∠AOB is a straight angle.

⇒ ∠AOB = 180° 

⇒ ∠AOC + ∠COD + ∠BOD = 180° 

⇒ (3x – 7)° + 55° + (x + 20)° = 180° 

⇒ 4x + 68° = 180° 

⇒ 4x = 112° 

⇒ x = 28° 

Thus, ∠AOC = (3x – 7)° = 3(28°) – 7° = 84° – 7° = 77° 

And, ∠BOD = (x + 20)° = 28° + 20° = 48° Question 3

In the given figure, AOB is a straight line. Find the value of x. Hence, find AOC, COD and BOD.

Solution 3

Since BOD and DOA from a linear pair of angles.

 BOD + DOA = 180^o

 BOD + DOC + COA = 180^o

 x^o + (2x – 19)^o + (3x + 7)^o = 180^o

 6x – 12 = 180

 6x = 180 + 12 = 192

 x = 32

 AOC = (3x + 7)^o = (3  32 + 7)^o = 103^o

 COD = (2x – 19)^o = (2  32 – 19)^o = 45^o

and BOD = x^o = 32^o
Question 4

In the given figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.

Solution 4

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15

But x + y + z = 180^o

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5.

If the sum of angles is 180^o, then, measure of  

And, if the total sum of the measures is 15, then the measure of y is 4.

If the sum of the angles is 180^o, then, measure of 

And z = 180^o – x – y

= 180^o – 60^o – 48^o

= 180^o – 108^o = 72^o

 x = 60, y = 48 and z = 72.
Question 5

In the given figure, what value of x will make AOB, a straight line?

Solution 5

AOB will be a straight line, if two adjacent angles form a linear pair.

BOC + AOC = 180^o

 (4x – 36)^o + (3x + 20)^o = 180^o

 4x – 36 + 3x + 20 = 180

 7x – 16 = 180^o

 7x = 180 + 16 = 196

 The value of x = 28.Question 6

Two lines AB and CD intersect at O. If AOC = 50^o, find AOD, BOD and BOC.

Solution 6

Since AOC and AOD form a linear pair.

AOC + AOD = 180^o

 50^o + AOD = 180^o

 AOD = 180^o – 50^o = 130^o

AOD and BOC are vertically opposite angles.

 AOD = BOC

 BOC = 130^o

BOD and AOC are vertically opposite angles.

BOD = AOC

 BOD = 50^oQuestion 7

In the given figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the values of x,y,z and t.

Solution 7

Since COE and DOF are vertically opposite angles, we have,

COE = DOF

 z = 50^o

Also BOD and COA are vertically opposite angles.

So, BOD = COA

 t = 90^o

As COA and AOD form a linear pair,

COA + AOD = 180^o

 COA + AOF + FOD = 180^o [t = 90^o]

 t + x + 50^o = 180^o

 90^o + x^o + 50^o = 180^o

 x + 140 = 180

 x = 180 – 140 = 40

Since EOB and AOF are vertically opposite angles

So, EOB = AOF

 y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90Question 8

In the given figure, three coplanar lines AB,CD and EF intersect at a point O. Find the value of x. Hence, find AOD, COE and AOE.

Solution 8

Since COE and EOD form a linear pair of angles.

 COE + EOD = 180^o

 COE + EOA + AOD = 180^o

 5x + EOA + 2x = 180

 5x + BOF + 2x = 180

[EOA and BOF are vertically opposite angles so, EOA = BOF]

 5x + 3x + 2x = 180

 10x = 180

 x = 18

Now AOD = 2x^o = 2  18^o = 36^o

COE = 5x^o = 5  18^o = 90^o

and, EOA = BOF = 3x^o = 3  18^o = 54^oQuestion 9

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each one of these angles.Solution 9

Let the two adjacent angles be 5x and 4x.

Now, since these angles form a linear pair.

So, 5x + 4x = 180^o

 9x = 180^o

 The required angles are 5x = 5x = 5  20^o = 100^o

and 4x = 4  20^o = 80^oQuestion 10

If two straight lines intersect each other in such a way that one of the angles formed measures 90^o, show that each of the remaining angles measures 90^o.Solution 10

Let two straight lines AB and CD intersect at O and let AOC = 90^o.

Now, AOC = BOD [Vertically opposite angles]

 BOD = 90^o

Also, as AOC and AOD form a linear pair.

 90^o + AOD = 180^o

 AOD = 180^o – 90^o = 90^o

Since, BOC = AOD [Verticallty opposite angles]

 BOC = 90^o

Thus, each of the remaining angles is 90^o.Question 11

Two lines AB and CD intersect at a point O such that BOC +AOD = 280^o, as shown in the figure. Find all the four angles.

Solution 11

Since, AOD and BOC are vertically opposite angles.

AOD = BOC

Now, AOD + BOC = 280^o [Given]

 AOD + AOD = 280^o

 2AOD = 280^o

 AOD = 

 BOC = AOD = 140^o

As, AOC and AOD form a linear pair.

So, AOC + AOD = 180^o

 AOC + 140^o = 180^o

 AOC = 180^o – 140^o = 40^o

Since, AOC and BOD are vertically opposite angles.

AOC = BOD

 BOD = 40^o

 BOC = 140^o, AOC = 40^o , AOD = 140^o and BOD = 40^o.Question 12

Two lines AB and CD intersect each other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.

Solution 12

Let ∠AOC = 5x and ∠AOD = 7x

Now, ∠AOC + ∠AOD = 180° (linear pair of angles)

⇒ 5x + 7x = 180° 

⇒ 12x = 180° 

⇒ x = 15° 

⇒ ∠AOC = 5x = 5(15°) = 75° and ∠AOD = 7x = 7(15°) = 105° 

Now, ∠AOC = ∠BOD (vertically opposite angles)

⇒ ∠BOD = 75° 

Also, ∠AOD = ∠BOC (vertically opposite angles)

⇒ ∠BOC = 105° Question 13

In the given figure, three lines AB, CD and EF intersect at a point O such that ∠AOE = 35° and ∠BOD = 40°. Find the measure of ∠AOC, ∠BOF, ∠COF and ∠DOE.

Solution 13

∠BOD = 40° 

⇒ AOC = ∠BOD = 40° (vertically opposite angles)

∠AOE = 35° 

⇒ ∠BOF = ∠AOE = 35° (vertically opposite angles)

∠AOB is a straight angle.

⇒ ∠AOB = 180° 

⇒ ∠AOE + ∠EOD + ∠BOD = 180° 

⇒ 35° + ∠EOD + 40° = 180° 

⇒ ∠EOD + 75° = 180° 

⇒ ∠EOD = 105° 

Now, ∠COF = ∠EOD = 105° (vertically opposite angles)Question 14

In the given figure, the two lines AB and CD intersect at a point O such that ∠BOC = 125°. Find the values of x, y and z.

Solution 14

∠AOC + ∠BOC = 180° (linear pair of angles)

⇒ x + 125 = 180° 

⇒ x = 55° 

Now, ∠AOD = ∠BOC  (vertically opposite angles)

⇒ y = 125° 

Also, ∠BOD = ∠AOC (vertically opposite angles)

⇒ z = 55° Question 15

If two straight lines intersect each other then prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.Solution 15

Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the BOD. OF is a ray opposite to ray OE.

To Prove: AOF = COF

Proof : Since  are two opposite rays,  is a straight line passing through O.

 AOF = BOE

and COF = DOE

[Vertically opposite angles]

But BOE = DOE (Given)

 AOF = COF

Hence, proved.Question 16

Prove that the bisectors of two adjacent supplementary angles include a right angle.Solution 16

Given: is the bisector of BCD and is the bisector of ACD.

To Prove: ECF = 90^o

Proof: Since ACD and BCD forms a linear pair.

ACD + BCD = 180^o

ACE + ECD + DCF + FCB = 180^o

 ECD + ECD + DCF + DCF = 180^o

because ACE = ECD

and DCF = FCB

 2(ECD) + 2 (CDF) = 180^o

 2(ECD + DCF) = 180^o

 ECD + DCF = 

 ECF = 90^o (Proved)

Exercise Ex. 7C

Question 1

In the given figure, l ∥ m and a transversal t cuts them. If ∠1 = 120°, find the measure of each of the remaining marked angles.

Solution 1

Given, ∠1 = 120° 

Now, ∠1 + ∠2 = 180° (linear pair)

⇒ 120° + ∠2 = 180° 

⇒ ∠2 = 60° 

∠1 = ∠3  (vertically opposite angles)

⇒ ∠3 = 120° 

Also, ∠2 = ∠4  (vertically opposite angles)

⇒ ∠4 = 60° 

Line l ∥ line m and line t is a transversal.

⇒ ∠5 = ∠1 = 120° (corresponding angles)

 ∠6 = ∠2 = 60° (corresponding angles)

 ∠7 = ∠3 = 120° (corresponding angles)

 ∠8 = ∠4 = 60° (corresponding angles)Question 2

In the given figure, l ∥ m and a transversal t cuts them. If ∠7 = 80°, find the measure of each of the remaining marked angles.

Solution 2

Given, ∠7 = 80° 

Now, ∠7 + ∠8 = 180° (linear pair)

⇒ 80° + ∠8 = 180° 

⇒ ∠8 = 100° 

∠7 = ∠5 (vertically opposite angles)

⇒ ∠5 = 80° 

Also, ∠6 = ∠8 (vertically opposite angles)

⇒ ∠6 = 100° 

Line l ∥ line m and line t is a transversal.

⇒ ∠1 = ∠5 = 80°  (corresponding angles)

 ∠2 = ∠6 = 100° (corresponding angles)

 ∠3 = ∠7 = 80°  (corresponding angles)

 ∠4 = ∠8 = 100° (corresponding angles) Question 3

In the given figure, l ∥ m and a transversal t cuts them. If ∠1 : ∠2 = 2 : 3, find the measure of each of the marked angles.

Solution 3

Given, ∠1 : ∠2 = 2 : 3

Now, ∠1 + ∠2 = 180° (linear pair)

⇒ 2x + 3x = 180° 

⇒ 5x = 180° 

⇒ x = 36° 

⇒ ∠1 = 2x = 72° and ∠2 = 3x = 108° 

∠1 = ∠3 (vertically opposite angles)

⇒ ∠3 = 72° 

Also, ∠2 = ∠4  (vertically opposite angles)

⇒ ∠4 = 108° 

Line l ∥ line m and line t is a transversal.

⇒ ∠5 = ∠1 = 72°  (corresponding angles)

 ∠6 = ∠2 = 108° (corresponding angles)

 ∠7 = ∠3 = 72°  (corresponding angles)

 ∠8 = ∠4 = 108° (corresponding angles) Question 4

For what value of x will the lines l and m be parallel to each other?

Solution 4

Lines l and m will be parallel if 3x – 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

3x – 2x = 10 + 20

x = 30Question 5

For what value of x will the lines l and m be parallel to each other?

*Question modified, back answer incorrect.Solution 5

For lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be equal.

⇒ (3x + 5)° = 4x° 

⇒ x = 5° Question 6

In the given figure, AB || CD and BC || ED. Find the value of x.

Solution 6

Since AB || CD and BC is a transversal.

So, BCD = ABC = x^o     [Alternate angles]

As BC || ED and CD is a transversal.

BCD + EDC = 180^o    

  BCD + 75^o =180^o

 BCD = 180^o – 75^o = 105^o 

 ABC = 105^o                 [since BCD = ABC]

 x^o = ABC = 105^o

Hence, x = 105.
Question 7

In the given figure, AB || CD || EF. Find the value of x.

Solution 7

Since AB || CD and BC is a transversal.

So, ABC = BCD                [atternate interior angles]

70^o = x^o + ECD(i)

Now, CD || EF and CE is transversal.

So,ECD + CEF = 180^o    [sum of consecutive interior angles is 180^o]

ECD + 130^o = 180^o

ECD = 180^o – 130^o = 50^o

Putting ECD = 50^o in (i) we get,

70^o = x^o + 50^o

x = 70 – 50 = 20Question 8

In the give figure, AB ∥ CD. Find the values of x, y and z.

Solution 8

AB ∥ CD and EF is transversal.

⇒ ∠AEF = ∠EFG (alternate angles)

Given, ∠AEF = 75° 

⇒ ∠EFG = y = 75° 

Now, ∠EFC + ∠EFG = 180° (linear pair)

⇒ x + y = 180° 

⇒ x + 75° = 180° 

⇒ x = 105° 

∠EGD = ∠EFG + ∠FEG (Exterior angle property)

⇒ 125° = y + z

⇒ 125° = 75° + z

⇒ z = 50° 

Thus, x = 105°, y = 75° and z = 50° Question 9(i)

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9(i)

Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So,GED = EDC = 65^o[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,BEG = ABE = 35^o[Alternate interior angles]

So,DEB = x^o

BEG + GED = 35^o + 65^o = 100^o.

Hence, x = 100.Question 9(ii)

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9(ii)

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

CDO + FOD = 180^o

[sum of consecutive interior angles is 180^o]

25^o + FOD = 180^o

FOD = 180^o – 25^o = 155^o

As OF || CD and AB || CD [Given]

Thus, OF || AB and OB is a transversal.

So,ABO + FOB = 180^o    [sum of consecutive interior angles is 180^o]

55^o + FOB = 180^o

FOB = 180^o – 55^o = 125^o

Now, x^o = FOB + FOD = 125^o + 155^o = 280^o.

Hence, x = 280.Question 9(iii)

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9(iii)

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

FEC + ECD = 180^o

[sum of consecutive interior angles is 180^o]

FEC + 124^o = 180^o

FEC = 180^o – 124^o = 56^o

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So,BAE + FEA = 180^o

[sum of consecutive interior angles is 180^o]

116^o + FEA = 180^o

FEA = 180^o – 116^o = 64^o

Thus,x^o = FEA + FEC

= 64^o + 56^o = 120^o.

Hence, x = 120.Question 10

In the given figure, AB || CD. Find the value of x.

Solution 10

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, GCE = CEA = 20^o            [Alternate angles]

 DCG = 130^o – GCE

= 130^o – 20^o = 110^o

Also, we have AB || CD and FG is a transversal.

So, BFC = DCG = 110^o          [Corresponding angles]

As, FG || AE, AF is a transversal.

BFG = FAE                           [Corresponding angles]

 x^o = FAE = 110^o.

Hence, x = 110Question 11

In the given figure, AB || PQ. Find the values of x and y.

Solution 11

Since AB || PQ and EF is a transversal.

So, CEB = EFQ                 [Corresponding angles]

 EFQ = 75^o

 EFG + GFQ = 75^o

 25^o + y^o = 75^o

 y = 75 – 25 = 50

Also, BEF + EFQ = 180^o   [sum of consecutive interior angles is 180^o]       BEF = 180^o – EFQ

           = 180^o – 75^o

  BEF = 105^o

 FEG + GEB = BEF = 105^o

 FEG = 105^o – GEB = 105^o – 20^o = 85^o

In EFG we have,

x^o + 25^o + FEG = 180^o

Hence, x = 70.
Question 12

In the given figure, AB || CD. Find the value of x.

Solution 12

Since AB || CD and AC is a transversal.

So, BAC + ACD = 180^o   [sum of consecutive interior angles is 180^o]

 ACD = 180^o – BAC

= 180^o – 75^o = 105^o

ECF = ACD                     [Vertically opposite angles]

 ECF = 105^o

Now in CEF,

ECF + CEF + EFC =180^o 105^o + x^o + 30^o = 180^o

 x = 180 – 30 – 105 = 45

_{Hence, x = 45.}
Question 13

In the given figure, AB || CD. Find the value of x.

Solution 13

Since AB || CD and PQ a transversal.

So, PEF = EGH [Corresponding angles]

 EGH = 85^o

EGH and QGH form a linear pair.

So, EGH + QGH = 180^o

 QGH = 180^o – 85^o = 95^o

Similarly, GHQ + 115^o = 180^o

 GHQ = 180^o – 115^o = 65^o

In GHQ, we have,

x^o + 65^o + 95^o = 180^o

 x = 180 – 65 – 95 = 180 – 160

 x = 20
Question 14

In the given figure, AB || CD. Find the values of x, y and z.

Solution 14

Since AB || CD and BC is a transversal.

So, ABC = BCD

 x = 35

Also, AB || CD and AD is a transversal.

So, BAD = ADC

 z = 75

In ABO, we have,

 x^o + 75^o + y^o = 180^o

 35 + 75 + y = 180

 y = 180 – 110 = 70

 x = 35, y = 70 and z = 75.
Question 16

In the given figure, AB || CD. Prove that p + q – r = 180.

Solution 16

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

KFG = FGD = r^o (i)

[alternate angles]

Again AE || KF, and EF is a transversal.

So,AEF + KFE = 180^o

KFE = 180^o – p^o (ii)

Adding (i) and (ii) we get,

KFG + KFE = 180 – p + r

EFG = 180 – p + r

q = 180 – p + r

i.e.,p + q – r = 180Question 17

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Solution 17

PRQ = x^o = 60^o            [vertically opposite angles]

Since EF || GH, and RQ is a transversal.

So, x = y                   [Alternate angles]

 y = 60

AB || CD and PR is a transversal.

So,          [Alternate angles]

      [since ]

x + QRD = 110^o

 QRD = 110^o – 60^o = 50^o

In QRS, we have,

QRD + t^o + y^o = 180^o

 50 + t + 60 = 180

 t = 180 – 110 = 70

Since, AB || CD and GH is a transversal

So, z^o = t^o = 70^o [Alternate angles]

 x = 60 , y = 60, z = 70 and t = 70
Question 18

In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF = 90°.

Solution 18

AB ∥ CD and a transversal t cuts them at E and F respectively.

⇒ ∠BEF + ∠DFE = 180° (interior angles)

⇒ ∠GEF + ∠GFE = 90° ….(i)

Now, in ΔGEF, by angle sum property

∠GEF + ∠GFE + ∠EGF = 180° 

⇒ 90° + ∠EGF = 180° ….[From (i)]

⇒ ∠EGF = 90° Question 19

In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP ∥ FQ.

Solution 19

Since AB ∥ CD and t is a transversal, we have

∠AEF = ∠EFD (alternate angles)

⇒ ∠PEF = ∠EFQ

But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.

∴ EP ∥ FQQuestion 20

In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC = ∠DEF.

Solution 20

Construction: Produce DE to meet BC at Z.

Now, AB ∥ DZ and BC is the transversal.

⇒ ∠ABC = ∠DZC (corresponding angles) ….(i)

Also, EF ∥ BC and DZ is the transversal.

⇒ ∠DZC = ∠DEF (corresponding angles) ….(ii)

From (i) and (ii), we have

∠ABC = ∠DEF Question 21

In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC + ∠DEF = 180°.

Solution 21

Construction: Produce ED to meet BC at Z.

Now, AB ∥ EZ and BC is the transversal.

⇒ ∠ABZ + ∠EZB = 180° (interior angles)

⇒ ∠ABC + ∠EZB = 180° ….(i)

Also, EF ∥ BC and EZ is the transversal.

⇒ ∠BZE = ∠ZEF (alternate angles)

⇒ ∠BZE = ∠DEF ….(ii)

From (i) and (ii), we have

∠ABC + ∠DEF = 180° Question 22

In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Solution 22

Let the normal to mirrors m and n intersect at P.

Now, OB ⊥ m, OC ⊥ n and m ⊥ n.

⇒ OB ⊥ OC

⇒ ∠APB = 90° 

⇒ ∠2 + ∠3 = 90° (sum of acute angles of a right triangle is 90°)

By the laws of reflection, we have

∠1 = ∠2 and ∠4 = ∠3 (angle of incidence = angle of reflection)

⇒ ∠1 + ∠4 = ∠2 + ∠3 = 90° 

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180° 

⇒ ∠CAB + ∠ABD = 180° 

But, ∠CAB and ∠ABD are consecutive interior angles formed, when the transversal AB cuts CA and BD.

∴ CA ∥ BD Question 23

In the figure given below, state which lines are parallel and why?

Solution 23

In the given figure,

∠BAC = ∠ACD = 110° 

But, these are alternate angles when transversal AC cuts AB and CD.

Hence, AB ∥ CD. Question 24

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.Solution 24

Let the two parallel lines be m and n.

Let p ⊥ m.

⇒ ∠1 = 90° 

Let q ⊥ n.

⇒ ∠2 = 90° 

Now, m ∥ n and p is a transversal.

⇒ ∠1 = ∠3 (corresponding angles)

⇒ ∠3 = 90° 

⇒ ∠3 = ∠2 (each 90°)

But, these are corresponding angles, when transversal n cuts lines p and q.

∴ p ∥ q. 

Hence, two lines which are perpendicular to two parallel lines, are parallel to each other. 

Exercise MCQ

Question 1

Which of the following is not a criterion for congruence of triangles?

(a) SSA

(b) SAS

(c) ASA

(d)SSSSolution 1

Correct option: (a)

SSA is not a criterion for congruence of triangles.Question 2

If AB = QR, BC = RP and CA = PQ, then which of the following holds?

(a) ∆ABC ≅ ∆PQR

(b) ∆CBA ≅ ∆PQR

(c) ∆CAB ≅ ∆PQR

(d) ∆BCA ≅ ∆PQRSolution 2

Correct option: (c)

Question 3

If ∆ABC ≅ ∆PQR then which of the following is not true?

(a) BC = PQ

(b) AC = PR

(c) BC = QR

(d) AB = PQSolution 3

Question 4

In Δ ABC, AB = AC and ∠B = 50°. Then, ∠A = ?

(a) 40° 

(b) 50° 

(c) 80° 

(d) 130° Solution 4

Correct option: (c)

In ΔABC,

AB = AC

⇒ ∠C = ∠B (angles opposite to equal sides are equal)

⇒ ∠C = 50° 

Now, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 50° + 50° = 180° 

⇒ ∠A = 80° Question 5

In Δ ABC, BC = AB and ∠B = 80°. Then, ∠A = ?

(a) 50° 

(b) 40° 

(c) 100° 

(d) 80° Solution 5

Correct option: (a)

In ΔABC,

BC = AB

⇒ ∠A = ∠C (angles opposite to equal sides are equal)

Now, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 80° + ∠A = 180° 

⇒ 2∠A = 100° 

⇒ ∠A = 50°  Question 6

In ΔABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?

(a) 4 cm

(b) 5 cm

(c) 8 cm

(d) 2.5 cmSolution 6

Correct option: (a)

In ΔABC,

∠C = ∠A

⇒ AB = BC (sides opposite to equal angles are equal)

⇒ AB = 4 cm Question 7

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be

(a) 6 cm

(b) 6.5 cm

(c) 5.5 cm

(d) 6.3 cmSolution 7

Correct option: (b)

The sum of any two sides of a triangle is greater than the third side.

Since, 4 cm + 2.5 cm = 6.5 cm

The length of third side of a triangle cannot be 6.5 cm. Question 8

In ΔABC, if ∠C > ∠B, then

(a) BC > AC

(b) AB > AC

(c) AB < AC

(d) BC < ACSolution 8

Correct option: (b)

We know that in a triangle, the greater angle has the longer side opposite to it.

In ΔABC,

∠C > ∠B

⇒ AB >AC  Question 9

It is given that ∆ABC ≅ ∆FDE in which AB = 5 cm, ∠B = 40^o, ∠A = 80^o and FD = 5 cm. Then which of the following is true?

(a) ∠D = 60^o

(b) ∠E = 60^o

(c) ∠F = 60^o

(d) ∠D = 80^oSolution 9

Question 10

In ∆ABC, ∠A = 40^o and ∠B = 60^o. Then the longest side of ∆ABC is

(a) BC

(b) AC

(c) AB

(d) Cannot be determinedSolution 10

Question 11

In the given figure AB > AC. Then, which of the following is true?

(a) AB < AD

(b) AB = AD

(c) AB > AD

(d) Cannot be determined

Solution 11

Correct option: (c)

Question 12

In the given figure AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then

(a) OB = OC

(b) OB > OC

(c) OB < OC

Solution 12

Question 13

In the given figure, AB = AC and OB = OC. Then, ∠ABO : ∠ACO = ?

(a) 1 :1

(b) 2 : 1

(c) 1 :2

(d) None of these 

Solution 13

Question 14

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is

(a) Equilateral

(b) Isosceles

(c) Scalene

(d) Right-angledSolution 14

Question 15

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have

(a) ∠A = ∠D

(b) ∠B = ∠E

(c) ∠C = ∠F

(d) None of these

Solution 15

Question 16

In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have

(a) AB = DF

(b) AC = DE

(c) BC = EF

(d) ∠A = ∠D

Solution 16

Question 17

In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are

(a) Isosceles but not congruent

(b) Isosceles but congruent

(c) Congruent but not isosceles

(d) Neither congruent nor isosceles

Solution 17

Question 18

Which is true ?

(a) A triangle can have two right angles.

(b) A triangle can have two obtuse angles.

(c) A triangle can have two acute angles.

(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Solution 18

Question 19

Fill in the blanks with

(a) (Sum of any two sides of a triangle)……(the third side)

(b) (Difference of any two sides of a triangle)…..(the third side)

(c) (Sum of three altitudes of a triangle) ……. (sum of its three sides

(d) (Sum of any two sides of a triangle)….. (twice the median to the 3^rd side)

(e) (Perimeter of a triangle)……(sum of its medians)Solution 19

Question 20

Fill in the blanks

(a) Each angle of an equilateral triangles measures …….

(b) Medians of an equilateral triangle are ……….

(c) In a right triangle the hypotenuse is the ….. side

(d) Drawing a ∆ABC with AB = 3cm, BC= 4 cm and CA = 7 cm is ……..Solution 20

Exercise Ex. 9B

Question 1(i)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

5 cm, 4 cm, 9 cmSolution 1(i)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 5 cm and 4 cm, is not greater than the third side, 9 cm. Question 1(ii)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

8 cm, 7 cm, 4 cmSolution 1(ii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(iii)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

10 cm, 5 cm, 6 cmSolution 1(iii)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(iv)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

2.5 cm, 5 cm, 7 cmSolution 1(iv)

Yes, it is not possible to construct a triangle with lengths of its sides given because the sum of any two sides is greater than the third side. Question 1(v)

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

3 cm, 4 cm, 8 cmSolution 1(v)

No, it is not possible to construct a triangle with lengths of its sides given because the sum of two sides, 3 cm and 4 cm, is not greater than the third side, 8 cm. Question 2

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.Solution 2

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 50° + 60° + ∠C = 180° 

⇒ ∠C = 70° 

Thus, we have

∠A < ∠B < ∠C

⇒ BC < AC < AB

Hence, the longest side is AB and the shortest side is BC. Question 3(iii)

In ΔABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?Solution 3(iii)

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 100° + ∠B + 50° = 180° 

⇒ ∠B = 30° 

Thus, we have

∠B < ∠C < ∠A

⇒ AC < AB < BC

Hence, the shortest side is AC. Question 3(i)

In ABC, if A = 90^o, which is the longest side?Solution 3(i)

Question 3(ii)

In ABC, if A = B = 45^o, name the longest side.Solution 3(ii)

Question 4

In ABC, side AB is produced to D such that BD = BC. If B = 60^o and A = 70^o, prove that (i) AD > CD and (ii) AD > AC.

Solution 4

Question 5

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution 5

In ΔAOB,

∠B < ∠A

⇒ AO < BO ….(i)

In ΔCOD,

∠C < ∠D

⇒ DO < CO ….(ii)

Adding (i) and (ii),

AO + DO < BO + CO

⇒ AD < BC Question 6

AB and CD are respectively the smallest and largest sides of quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Solution 6

Construction: Join AC and BD.

In ΔABC,

BC > AB

⇒ ∠BAC > ∠ACB ….(i)

In ΔACD,

CD > AD

⇒ ∠CAD > ∠ACD ….(ii)

Adding (i) and (ii), we get

∠BAC + ∠CAD > ∠ACB + ∠ACD

⇒ ∠A > ∠C 

In ΔADB,

AD > AB

⇒ ∠ABD > ∠ADB ….(iii)

In ΔBDC,

CD > BC

⇒ ∠CBD > ∠BDC ….(iv)

Adding (iii) and (iv), we get

∠ABD + ∠CBD > ∠ADB + ∠BDC

⇒ ∠B > ∠D Question 7

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) > (AC + BD).Solution 7

In ΔABC,

AB + BC > AC  ….(i)

In ΔACD,

DA + CD > AC  ….(ii)

In ΔADB,

DA + AB > BD  ….(iii)

In ΔBDC,

BC + CD > BD  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AB + BC + DA + CD + DA + AB + BC + CD > AC + AC + BD + BD

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)

⇒ AB + BC + CD + DA > AC + BD Question 8

In a quadrilateral ABCD, show that

(AB + BC + CD + DA) < 2(BD + AC).Solution 8

In ΔAOB,

AO + BO > AB  ….(i)

In ΔBOC,

BO + CO > BC  ….(ii)

In ΔCOD,

CO + DO > CD  ….(iii)

In ΔAOD,

DO + AO > DA  ….(iv)

Adding (i), (ii), (iii) and (iv), we get

AO + BO + BO + CO + CO + DO + DO + AO > AB + BC + CD + DA

⇒ 2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

⇒ 2AC + 2BD > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA

⇒ AB + BC + CD + DA < 2(AC + BD) Question 9

In ABC, B = 35^o, C = 65^o and the bisector of BAC meets BC in X. Arrange AX, BX and CX in descending order.

Solution 9

Question 10

In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

Solution 10

In ΔPQR,

PQ > PR

⇒ ∠PRQ > ∠PQR

⇒ ∠SRQ > ∠SQR

⇒ SQ > SRQuestion 11

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.Solution 11

In ΔABC,

AB = AC

⇒ ∠ABC = ∠ACB ….(i)

Now, ∠ABC = ∠ABD + ∠DBC

⇒ ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC [From (i)]

⇒ ∠DCB > ∠DBC

⇒ BD > CD

i.e. CD < BD Question 12

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than  of a right angle.Solution 12

Let PQR be the required triangle.

Let PR be the longest side.

Then, PR > PQ

⇒ ∠Q > ∠R ….(i)

Also, PR > QR

⇒ ∠Q > ∠P ….(ii)

Adding (i) and (ii), we get

2∠Q > ∠R + ∠P

⇒ 2∠Q + ∠Q > ∠P + ∠Q + ∠R (adding ∠Q to both sides)

⇒ 3∠Q > 180° 

⇒ ∠Q > 60° 

Question 13(i)

In the given figure, prove that CD + DA + AB > BC

Solution 13(i)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AC + AB > BC ….(ii)

Adding (i) and (ii), we get

CD + DA + AC + AB > AC + BC

Subtracting AC from both sides, we get

CD + DA + AB > BC Question 13(ii)

In the given figure, prove that

CD + DA + AB + BC > 2AC.

Solution 13(ii)

In ΔCDA,

CD + DA > AC ….(i)

In ΔABC,

AB + BC > AC ….(ii)

Adding (i) and (ii), we get

CD + DA + AB + BC > AC + AC

⇒ CD + DA + AB + BC > 2AC Question 14(i)

If O is a point within ABC, show that:

AB + AC > OB + OCSolution 14(i)

Given : ABC is a triangle and O is appoint insideit.

To Prove : (i) AB+AC > OB +OCQuestion 14(ii)

If O is a point within ABC, show that:

AB + BC + CA > OA + OB + OCSolution 14(ii)

AB+BC+CA > OA+OB+OCQuestion 14(iii)

If O is a point within ABC, show that:

OA + OB + OC > (AB + BC + CA)Solution 14(iii)

OA+OB+OC> (AB+BC+CA)

Proof:

(i)InABC,

AB+AC>BC.(i)

And in , OBC,

OB+OC>BC.(ii)

Subtracting (i) from (i) we get

(AB+AC)-(OB+OC)> (BC-BC)

i.e.AB+AC>OB+OC

(ii)AB+AC> OB+OC[proved in (i)]

Similarly,AB+BC > OA+OC

AndAC+BC> OA +OB

Addingboth sides of these three inequalities, we get

(AB+AC) +(AC+BC) +(AB+BC)>OB+OC+OA+OB+OA+OC

i.e.2(AB+BC+AC)> 2(OA+OB+OC)

Therefore, we have

AB+BC+AC > OA+OB+OC

(iii)InOAB

OA+OB > AB(i)

InOBC,

OB+OC > BC(ii)

And, in OCA,

OC+OA>CA

Adding (i), (ii) and (iii)we get

(OA+OB) + (OB+OC)+(OC+OA)> AB+BC+CA

i.e2(OA+OB+OC) > AB+BC+CA

OA+OB+OC> (AB+BC+CA)Question 15

In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.

Solution 15

Construction: Mark a point S on BC such that BD = SD. Join AS.

In ΔADB and ΔADS,

BD = SD (by construction)

∠ADB = ∠ADS   (Each equal to 90°)

AD = AD (common)

∴ ΔADB ≅ ΔADS (by SAS congruence criterion)

⇒ AB = AS (c.p.c.t.)

Now, in ΔABS,

AB = AS

⇒ ∠ASB = ∠ABS ….(i)(angles opposite to equal sides are equal)

In ΔACS,

∠ASB > ∠ACS ….(ii)

From (i) and (ii), we have

∠ABS > ∠ACS

⇒ ∠ABC > ∠ACB

⇒ AC > ABQuestion 16

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.

Solution 16

In ΔABC,

AB + AC > BC

⇒ AB + AC >BD + DC

⇒ AB + AC >BD + DE ….(i) [since CD = DE]

In ΔBED,

BD + DE > BE ….(ii)

From (i) and (ii), we have

AB + AC > BE

Exercise Ex. 9A

Question 1

In the given figure, AB ∥ CD and O is the midpoint of AD.

Show that (i) Δ AOB ≅ Δ DOC (ii) O is the midpoint of BC.

Solution 1

(i) In ΔAOB and ΔDOC,

∠BAO = ∠CDO (AB ∥ CD, alternate angles)

AO = DO (O is the mid-point of AD)

∠AOB = ∠DOC (vertically opposite angles)

∴ ΔAOB ≅ ΔDOC (by ASA congruence criterion)

(ii) Since ΔAOB ≅ ΔDOC,

BO = CO (c.p.c.t.)

⇒ O is the mid-point of BC.Question 2

In the given figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisect AB.

Solution 2

In ΔAOD and ΔBOC,

∠AOD = ∠BOC (vertically opposite angles)

∠DAO = ∠CBO (Each 90°)

AD = BC (given)

∴ ΔAOD ≅ BOC (by AAS congruence criterion)

⇒ AO = BO (c.p.c.t.)

⇒ CD bisects AB. Question 3

In the given figure, two parallels lines l and m are intersected by two parallels lines p and q. Show that Δ ABC ≅ Δ CDA.

Solution 3

In ΔABC and ΔCDA

∠BAC = ∠DCA  (alternate interior angles for p ∥ q)

AC = CA  (common)

∠BCA = ∠DAC (alternate interior angles for l ∥ m)

∴ ΔABC ≅ ΔCDA (by ASA congruence rule)Question 4

AD is an altitude of an isosceles ΔABC in which AB = AC.

Show that (i) AD bisects BC, (ii) AD bisects ∠A.

Solution 4

(i) In ΔBAD and ΔCAD

∠ADB = ∠ADC  (Each 90° as AD is an altitude)

AB = AC (given)

AD = AD (common)

∴ ΔBAD ≅ ΔCAD (by RHS Congruence criterion)

⇒ BD = CD (c.p.c.t.)

Hence AD bisects BC.

(ii) Also, ∠BAD = ∠CAD (c.p.c.t.)

 Hence, AD bisects ∠A.Question 5

In the given figure, BE and CF are two equal altitudes of ΔABC.

Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

Solution 5

(i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90°)

BE = CF (given)

∠BAE = ∠CAF (common ∠A)

∴ ΔABE ≅ ACF (by ASA congruence criterion)

(ii) Since ΔABE ≅ ΔACF,

AB = AC (c.p.c.t.)Question 6

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABE ≅ ΔACE

(iii) AE bisects ∠A as well as ∠D

(iv) AE is the perpendicular bisector of BC.

Solution 6

(i) In ΔABD and ΔACD,

AB = AC  (equal sides of isosceles ΔABC)

DB = DC  (equal sides of isosceles ΔDBC)

AD = AD (common)

∴ ΔABD ≅ ACD (by SSS congruence criterion)

(ii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE ….(1)

Now, in ΔABE and ΔACE

AB = AC (equal sides of isosceles ΔABC)

∠BAE = ∠CAE [From (1)]

AE = AE (common) 

∴ ΔABE ≅ ACE (by SAS congruence criterion)

(iii) Since ΔABD ≅ ΔACD,

∠BAD = ∠CAD (c.p.c.t.)

⇒ ∠BAE = ∠CAE 

Thus, AE bisects ∠A.

In ΔBDE and ΔCDE,

BD = CD (equal sides of isosceles ΔABC)

BE = CE (c.p.c.t. since ΔABE ≅ ACE) 

DE = DE (common)

∴ ΔBDE ≅ CDE (by SSS congruence criterion)

⇒ ∠BDE = ∠CDE (c.p.c.t.)

Thus, DE bisects ∠D, i.e., AE bisects ∠D.

Hence, AE bisects ∠A as well as ∠D.

(iv) Since ΔBDE ≅ ΔCDE,

BE = CE and ∠BED = ∠CED (c.p.c.t.)

⇒ BE = CE and ∠BED = ∠CED = 90° (since ∠BED and ∠CED form a linear pair) 

⇒ DE is the perpendicular bisector of BC.

⇒ AE is the perpendicular bisector of BC.Question 7

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Solution 7

Question 8

In the given figure, line l is the bisector of an angle ∠A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠A, show that

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Solution 8

(i) In ΔAPB and ΔAQB,

∠APB = ∠AQC (Each 90°)

∠BAP = ∠BAQ (line l is the bisector of ∠A)

AB = AB (common)

∴ ΔAPB ≅ AQB (by AAS congruence criterion)

(ii) Since ΔAPB ≅ ΔAQB,

BP = BQ (c.p.c.t.) Question 9

ABCD is a quadrilateral such that diagonal AC bisect the angles ∠A and ∠C. Prove that AB = AD and CB = CD.Solution 9

In ΔABC and ΔADC,

∠BAC = ∠DAC (AC bisects ∠A)

AC = AC (common)

∠BCA = ∠DCA (AC bisects ∠C)

∴ ΔABC ≅ ADC (by ASA congruence criterion)

⇒ AB = AD and CB = CD (c.p.c.t.) Question 10

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersect the side AB at D. Prove that AC + AD = BC.Solution 10

Construction: Draw DE ⊥ BC.

In ΔDAC and ΔDEC,

∠DAC = ∠DEC (Each 90°)

∠DCA = ∠DCE (CD bisects ∠C)

CD = CD (common)

∴ ΔDAC ≅ ΔDEC (by AAS congruence criterion)

⇒ DA = DE (c.p.c.t.) ….(i)

and AC = EC (c.p.c.t.)  ….(ii)

Given, AB = AC

⇒ ∠B = ∠C (angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠A + ∠B + ∠C = 180° 

⇒ 90° + ∠B + ∠B = 180° 

⇒ 2∠B = 90° 

⇒ ∠B = 45° 

In ΔBED,

∠BDE + ∠B = 90° (since ∠BED = 90°)

⇒ ∠BDE + 45° = 90° 

⇒ ∠BDE = 45° 

⇒ ∠BDE = ∠DBE = 45° 

⇒ DE = BE ….(iii)

From (i) and (iii),

DA = DE = BE ….(iv) 

Now, BC = BE + EC

⇒ BC = DA + AC [From (ii) and (iv)

⇒ AC + AD = BCQuestion 11

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX (ii) AX = BX.

Solution 11

Question 12

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

Solution 12

Question 13

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

Solution 13

Question 14

In the given figure, ABCD is square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

Solution 14

Question 15

In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Solution 15

ΔOAB is an equilateral triangle.

⇒ ∠OAB = ∠OBA = AOB = 60° 

ABCD is a square.

⇒ ∠A = ∠B = ∠C = ∠D = 90° 

Now, ∠A = ∠DAO + ∠OAB

⇒ 90° = ∠DAO + 60° 

⇒ ∠DAO = 90° – 60° = 30° 

Similarly, ∠CBO = 30° 

In ΔOAD and ΔOBC,

AD = BC  (sides of a square ABCD)

∠DAO = ∠CBO = 30° 

OA = OB (sides of an equilateral ΔOAB)

∴ ΔOAD ≅ ΔOBC (by SAS congruence criterion)

⇒ OD = OC (c.p.c.t.)

Hence, ΔOCD is an isosceles triangle.Question 16

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ABC such that AX = AY. Prove that CX = BY.

Solution 16

Question 17

In ABC, D is the midpoint of BC. If DL AB and DM AC such that DL = DM, prove that AB = AC.

Solution 17

Question 18

In ABC, AB = AC and the bisectors of _B and C meet at a point O. Prove that BO = CO and the ray AO is the bisector A.

Solution 18

Question 19

The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.Solution 19

Construction: Join AN and BN.

In ΔANM and ΔBNM

AM = BM (M is the mid-point of AB)

∠AMN = ∠BMN (Each 90°)

MN = MN (common)

∴ ΔANM ≅ ΔBNM (by SAS congruence criterion)

⇒ AN = BN (c.p.c.t.) ….(i)

And, ∠ANM = ∠BNM (c.p.c.t.)

⇒ 90° – ∠ANM = 90° – ∠BNM

⇒ ∠AND = ∠BNC ….(ii) 

In ΔAND and DBNC,

AN = BN [From (i)]

∠AND = ∠BNC [From (ii)]

DN = CN (N is the mid-point of DC)

∴ ΔAND ≅ ΔBNC (by SAS congruence criterion)

⇒ AD = BC (c.p.c.t.)Question 20

The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.Solution 20

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

Now, by exterior angle property,

∠MOC = ∠OBC + ∠OCB

⇒ ∠MOC = 2∠OBC [From (i)]

⇒ ∠MOC = ∠ABC (OB is the bisector of ∠ABC)  Question 21

The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ABC is equal to ∠BOC.Solution 21

In ΔABC, AB = AC

⇒ ∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB ….(i)

In ΔBOC, by angle sum property,

∠BOC + ∠OBC + ∠OCB = 180° 

⇒ ∠BOC + 2∠OBC = 180° [From (i)]

⇒ ∠BOC + ∠ABC = 180° 

⇒ ∠BOC + (180° – ∠ABP) = 180° (∠ABC and ∠ABP form a linear pair)

⇒ ∠BOC + 180° – ∠ABP = 180° 

⇒ ∠BOC – ∠ABP = 0

⇒ ∠BOC = ∠ABP Question 22

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.Solution 22

AB ∥ PQ and BP is a transversal.

⇒ ∠ABP = ∠BPQ (alternate angles) ….(i)

BP is the bisector of ∠ABC.

⇒ ∠ABP = ∠PBC

⇒ ∠ABP = ∠PBQ ….(ii)

From (i) and (ii), we have

∠BPQ = ∠PBQ

⇒ PQ = BQ (sides opposite to equal angles are equal)

⇒ ΔBPQ is an isosceles triangle.Question 23

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Solution 23

To prove that the image is as far behind the mirror as the object is in front of the mirror, we need to prove that AT = BT. 

We know that angle of incidence = angle of reflection.

⇒ ∠ACN = ∠DCN ….(i)

AB ∥ CN and AC is the transversal.

⇒ ∠TAC = ∠ACN (alternate angles) ….(ii)

Also, AB ∥ CN and BD is the transversal.

⇒ ∠TBC = ∠DCN (corresponding angles) ….(iii)

From (i), (ii) and (iii),

∠TAC = ∠TBC ….(iv)

In ΔACT and ΔBCT,

∠TAC = ∠TBC [From (iv)]

∠ATC = ∠BTC (Each 90°)

CT = CT (common)

∴ ΔACT ≅ ΔBCT (by AAS congruence criterion)

⇒ AT = BT (c.p.c.t.)Question 24

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Solution 24

_{Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM.}

Question 25

In a ΔABC, D is the midpoint of side AC such that BD = . Show that ∠ABC is a right angle.Solution 25

D is the mid-point of AC.

⇒ AD = CD =  

Given, BD =  

⇒ AD = CD = BD 

Consider AD = BD

⇒ ∠BAD = ∠ABD (i)(angles opposite to equal sides are equal)

Consider CD = BD

⇒ ∠BCD = ∠CBD (ii)(angles opposite to equal sides are equal)

In ΔABC, by angle sum property,

∠ABC + ∠BAC + ∠BCA = 180° 

⇒ ∠ABC + ∠BAD + ∠BCD = 180° 

⇒ ∠ABC + ∠ABD + ∠CBD = 180° [From (i) and (ii)]

⇒ ∠ABC + ∠ABC = 180° 

⇒ 2∠ABC = 180° 

⇒ ∠ABC = 90° 

Hence, ∠ABC is a right angle.Question 26

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.” Is the statement true? Why?Solution 26

The given statement is not true.

Two triangles are congruent if two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle. Question 27

“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.” Is the statement true? Why?Solution 27

The given statement is not true.

Two triangles are congruent if two angles and the included side of one triangle are equal to corresponding two angles and the included angle of another triangle. 

Exercise MCQ

Question 1

If (x + 1) is a factor of the polynomial (2x² + kx) then the value of k is

(a) -2

(b) -3

(c) 2

(d) 3Solution 1

Correct option: (c)

Let p(x) = 2x² + kx

Since (x + 1) is a factor of p(x),

P(-1) = 0

⇒ 2(-1)² + k(-1) = 0

⇒ 2 – k = 0

⇒ k = 2Question 2

The value of (249)² – (248)² is

(a) 1²

(b) 477

(c) 487

(d) 497Solution 2

Correct option: (d)

(249)² – (248)²

= (249 + 248)(249 – 248)

= 497 × 1

= 497Question 3

Solution 3

Question 4

If a + b + c = 0, then a³ + b³ + c³= ?

(a) 0

(b) abc 

(c) 2abc

(d)3abcSolution 4

Question 5

(a) 0

(b) 

(c) 

(d) Solution 5

Correct option: (c)

Question 6

The coefficient of x in the expansion of (x + 3)³ is

(a) 1

(b) 9

(c) 18

(d)27Solution 6

Question 7

Which of the following is a factor of (x + y)³ – (x³ + y³)?

(a) x² + y² + 2xy

(b) x² + y² – xy

(c) xy²

(d) 3xySolution 7

Correct option: (d)

(x + y)³ – (x³ + y³)

= (x + y)³ – [(x + y)(x² – xy + y²)]

= (x + y)[(x + y)² – (x² – xy + y²)]

= (x + y)[x² + y² + 2xy – x² + xy – y²]

= (x + y)(3xy) Question 8

One of the factors of (25x² – 1) + (1 + 5x)² is

(a) 5 +x

(b) 5 – x

(c) 5x – 1

(d) 10xSolution 8

Correct option: (d)

(25x² – 1) + (1 + 5x)²

= [(5x)² – (1)²] + (1 + 5x)²

= (5x – 1)(5x + 1) + (1 + 5x)²

= (1 + 5x)[(5x – 1) + (1 + 5x)]

= (1 + 5x)(5x – 1 + 1 + 5x)

= (1 + 5x)(10x)Question 9

If (x + 5) is a factor of p(x) = x³ – 20x + 5k then k = ?

(a) -5

(b) 5

(c) 3

(d) -3Solution 9

Correct option: (b)

Since (x + 5) is a factor of p(x) = x³ – 20x + 5k,

p(-5) = 0

⇒ (-5)³ – 20(-5) + 5k = 0

⇒ -125 + 100 + 5k = 0

⇒ -25 + 5k = 0

⇒ 5k = 25

⇒ k = 5 Question 10

If (x + 2) and (x – 1) are factors of (x³ + 10x² + mx + n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19Solution 10

Question 11

104 ⨯ 96 =?

(a) 9894

(b) 9984

(c) 9684

(d)9884Solution 11

Question 12

305 ⨯ 308 = ?

(a) 94940

(b) 93840

(c) 93940

(d)94840Solution 12

Question 13

207 ⨯ 193 = ?

(a) 39851

(b) 39951

(c) 39961

(d)38951Solution 13

Question 14

4a² + b² + 4ab + 8a + 4b + 4 =?

(a) (2a + b + 2)²

(b) (2a – b + 2)²

(c) (a + 2b + 2)²

(d)None of theseSolution 14

Question 15

(x² – 4x – 21) =?

(a) (x – 7)(x – 3)

(b) (x + 7)(x – 3)

(c) (x – 7)(x + 3)

(d)none of theseSolution 15

Question 16

(4x² + 4x -3) = ?

(a) (2x – 1)(2x – 3)

(b) (2x + 1)(2x – 3)

(c) (2x + 3)(2x – 1)

(d)none of theseSolution 16

Question 17

6x² + 17x + 5 =?

(a) (2x +3)(3x + 5)

(b) (2x + 5)(3x + 1)

(c) (6x + 5)(x + 1)

(d)none of theseSolution 17

Question 18

(x + 1) is a factor of the polynomial

(a) x³ – 2x² + x + 2

(b) x³ + 2x² + x – 2

(c) x³ + 2x² – x – 2

(d)x³ + 2x² – x + 2Solution 18

Question 19

3x³ + 2x² + 3x + 2 =?

(a) (3x – 2)(x² – 1)

(b) (3x – 2)(x² + 1)

(c) (3x + 2)(x² – 1)

(d)(3x + 2)(x² + 1)Solution 19

Question 20

(a) 1

(b) 0

(c) -1

(d)3Solution 20

Question 21

If x + y + z =9 and xy + yz + zx = 23, then the value of (x³ + y³ + z³ – 3xyz) = ?

(a) 108

(b) 207

(c) 669

(d)729Solution 21

Question 22

then (a³ – b³) = ?

(a) -3

(b) -2

(c) -1

(d) 0Solution 22

Correct option: (d)

Exercise Ex. 3C

Question 1

Factorise:

x² + 11x + 30Solution 1

x² + 11x + 30

= x² + 6x + 5x + 30

= x (x + 6) + 5 (x + 6)

= (x + 6) (x + 5).Question 2

Factorise:

x² + 18x + 32Solution 2

x² + 18x + 32

= x² + 16x + 2x + 32

= x (x + 16) + 2 (x + 16)

= (x + 16) (x + 2).Question 3

Factorise:

x² + 20x – 69Solution 3

x² + 20x – 69

= x² + 23x – 3x – 69

= x(x + 23) – 3(x + 23)

= (x + 23)(x – 3) Question 4

Factorise:

x² + 19x – 150Solution 4

x² + 19x – 150

= x² + 25x – 6x – 150

= x(x + 25) – 6(x + 25)

= (x + 25)(x – 6) Question 5

Factorise:

x² + 7x – 98Solution 5

x² + 7x – 98

= x² + 14x – 7x – 98

= x(x + 14) – 7(x + 14)

= (x + 14)(x – 7) Question 6

Factorise:

Solution 6

Question 7

Factorise:

x² – 21x + 90Solution 7

x² – 21x + 90

= x² – 6x – 15x + 90

= x(x – 6) – 15(x – 6)

= (x – 6)(x – 15) Question 8

Factorise:

x² – 22x + 120Solution 8

x² – 22x + 120

= x² – 10x – 12x + 120

= x(x – 10) – 12(x – 10)

= (x – 10)(x – 12) Question 9

Factorise:

x² – 4x + 3Solution 9

x² – 4x + 3

= x² – 3x – x + 3

= x(x – 3) – 1(x – 3)

= (x – 3)(x – 1) Question 10

Factorise:

Solution 10

Question 11

Factorise:

Solution 11

Question 12

Factorise:

Solution 12

Question 13

Factorise:

Solution 13

Question 14

Factorise:

x² – 24x – 180Solution 14

x² – 24x – 180

= x² – 30x + 6x – 180

= x(x – 30) + 6(x – 30)

= (x – 30)(x + 6) Question 15

Factorise:

z² – 32z – 105Solution 15

z² – 32z – 105

= z² – 35z + 3z – 105

= z (z – 35) + 3 (z – 35)

= (z – 35) (z + 3)Question 16

Factorise:

x² – 11x – 80Solution 16

x² – 11x – 80

= x² – 16x + 5x – 80

= x (x – 16) + 5 (x – 16)

= (x – 16) (x + 5).Question 17

Factorise:

6 – x – x²Solution 17

6 – x – x²

= 6 + 2x – 3x – x²

= 2(3 + x) – x (3 + x)

= (3 + x) (2 – x).Question 18

Factorise:

Solution 18

Question 19

Factorise:

40 + 3x – x²Solution 19

40 + 3x – x²

= 40 + 8x – 5x – x²

= 8 (5 + x) -x (5 + x)

= (5 + x) (8 – x).Question 20

Factorise:

x² – 26x + 133Solution 20

x² – 26x + 133

= x² – 19x – 7x + 133

= x(x – 19) – 7(x – 19)

= (x – 19)(x – 7) Question 21

Factorise:

Solution 21

Question 22

Factorise:

Solution 22

Question 23

Factorise:

Solution 23

Question 24

Factorise:

Solution 24

Question 25

Factorise:

x² – x – 156Solution 25

x² – x – 156

= x² – 13x + 12x – 156

= x (x – 13) + 12 (x – 13)

= (x – 13) (x + 12).Question 27

Factorise:

9x² + 18x + 8Solution 27

9x² + 18x + 8

= 9x² + 12x + 6x + 8

= 3x (3x+ 4) +2 (3x + 4)

= (3x + 4) (3x + 2).Question 28

Factorise:

6x² + 17x + 12Solution 28

6x² + 17x + 12

= 6x² + 9x + 8x + 12

= 3x (2x + 3) + 4(2x + 3)

= (2x + 3) (3x + 4).Question 29

Factorise:

18x² + 3x – 10Solution 29

18x² + 3x – 10

= 18x² – 12x + 15x – 10

= 6x (3x – 2) + 5 (3x – 2)

= (6x + 5) (3x – 2).Question 30

Factorise:

2x² + 11x – 21Solution 30

2x² + 11x – 21

= 2x² + 14x – 3x – 21

= 2x (x + 7) – 3 (x + 7)

= (x + 7) (2x – 3).Question 31

Factorise:

15x² + 2x – 8Solution 31

15x² + 2x – 8

= 15x² – 10x + 12x – 8

= 5x (3x – 2) + 4 (3x – 2)

= (3x – 2) (5x + 4).Question 32

Factorise:

21x² + 5x – 6Solution 32

21x² + 5x – 6

= 21x² + 14x – 9x – 6

= 7x(3x + 2) – 3(3x + 2)

= (3x + 2)(7x – 3) Question 33

Factorise:

24x² – 41x + 12Solution 33

24x² – 41x + 12

= 24x² – 32x – 9x + 12

= 8x (3x – 4) – 3 (3x – 4)

= (3x – 4) (8x – 3).Question 34

Factorise:

3x² – 14x + 8Solution 34

3x² – 14x + 8

= 3x² – 12x – 2x +8

= 3x (x – 4) – 2(x – 4)

= (x – 4) (3x – 2).Question 35

Factorise:

2x² + 3x – 90Solution 35

2x² + 3x – 90

= 2x² – 12x + 15x – 90

= 2x (x – 6) + 15 (x – 6)

= (x – 6) (2x + 15).Question 37

Factorise:

Solution 37

Question 38

Factorise:

Solution 38

Question 39

Factorise:

Solution 39

Question 40

Factorise:

Solution 40

Question 41

Factorise:

Solution 41

Question 42

Factorise:

Solution 42

Question 43

Factorise:

Solution 43

Question 44

Factorise:

15x² – x – 28Solution 44

15x² – x – 28

= 15x² + 20x – 21x – 28

= 5x (3x + 4) – 7 (3x + 4)

= (3x + 4) (5x – 7).Question 45

Factorise:

6x² – 5x – 21Solution 45

6x² – 5x – 21

= 6x² + 9x – 14x – 21

= 3x (2x + 3) – 7 (2x + 3)

= (3x – 7) (2x + 3).Question 46

Factorise:

2x² – 7x – 15Solution 46

2x² – 7x – 15

= 2x² – 10x + 3x – 15

= 2x (x – 5) + 3 (x – 5)

= (x – 5) (2x + 3).Question 47

Factorise:

5x² – 16x – 21Solution 47

5x² – 16x – 21

= 5x² + 5x – 21x – 21

= 5x (x + 1) -21 (x + 1)

= (x + 1) (5x – 21).Question 48

Factorise:

6x² – 11x – 35Solution 48

6x² – 11x – 35

= 6x² – 21x + 10x – 35

= 3x(2x – 7) + 5(2x – 7)

= (2x – 7)(3x + 5) Question 49

Factorise:

9x² – 3x – 20Solution 49

9x² – 3x – 20

= 9x² – 15x + 12x – 20

= 3x(3x – 5) + 4(3x – 5)

= (3x – 5)(3x + 4) Question 50

Factorise:

10x² – 9x – 7Solution 50

10x² – 9x – 7

= 10x² + 5x – 14x – 7

= 5x (2x + 1) – 7 (2x+ 1)

= (2x + 1) (5x – 7).Question 51

Factorise:

x² – 2x + Solution 51

x² – 2x + 

Question 52

Factorise:

Solution 52

Question 53

Factorise:

Solution 53

Question 54

Factorise:

Solution 54

Question 55

Factorise:

Solution 55

Question 56

Factorise:

Solution 56

Question 57

Factorise:

Solution 57

Question 58

Factorise:

Solution 58

Question 59

Factorise:

Solution 59

Letx + y = z

Then, 2 (x+ y)² – 9 (x + y) – 5

Now, replacing z by (x + y), we get

Question 60

Factorise:

9 (2a – b)² – 4 (2a – b) – 13Solution 60

Let2a – b = c

Then,9 (2a – b)² – 4 (2a – b) -13

Now, replacing c by (2a – b) , we get

9 (2a – b)² – 4 (2a – b) – 13

Question 61

Factorise:

7 (x – 2y)² – 25 (x – 2y) + 12Solution 61

Let x – 2y = z

Then, 7 (x – 2y)² – 25 (x – 2y) + 12

Now replace z by (x – 2y), we get

7 (x – 2y)² – 25 (x – 2y) + 12

Question 62

Factorise:

Solution 62

Question 63

Factorise:

Solution 63

Question 64

Factorise:

(a + 2b)² + 101(a + 2b) + 100Solution 64

Given equation: (a + 2b)² + 101(a + 2b) + 100

Let (a + 2b) = x

Then, we have

x² + 101x + 100

= x² + 100x + x + 100

= x(x + 100) + 1(x + 100)

= (x + 100)(x + 1)

= (a + 2b + 100)(a + 2b + 1) Question 65

Factorise:

Solution 65

Let x² = y

Then, 4x⁴ + 7x² – 2

Now replacing y by x², we get

Question 66

Evaluate {(999)² – 1}.Solution 66

{(999)² – 1}

= {(999)² – (1)²}

= {(999 – 1)(999 + 1)}

= 998 × 1000

= 998000 

Exercise Ex. 3D

Question 1(i)

Expand:

(a + 2b + 5c)²Solution 1(i)

We know:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(i) (a + 2b + 5c)²

= (a)² + (2b)² + (5c)² + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)

= a² + 4b² + 25c² + 4ab + 20bc + 10acQuestion 1(ii)

Expand:

(2a – b + c)²Solution 1(ii)

We know:

(2a – b + c)²

= (2a)² + (-b)² + (c)² + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)

= 4a² + b² + c² – 4ab – 2bc + 4ac.Question 1(iii)

Expand:

(a – 2b – 3c)²Solution 1(iii)

We know:

(a – 2b – 3c)²

= (a)² + (-2b)² + (-3c)²+ 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)

= a² + 4b² + 9c² – 4ab + 12bc – 6ac.Question 2(i)

Expand:

(2a – 5b – 7c)²Solution 2(i)

We know:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(2a – 5b – 7c)²

= (2a)² + (-5b)² + (-7c)² + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)

= 4a² + 25b² + 49c² – 20ab + 70bc – 28ac.Question 2(ii)

Expand:

(ii) (-3a + 4b – 5c)²Solution 2(ii)

(-3a + 4b – 5c)²

= (-3a)² + (4b)² + (-5c)² + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)

= 9a² + 16b² + 25c² – 24ab – 40bc + 30ac.Question 2(iii)

Expand:

Solution 2(iii)

= Question 3

Factorise: 4x² + 9y² + 16z² + 12xy – 24yz – 16xz.Solution 3

4x² + 9y² + 16z² + 12xy – 24yz – 16xz

= (2x)² + (3y)² + (-4z)² + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)

= (2x + 3y – 4z)²Question 4

Factorise: 9x² + 16y² + 4z² – 24xy + 16yz – 12xzSolution 4

9x² + 16y² + 4z² – 24xy + 16yz – 12xz

= (-3x)² + (4y)² + (2z)² + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)

= (-3x + 4y + 2z)².Question 5

Factorise: 25x² + 4y² + 9z² – 20xy – 12yz + 30xz.Solution 5

25x² + 4y² + 9z² – 20xy – 12yz + 30xz

= (5x)² + (-2y)² + (3z)² + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)

= (5x – 2y + 3z)²Question 6

16x² + 4y² + 9z² – 16xy – 12yz + 24xzSolution 6

16x² + 4y² + 9z² – 16xy – 12yz + 24xz

= (4x)² + (-2y)² + (3z)² + 2(4x)(-2y) + 2(-2y)(3z) + 2(4x)(3z)

= [4x + (-2y) + 3z]²

= (4x – 2y + 3z)²

= (4x – 2y + 3z)(4x – 2y + 3z) Question 7(i)

Evaluate

(99)²Solution 7(i)

(99)²

= (100 – 1)²

= (100)² – 2(100) (1) + (1)²

= 10000 – 200 + 1

= 9801.Question 7(ii)

Evaluate

(995)²Solution 7(ii)

(995)²

= (1000 – 5)²

= (1000)² + (5)² – 2(1000)(5)

= 1000000 + 25 – 10000

= 990025 Question 7(iii)

Evaluate

(107)²Solution 7(iii)

(107)²

= (100 + 7)²

= (100)² + (7)² + 2(100)(7)

= 10000 + 49 + 1400

= 11449 

Exercise Ex. 3E

Question 1(i)

Expand:

(3x + 2)³Solution 1(i)

(3x + 2)³

= (3x)³ + (2)³ + 3 3x 2 (3x + 2)

= 27x³ + 8 + 18x (3x + 2)

= 27x³ + 8 + 54x² + 36x.Question 1(ii)

Expand:

Solution 1(ii)

Question 1(iii)

Expand

Solution 1(iii)

Question 2(i)

Expand

(5a – 3b)³Solution 2(i)

(5a – 3b)³

= (5a)³ – (3b)³ – 3(5a)(3b)(5a – 3b)

= 125a³ – 27b³ – 45ab(5a – 3b)

= 125a³ – 27b³ – 225a²b + 135ab² Question 2(ii)

Expand

Solution 2(ii)

Question 2(iii)

Expand

Solution 2(iii)

Question 3

Factorise 

8a³ + 27b³ + 36a²b + 54ab²Solution 3

8a³ + 27b³ + 36a²b + 54ab²

= (2a)³ + (3b)³ + 3 × (2a)² × (3b) + 3 × (2a) × (3b)²

= (2a + 3b)³

= (2a + 3b)(2a + 3b)(2a + 3b) Question 4

Factorise 

64a³ – 27b³ – 144a²b + 108ab²Solution 4

64a³ – 27b³ – 144a²b + 108ab²

= (4a)³ – (3b)³ – 3 × (4a)² × (3b) + 3 × (4a) × (3b)²

= (4a – 3b)³

= (4a – 3b)(4a – 3b)(4a – 3b) Question 5

Factorise 

Solution 5

Question 6

Factorise 

125x³ – 27y³ – 225x²y + 135xy²Solution 6

125x³ – 27y³ – 225x²y + 135xy²

= (5x)³ – (3y)³ – 3 × (5x)² × (3y) + 3 × (5x) × (3y)²

= (5x – 3y)³

= (5x – 3y)(5x – 3y)(5x – 3y)  Question 7

Factorise 

a³x³ – 3a²bx² + 3ab²x – b³Solution 7

a³x³ – 3a²bx² + 3ab²x – b³

= (ax)³ – 3 × (ax)² × b + 3 × (ax) × (b)² – (b)³

= (ax – b)³

= (ax – b)(ax – b)(ax – b)  Question 8

Factorise 

Solution 8

Question 9

Factorise 

a³ – 12a(a – 4) – 64Solution 9

a³ – 12a(a – 4) – 64

= a³ – 3 × a × 4(a – 4) – (4)³

= (a – 4)³

= (a – 4)(a – 4)(a – 4) Question 10(i)

Evaluate

(103)³Solution 10(i)

(103)³

= (100 + 3)³

= (100)³ + (3)³ + 3 × 100 × 3 × (100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727 Question 10(ii)

Evaluate

(99)³Solution 10(ii)

(99)³

= (100 – 1)³

= (100)³ – (1)³ – 3 × 100 × 1 × (100 – 1)

= 1000000 – 1 – 300(99)

= 1000000 – 1 – 29700

= 970299  

Exercise Ex. 3F

Question 1

Factorise:

x³ + 27Solution 1

x³ + 27

= x³ + 3³

= (x + 3) (x² – 3x + 9)

Question 2

Factorise 

27a³ + 64b³Solution 2

27a³ + 64b³

= (3a)³ + (4b)³

= (3a + 4b)[(3a)² – (3a)(4b) + (4b)²]

= (3a + 4b)(9a² – 12ab + 16b²) Question 3

Factorise:

125a³ + Solution 3

125a³ + 

We know that 

Let us rewrite

Question 4

Factorise:

216x³ + Solution 4

216x³ + 

We know that 

Let us rewrite

Question 5

Factorise:

16x⁴ + 54xSolution 5

16x ⁴ + 54x

= 2x (8x ³ + 27)

= 2x [(2x)³ + (3)³]

= 2x (2x + 3) [(2x)² – 2x3 + 3²]    

=2x(2x+3)(4x² -6x +9)
Question 6

Factorise:

7a³ + 56b³Solution 6

7a³ + 56b³

= 7(a³ + 8b³)

= 7 [(a)³ + (2b)³]

= 7 (a + 2b) [a² – a 2b + (2b)²]    

= 7 (a + 2b) (a² – 2ab + 4b²).Question 7

Factorise:

x⁵ + x²Solution 7

x⁵ + x²

= x²(x³ + 1)

= x² (x + 1) [(x)² – x 1 + (1)²]     

= x² (x + 1) (x² – x + 1).Question 8

Factorise:

a³ + 0.008Solution 8

a³ + 0.008

= (a)³ + (0.2)³

= (a + 0.2) [(a)² – a 0.2 + (0.2)²]     

= (a + 0.2) (a² – 0.2a + 0.04).Question 9

Factorise:

1 – 27x³Solution 9

1 – 27x³

= (1)³ – (3x)³

= (1 – 3x) [(1)² + 1 3x + (3x)²]       

= (1 – 3x) (1 + 3x + 9x²).Question 10

Factorise 

64a³ – 343Solution 10

64a³ – 343

= (4a)³ – (7)³

= (4a – 7)[(4a)² + (4a)(7) + (7)²]

= (4a – 7)(16a² + 28a + 49) Question 11

Factorise:

x³ – 512Solution 11

x³ – 512

= (x)³ – (8)³

= (x – 8) [(x)² + x 8 + (8)²]       

= (x – 8) (x² + 8x + 64).Question 12

Factorise:

a³ – 0.064Solution 12

a³ – 0.064

= (a)³ – (0.4)³

= (a- 0.4) [(a)² + a 0.4 + (0.4)²]     

= (a – 0.4) (a² + 0.4 a + 0.16).Question 13

Factorise:

Solution 13

We know that 

Let us rewrite

Question 14

Factorise 

Solution 14

Question 15

Factorise:

x – 8xy³Solution 15

x – 8xy³

= x (1 – 8y³)

= x [(1)³ – (2y)³]

= x (1 – 2y) [(1)² + 1 2y + (2y)²]    

= x (1 – 2y) (1 + 2y + 4y²).Question 16

Factorise:

32x⁴ – 500xSolution 16

32x⁴ – 500x

= 4x (8x³ – 125)

= 4x [(2x)³ – (5)³]

= 4x [(2x – 5) [(2x)² + 2x 5 + (5)²]      

= 4x (2x – 5) (4x² + 10x + 25).Question 17

Factorise:

3a⁷b – 81a⁴b⁴Solution 17

3a⁷b – 81a⁴b⁴

= 3a⁴b (a³ – 27b³)

= 3a⁴b [(a)³ – (3b)³]

= 3a⁴b (a – 3b) [(a)² + a 3b + (3b)²]      

= 3a⁴b (a – 3b) (a² + 3ab + 9b²).Question 18

Factorise 

x⁴y⁴ – xySolution 18

x⁴y⁴ – xy

= xy(x³y³ – 1)

= xy[(xy)³ – (1)³]

= xy{(xy – 1)[(xy)² + (xy)(1) + (1)²]}

= xy(xy – 1)(x²y² + xy + 1) Question 19

Factorise 

8x²y³ – x⁵Solution 19

8x²y³ – x⁵

= x² (8y³ – x³)

= x² [(2y)³ – x³]

= x² [(2y – x)[(2y)² + (2y)(x) + x²]

= x² (2y – x)(4y² + 2xy + x²) Question 20

Factorise 

1029 – 3x³Solution 20

1029 – 3x³

= 3(343 – x³)

= 3[(7)³ – x³]

= 3[(7 – x)(7² + 7x + x²)]

= 3(7 – x)(49 + 7x + x²) Question 21

Factorise:

x⁶ – 729Solution 21

x⁶ – 729

= (x²)³ – (9)³

= (x² – 9) [(x²)² + x² 9 + (9)²]      

= (x² – 9) (x⁴ + 9x² + 81)

= (x + 3) (x – 3) [(x² + 9)² – (3x)²]

= (x + 3) (x – 3) (x² + 3x + 9) (x² – 3x + 9).Question 22

Factorise 

x⁹ – y⁹Solution 22

x⁹ – y⁹

= (x³)³ – (y³)³ 

= [(x³ – y³)][(x³)² + x³y³ + (y³)²]

= [(x – y)(x² + xy + y²)(x⁶ + x³y³ + y⁶)  Question 23

Factorise:

(a + b)³ – (a – b)³Solution 23

We know that, 

Therefore,

(a + b)³ – (a – b)³

= [a + b – (a – b)] [ (a + b)² + (a + b) (a – b) + (a – b)²] 

= (a + b – a + b) [ a² + b² + 2ab + a² – b² + a² + b² – 2ab]

= 2b (3a² + b²).Question 24

Factorise:

8a³ – b³ – 4ax + 2bxSolution 24

8a³ – b³ – 4ax + 2bx

= 8a³ – b³ – 2x (2a – b)

= (2a)³ – (b)³ – 2x (2a – b)

= (2a – b) [(2a)² + 2a b + (b)²] – 2x (2a – b)    

= (2a – b) (4a² + 2ab + b²) – 2x (2a – b)

= (2a – b) (4a² + 2ab + b² – 2x).Question 25

Factorise:

a³ + 3a²b + 3ab² + b³ – 8Solution 25

Question 26

Factorise:

Solution 26

We know that 

Question 27

Factorise:

2a³ + 16b³ – 5a – 10bSolution 27

2a³ + 16b³ – 5a – 10b

= 2 (a³ + 8b³) – 5 (a + 2b)

= 2 [(a)³ + (2b)³] – 5 (a + 2b)

= 2 (a + 2b) [(a)² – a 2b + (2b)² ] – 5 (a + 2b)     

= (a+ 2b) [2(a² – 2ab + 4b²) – 5]Question 28

Factorise 

a⁶ + b⁶Solution 28

a⁶ + b⁶

= (a²)³ + (b²)³ 

= (a² + b²)[(a²)² – (a²b²) + (b²)²]

= (a² + b²)(a⁴ – a²b² + b⁴) Question 29

Factorise 

a¹² – b¹²Solution 29

a¹² – b¹²

= (a⁶)² – (b⁶)² 

= (a⁶ – b⁶)(a⁶ + b⁶)

= [(a³)² – (b³)²][(a²)³ + (b²)³]

= (a³ – b³)(a³ + b³)[(a² + b²)(a⁴ – a²b² + b⁴)] 

= (a – b)(a² + ab + b²)(a + b)(a² – ab + b²)(a² + b²)(a⁴ – a²b² + b⁴)

= (a – b)(a + b)(a² + b²)(a² + ab + b²)(a² – ab + b²)(a⁴ – a²b² + b⁴)Question 30

Factorise 

x⁶ – 7x³ – 8Solution 30

Given equation is x⁶ – 7x³ – 8.

Putting x³ = y, we get

y² – 7y – 8

= y² – 8y + y – 8

= y(y – 8) + 1(y – 8)

= (y – 8)(y + 1)

= (x³ – 8)(x³ + 1)

= (x³ – 2³)(x³ + 1³)

= (x – 2)(x² + 2x + 4)(x + 1)(x² – x + 1)

= (x – 2)(x + 1)(x² + 2x + 4)(x² – x + 1) Question 31

Factorise 

x³ – 3x² + 3x + 7Solution 31

x³ – 3x² + 3x + 7

= x³ – 3x² + 3x – 1 + 8

= (x³ – 3x² + 3x – 1) + 8

= (x – 1)³ + 2³

= (x – 1 + 2)[(x – 1)² – (x – 1)(2) + 2²]

= (x + 1)(x² – 2x + 1 – 2x + 2 + 4)

= (x + 1)(x² – 4x + 7) Question 32

Factorise 

(x + 1)³ + (x – 1)³Solution 32

(x + 1)³ + (x – 1)³

= (x + 1 + x – 1)[(x + 1)² – (x + 1)(x – 1) + (x – 1)²]

= 2x(x² + 2x + 1 – x² + 1 + x² – 2x + 1)

= 2x(x² + 3)  Question 33

Factorise 

(2a + 1)³ + (a – 1)³Solution 33

(2a + 1)³ + (a – 1)³

= (2a + 1 + a – 1)[(2a + 1)² – (2a + 1)(a – 1) + (a – 1)²]

= (3a)[4a² + 4a + 1 – 2a² + 2a – a + 1 + a² – 2a + 1]

= 3a(3a² + 3a + 3)

= 9a(a² + a + 1)  Question 34

Factorise 

8(x + y)³ – 27(x – y)³Solution 34

8(x + y)³ – 27(x – y)³

= [2³ (x + y)³] – [3³ (x – y)³]

= [2(x + y) – 3(x – y)]{[2(x + y)]² + 2(x + y)3(x – y) + [3(x – y)]²}

= (2x + 2y – 3x + 3y){[4(x² + y² + 2xy)] + 6(x² – y²) + [9(x² + y² – 2xy]}

= (-x + 5y){4x² + 4y² + 8xy + 6x² – 6y² + 9x² + 9y² – 18xy}

= (-x + 5y)(19x² + 7y² – 10xy) Question 35

Factorise 

(x + 2)³ + (x – 2)³Solution 35

(x + 2)³ + (x – 2)³

= [(x + 2) + (x – 2)][(x + 2)² – (x + 2)(x – 2) + (x – 2)²]

=(2x)(x² + 4x + 4 – x² + 4 + x² – 4x + 4)

= 2x(x² + 12) Question 36

Factorise 

(x + 2)³ – (x – 2)³Solution 36

(x + 2)³ – (x – 2)³

= [(x + 2) – (x – 2)][(x + 2)² + (x + 2)(x – 2) + (x – 2)²]

= 4(x² + 4x + 4 + x² – 4 + x² – 4x + 4)

= 4(3x² + 4) Question 37

Prove that

Solution 37

Question 38

Prove that

Solution 38

Exercise Ex. 3G

Question 1

Find the product:

(x + y – z) (x² + y² + z² – xy + yz + zx)Solution 1

(x + y – z) (x² + y² + z² – xy + yz + zx)

= [x + y + (-z)] [(x)² + (y)² + (-z)² – (x) (y) – (y) (-z) – (-z) (x)]

= x³ + y³ – z³ + 3xyz.Question 2

Find the product.

(x – y – z)(x² + y² + z² + xy – yz + xz)Solution 2

(x – y – z)(x² + y² + z² + xy – yz + xz)

= x³ + xy² + xz² + x²y  – xyz + x²z – x²y – y³ – yz² – xy² + y²z – xyz – x²z – y²z – z³ – xyz + yz² – xz²

= x³ – y³ – z³ – xyz – xyz – xyz

= x³ – y³ – z³ – 3xyz Question 3

Find the product:

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)Solution 3

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= [x + (-2y) + 3] [(x)² + (-2y)² + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]

= (a + b + c) (a² + b² + c² – ab – bc – ca)

= a³ + b³ + c³ – 3abc

Where, x = a, (-2y) = b and 3 = c

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= (x)³ + (-2y)³ + (3)² – 3 (x) (-2y) (3)

= x³ – 8y³ + 27 + 18xy.Question 4

Find the product.

(3x – 5y + 4)(9x² + 25y² + 15xy + 20y – 12x + 16)

*Modified the questionSolution 4

(3x – 5y + 4)(9x² + 25y² + 15xy + 20y – 12x + 16)

= (3x + (-5y) + 4)[(3x)² + (-5y)² + (4)² – (3x)(-5y) – (-5y)(4) – (3x)(4)]

= (3x)³ + (-5y)³ + (4)³ – 3(3x)(-5y)(4)

= 27x³ – 125y³ + 64 + 180xy Question 5

Factorise:

125a³ + b³ + 64c³ – 60abcSolution 5

125a³ + b³ + 64c³ – 60abc

= (5a)³ + (b)³ + (4c)³ – 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)² + b² + (4c)² – (5a) (b) – (b) (4c) – (5a) (4c)]

[ a³ + b³ + c³ – 3abc = (a+ b + c) (a² + b² + c² – ab – bc – ca)]

= (5a + b + 4c) (25a² + b² + 16c² – 5ab – 4bc – 20ac).Question 6

Factorise:

a³ + 8b³ + 64c³ – 24abcSolution 6

a³ + 8b³ + 64c³ – 24abc

= (a)³ + (2b)³ + (4c)³ – 3 a 2b 4c

= (a + 2b + 4c) [a² + 4b² + 16c² – 2ab – 8bc – 4ca).Question 7

Factorise:

1 + b³ + 8c³ – 6bcSolution 7

1 + b³ + 8c³ – 6bc

= 1 + (b)³ + (2c)³ – 3 (b) (2c)

= (1 + b + 2c) [1 + b² + (2c)² – b – b 2c – 2c]

= (1 + b + 2c) (1 + b² + 4c² – b – 2bc – 2c).Question 8

Factorise:

216 + 27b³ + 8c³ – 108bcSolution 8

216 + 27b³ + 8c³ – 108bc

= (6)³ + (3b)³ + (2c)² – 3 6 3b 2c

= (6 + 3b + 2c) [(6)² + (3b)² + (2c)² – 6 3b – 3b 2c – 2c 6]

= (6 + 3b + 2c) (36 + 9b² + 4c² – 18b – 6bc – 12c).Question 9

Factorise:

27a³ – b³ + 8c³ + 18abcSolution 9

27a³ – b³ + 8c³ + 18abc

= (3a)³ + (-b)³ + (2c)³ + 3(3a) (-b) (2c)

= [3a + (-b) + 2c] [(3a)² + (-b)² + (2c)² – 3a (-b) – (-b) (2c) – (2c) (3a)]

= (3a – b + 2c) (9a² + b² + 4c² + 3ab + 2bc – 6ca).Question 10

Factorise:

8a³ + 125b³ – 64c³ + 120abcSolution 10

8a³ + 125b³ – 64c³ + 120abc

= (2a)³ + (5b)³ + (-4c)³ – 3 (2a) (5b) (-4c)

= (2a + 5b – 4c) [(2a)² + (5b)² + (-4c)² – (2a) (5b) – (5b) (-4c) – (-4c) (2a)]

= (2a + 5b – 4c) (4a² + 25b² + 16c² – 10ab + 20bc + 8ca).Question 11

Factorise:

8 – 27b³ – 343c³ – 126bcSolution 11

8 – 27b³ – 343c³ – 126bc

= (2)³ + (-3b)³ + (-7c)³ – 3(2) (-3b) (-7c)

= (2 – 3b – 7c) [(2)² + (-3b)² + (-7c)² – (2) (-3b) – (-3b) (-7c) – (-7c) (2)]

= (2 – 3b – 7c) (4 + 9b² + 49c² + 6b – 21bc + 14c).Question 12

Factorise:

125 – 8x³ – 27y³ – 90xySolution 12

125 – 8x³ – 27y³ – 90xy

= (5)³ + (-2x)³ + (-3y)³ – 3 (5) (-2x) (-3y)

= (5 – 2x – 3y) [(5)² + (-2x)² + (-3y)² – (5) (-2x) – (-2x) (-3y) – (-3y) (5)]

= (5 – 2x – 3y) (25 + 4x² + 9y² + 10x – 6xy + 15y).Question 13

Factorise:

Solution 13

Question 14

Factorise:

27x³ – y³ – z³ – 9xyzSolution 14

27x³ – y³ – z³ – 9xyz

= (3x)³ – y³ – z³ – 3(3x)(y)(z)

= (3x)³ + (-y)³ + (-z)³ – 3(3x)(-y)(-z)

= [3x + (-y) + (-z)][(3x)² + (-y)² + (-z)² – (3x)(-y) – (-y)(-z) – (3x)(-z)]

= (3x – y – z)(9x² + y² + z² + 3xy – yz + 3zx) Question 15

Factorise:

Solution 15

Question 16

Factorise:

Solution 16

Question 17

Factorise:

(a – b)³ + (b – c)³ + (c – a)³Solution 17

Putting (a – b) = x, (b – c) = y and (c – a) = z, we get,

(a – b)³ + (b – c)³ + (c – a)³

= x³ + y³ + z³, where (x + y + z) = (a – b) + (b – c) + (c – a) = 0

= 3xyz[ (x + y + z)= 0 (x³ + y³ + z³) = 3xyz]

= 3(a – b) (b – c) (c – a).Question 18

Factorise:

(a – 3b)³ + (3b – c)³ + (c – a)³Solution 18

Given equation is (a – 3b)³ + (3b – c)³ + (c – a)³

Now,

(a – 3b) + (3b – c) + (c – a)

= a – a – 3b + 3b – c + c

= 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(a – 3b)³ + (3b – c)³ + (c – a)³

= 3(a – 3b)(3b – c)(c – a)Question 19

Factorise:

(3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³Solution 19

We have:

(3a – 2b) + (2b – 5c) + (5c – 3a) = 0

So, (3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³

= 3(3a – 2b) (2b – 5c) (5c – 3a).Question 20

Factorise:

(5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³Solution 20

Given equation is (5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³

Now,

(5a – 7b) + (7b – 9c) + (9c – 5a)

= 5a – 7b + 7b – 9c + 9c – 5a

= 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(5a – 7b)³ + (7b – 9c)³ + (9c – 5a)³

= 3(5a – 7b)(7b – 9c)(9c – 5a) Question 21

Factorise:

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³Solution 21

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= [a (b – c)]³ + [b (c – a)]³ + [c (a – b)]³

Now, since, a (b – c) + b (c -a) + c (a – b)

= ab – ac + bc – ba + ca – bc = 0

So, a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= 3a (b – c) b (c – a) c (a – b)

= 3abc (a – b) (b – c) (c – a).Question 22(i)

Evaluate

(-12)³ + 7³ + 5³Solution 22(i)

Given equation is (-12)³ + 7³ + 5³

Now,

-12 + 7 + 5 = -12 + 12 = 0

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz

Hence,

(-12)³ + 7³ + 5³

= 3(-12)(7)(5)

= -1260 Question 22(ii)

Evaluate

(28)³ + (-15)³ + (-13)³Solution 22(ii)

Given equation is (28)³ + (-15)³ + (-13)³

Now,

28 + (-15) + (-13) = 28 – 28 = 0

We know that if x + y + z = 0, then x³ +  y³ + z³ = 3xyz

Hence,

(28)³ + (-15)³ + (-13)³

= 3(28)(-15)(-13)

= 16380 Question 23

Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a)Solution 23

L.H.S. = (a + b + c)³ – a³ – b³ – c³

= [(a + b) + c]³ – a³ – b³ – c³

= (a + b)³ + c³ + 3(a + b)c × [(a + b) + c] – a³ – b³ – c³

= a³ + b³ + 3ab(a + b) + c³ + 3(a + b)c × [(a + b) + c] – a³ – b³ – c³

= 3ab(a + b) + 3(a + b)c × [(a + b) + c]

= 3(a + b)[ab + c(a + b) + c²]

= 3(a + b)[ab + ac + bc + c²]

= 3(a + b)[a(b + c) + c(b + c)]

= 3(a + b)[(b + c)(a + c)]

= 3(a + b)(b + c)(c + a)

= R.H.S.Question 24

If a, b, c are all nonzero and a + b + c = 0, prove that

.Solution 24

Question 25

If a + b + c = 9 and a² + b² + c² = 35, find the value of (a³ + b³ + c³ – 3abc).Solution 25

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

= (a + b + c)[(a² + b² + c²) – (ab + bc + ca)] ….(i)

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (9)² = 35 + 2(ab + bc + ca)

⇒ 81 = 35 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 46

⇒ ab + bc + ca = 23

Substituting in (i), we get

a³ + b³ + c³ – 3abc = (9)[35 – 23] = 9 × 12 = 108 

Exercise Ex. 3B

Question 1

Factorise:

9x² – 16y²Solution 1

9x² – 16y²

= (3x)² – (4y)²

= (3x + 4y)(3x – 4y) Question 2

Factorise:

Solution 2

Question 3

Factorise:

81 – 16x²Solution 3

81 – 16x²

= (9)² – (4x)²

= (9 – 4x)(9 + 4x) Question 4

Factorise:

5 – 20x²Solution 4

5 – 20x²

= 5(1 – 4x²)

= 5[(1)² – (2x)²]

= 5[(1 – 2x)(1 + 2x)]

= 5(1 – 2x)(1 + 2x) Question 5

Factorise:

2x⁴ – 32Solution 5

2x⁴ – 32

= 2(x⁴ – 16)

= 2[(x²)² – (4)²]

= 2[(x² – 4)(x² + 4)]

= 2[(x² – 2²)(x² + 4)]

= 2[(x – 2)(x + 2)(x² + 4)]

= 2(x – 2)(x + 2)(x² + 4) Question 6

Factorise:

3a³b – 243ab³Solution 6

3a³b – 243ab³

= 3ab (a² – 81 b²)

= 3ab [(a)² – (9b)²]

= 3ab (a + 9b) (a – 9b)Question 7

Factorise:

3x³ – 48xSolution 7

3x³ – 48x

= 3x (x² – 16)

= 3x [(x)² – (4)²]

= 3x (x + 4) (x – 4)Question 8

Factorise:

27a² – 48b²Solution 8

27a² – 48b²

= 3 (9a² – 16b²)

= 3 [(3a)² – (4b)²]

= 3(3a + 4b) (3a – 4b)Question 9

Factorise:

x – 64x³Solution 9

x – 64x³

= x (1 – 64x²)

= x[(1)² – (8x)²]

= x (1 + 8x) (1 – 8x)Question 10

Factorise:

8ab² – 18a³Solution 10

8ab² – 18a³

= 2a (4b² – 9a²)

= 2a [(2b)² – (3a)²]

= 2a (2b + 3a) (2b – 3a)Question 11

Factorise:

150 – 6x²Solution 11

150 – 6x²

= 6 (25 – x²)

= 6 (5² – x²)

= 6 (5 + x) (5 – x)Question 12

Factorise:

2 – 50x²Solution 12

2 – 50x²

= 2 (1 – 25x²)

= 2 [(1)² – (5x)²]

= 2 (1 + 5x) (1 – 5x)Question 13

Factorise:

20x² – 45Solution 13

20x² – 45

= 5(4x² – 9)

= 5 [(2x)² – (3)²]

= 5 (2x + 3) (2x – 3)Question 14

Factorise:

(3a + 5b)² – 4c²Solution 14

(3a + 5b)² – 4c²

= (3a + 5b)² – (2c)²

= (3a + 5b – 2c)(3a + 5b + 2c) Question 15

Factorise:

a² – b² – a – bSolution 15

a² – b² – a – b

= a² – b² – (a + b)

= (a – b)(a + b) – (a + b)

= (a + b)(a – b – 1) Question 16

Factorise:

4a² – 9b² – 2a – 3bSolution 16

4a² – 9b² – 2a – 3b

= (2a)² – (3b)² – (2a + 3b)

= (2a – 3b)(2a + 3b) – (2a + 3b)

= (2a + 3b)(2a – 3b – 1) Question 17

Factorise:

a² – b² + 2bc – c²Solution 17

a² – b² + 2bc – c²

= a² – (b² – 2bc + c²)

= a² – (b – c)²

= [a – (b – c)][a + (b – c)]

= (a – b + c)(a + b – c) Question 18

Factorise:

4a² – 4b² + 4a + 1Solution 18

4a² – 4b² + 4a + 1

= (4a² + 4a + 1) – 4b²

= [(2a)² + 2 × 2a × 1 + (1)²] – (2b)²

= (2a + 1)² – (2b)²

= (2a + 1 – 2b)(2a + 1 + 2b)

= (2a – 2b + 1)(2a + 2b + 1) Question 19

Factorise:

a² + 2ab + b² – 9c²Solution 19

a² + 2ab + b² – 9c²

= (a + b)² – (3c)²

= (a + b + 3c) (a + b – 3c)Question 20

Factorise:

108a² – 3(b – c)²Solution 20

108a² – 3(b – c)²

= 3 [(36a² – (b -c)²]

= 3 [(6a)² – (b – c)²]

= 3 (6a + b – c) (6a – b + c)Question 21

Factorise:

(a + b)³ – a – bSolution 21

(a + b)³ – a – b

= (a + b)³ – (a + b)

= (a + b) [(a + b)² – 1²]

= (a + b) (a + b + 1) (a + b – 1)Question 22

Factorise:

x² + y² – z² – 2xySolution 22

x² + y² – z² – 2xy

= (x² + y² – 2xy) – z²

= (x – y)² – z²

= (x – y – z)(x – y + z) Question 23

Factorise:

x² + 2xy + y² – a² + 2ab – b²Solution 23

x² + 2xy + y² – a² + 2ab – b²

= (x² + 2xy + y²) – (a² – 2ab + b²)

= (x + y)² – (a – b)²

= [(x + y) – (a – b)][(x + y) + (a – b)]

= (x + y – a + b)(x + y + a – b) Question 24

Factorise:

25x² – 10x + 1 – 36y²Solution 24

25x² – 10x + 1 – 36y²

= (25x² – 10x + 1) – 36y²

= [(5x)² – 2(5x)(1) + (1)²] – (6y)²

= (5x – 1)² – (6y)²

= (5x – 1 – 6y)(5x – 1 + 6y) Question 25

Factorise:

a – b – a² + b²Solution 25

a – b – a² + b²

= (a – b) – (a² – b²)

= (a – b) – (a – b) (a + b)

= (a – b) (1 – a – b)Question 26

Factorise:

a² – 4ac + 4c² – b²Solution 26

a² – 4ac + 4c² – b²

= a² – 4ac + 4c² – b²

= a² – 2 a 2c + (2c)² – b²

= (a – 2c)² – b²

= (a – 2c + b) (a – 2c – b)Question 27

Factorise:

9 – a² + 2ab – b²Solution 27

9 – a² + 2ab – b²

= 9 – (a² – 2ab + b²)

= 3² – (a – b)²

= (3 + a – b) (3 – a + b)Question 28

Factorise:

x³ – 5x² – x + 5Solution 28

x³ – 5x² – x + 5

= x² (x – 5) – 1 (x – 5)

= (x – 5) (x² – 1)

= (x – 5) (x + 1) (x – 1)Question 29

Factorise:

1 + 2ab – (a² + b²)Solution 29

1 + 2ab – (a² + b²)

= 1 – (a² + b² – 2ab)

= (1)² – (a – b)²

= [1 – (a – b)][1 + (a – b)]

= (1 – a + b)(1 + a – b) Question 30

Factorise:

9a² + 6a + 1 – 36b²Solution 30

9a² + 6a + 1 – 36b²

= (9a² + 6a + 1) – 36b²

= [(3a)² + 2(3a)(1) + (1)²] – (6b)²

= (3a + 1)² – (6b)²

= (3a + 1 – 6b)(3a + 1 + 6b) Question 31

Factorise:

x² – y² + 6y – 9Solution 31

x² – y² + 6y – 9

= x² – (y² – 6y + 9)

= x² – (y² – 2 y 3 + 3²)

= x² – (y – 3)²

= [x + (y – 3)] [x – (y – 3)] 

= (x + y – 3) (x – y + 3)Question 32

Factorise:

4x² – 9y² – 2x – 3ySolution 32

4x² – 9y² – 2x – 3y

= (2x)² – (3y)² – (2x + 3y)

= (2x + 3y) (2x – 3y) – (2x + 3y) 

= (2x + 3y) (2x – 3y – 1)Question 33

Factorise:

9a² + 3a – 8b – 64b²Solution 33

9a² + 3a – 8b – 64b²

= 9a² – 64b² + 3a – 8b

= (3a)² – (8b)² + (3a – 8b)

= (3a + 8b) (3a – 8b) + (3a – 8b)

= (3a – 8b) (3a + 8b + 1)Question 34

Factorise:

Solution 34

Question 35

Factorise:

Solution 35

Question 36

Factorise:

Solution 36

Question 37

Factorise:

x⁸ – 1Solution 37

x⁸ – 1

= (x⁴)² – (1)²

= (x⁴ – 1)(x⁴ + 1)

= [(x²)² – (1)²)(x⁴ + 1)

= (x² – 1)(x² + 1)(x⁴ + 1)

= (x – 1)(x + 1)(x² + 1)[(x²)² + (1)² + 2x² – 2x²

= (x – 1)(x + 1)(x² + 1)[(x²)² + (1)² + 2x²) – 2x²

Question 38

Factorise:

16x⁴ – 1Solution 38

16x⁴ – 1

= (4x²)² – (1)²

= (4x² – 1)(4x² + 1)

= [(2x)² – (1)²](4x² + 1)

= (2x – 1)(2x + 1)(4x² + 1) Question 39

Factorise:

81x⁴ – y⁴Solution 39

81x⁴ – y⁴

= (9x²)² – (y²)²

= (9x² – y²)(9x² + y²)

= [(3x)² – y²](9x² + y²)

= (3x – y)(3x + y)(9x² + y²) Question 40

Factorise:

x⁴ – 625Solution 40

x⁴ – 625

= (x²)² – (25)²

= (x² + 25) (x² – 25)

= (x² + 25) (x² – 5²)

= (x² + 25) (x + 5) (x – 5)

Exercise Ex. 3A

Question 1

Factorise:

9x² + 12xySolution 1

9x² + 12xy = 3x (3x + 4y)Question 2

Factorise:

18x²y – 24xyzSolution 2

18x²y – 24xyz = 6xy (3x – 4z)Question 3

Factorise:

27a³b³ – 45a⁴b²Solution 3

27a³b³ – 45a⁴b² = 9a³b² (3b – 5a)Question 4

Factorise:

2a (x+ y) – 3b (x + y)Solution 4

2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)Question 5

Factorise:

2x (p² + q²) + 4y (p² + q²)Solution 5

2x (p² + q²) + 4y (p² + q²)

= (2x + 4y) (p² + q²)

= 2(x+ 2y) (p² + q²)Question 6

Factorise:

x (a – 5) + y (5 – a)Solution 6

x (a – 5) + y (5 – a)

= x (a – 5) + y (-1) (a – 5)

= (x – y) (a – 5)Question 7

Factorise:

4 (a + b) – 6 (a + b)²Solution 7

4 (a + b) – 6 (a + b)²

= (a + b) [4 – 6 (a + b)]

= 2 (a + b) (2 – 3a – 3b)

= 2 (a + b) (2 – 3a – 3b)Question 8

Factorise:

8 (3a – 2b)² – 10 (3a – 2b)Solution 8

8 (3a – 2b)² – 10 (3a – 2b)

= (3a – 2b) [8(3a – 2b) – 10]

= (3a – 2b) 2[4 (3a – 2b) – 5]

= 2 (3a – 2b) (12 a – 8b – 5)Question 9

Factorise:

x (x + y)³ – 3x²y (x + y)Solution 9

x (x + y)³ – 3x²y (x + y)

= x (x + y) [(x + y)² – 3xy]

= x (x + y) (x² + y² + 2xy – 3xy)

= x (x + y) (x² + y² – xy)Question 10

Factorise:

x³+ 2x² + 5x + 10Solution 10

x³+ 2x² + 5x + 10

= x² (x + 2) + 5 (x + 2)

= (x² + 5) (x + 2)Question 11

Factorise:

x² + xy – 2xz – 2yzSolution 11

x² + xy – 2xz – 2yz

= x (x + y) – 2z (x + y)

= (x+ y) (x – 2z)Question 12

Factorise:

a³b – a²b + 5ab – 5bSolution 12

a³b – a²b + 5ab – 5b

= a²b (a – 1) + 5b (a – 1)

= (a – 1) (a²b + 5b)

= (a – 1) b (a² + 5)

= b (a – 1) (a² + 5)Question 13

Factorise:

8 – 4a – 2a³ + a⁴Solution 13

8 – 4a – 2a³ + a⁴

= 4(2 – a) – a³ (2 – a)

= (2 – a) (4 – a³)Question 14

Factorise:

x³ – 2x²y + 3xy² – 6y³Solution 14

x³ – 2x²y + 3xy² – 6y³

= x² (x – 2y) + 3y² (x – 2y)

= (x – 2y) (x² + 3y²)Question 15

Factorise:

px – 5q + pq – 5xSolution 15

px + pq – 5q – 5x

= p(x + q) – 5 (q + x)

= (x + q) (p – 5)Question 16

Factorise:

x² + y – xy – xSolution 16

x² – xy + y – x

= x (x – y) – 1 (x – y)

= (x – y) (x – 1)Question 17

Factorise:

(3a – 1)² – 6a + 2Solution 17

(3a – 1)² – 6a + 2

= (3a – 1)² – 2 (3a – 1)

= (3a – 1) [(3a – 1) – 2]

= (3a – 1) (3a – 3)

= 3(3a – 1) (a – 1)Question 18

Factorise:

(2x – 3)² – 8x + 12Solution 18

(2x – 3)² – 8x + 12

= (2x – 3)² – 4 (2x – 3)

= (2x – 3) (2x – 3 – 4)

= (2x – 3) (2x – 7)Question 19

Factorise:

a³ + a – 3a² – 3Solution 19

a³ + a – 3a² – 3

= a(a² + 1) – 3 (a² + 1)

= (a – 3) (a² + 1)Question 20

Factorise:

3ax – 6ay – 8by + 4bxSolution 20

3ax – 6ay – 8by + 4bx

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)Question 21

Factorise:

abx² + a²x + b²x +abSolution 21

abx² + a²x + b²x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)Question 22

Factorise:

x³ – x² + ax + x – a – 1Solution 22

x³ – x² + ax + x – a – 1

= x³ – x² + ax – a + x – 1

= x² (x – 1) + a (x – 1) + 1 (x – 1)

= (x – 1) (x² + a + 1)Question 23

Factorise:

2x + 4y – 8xy – 1Solution 23

2x + 4y – 8xy – 1

= 2x – 1 – 8xy + 4y

= (2x – 1) – 4y (2x – 1)

= (2x – 1) (1 – 4y)Question 24

Factorise:

ab (x² + y²) – xy (a² + b²)Solution 24

ab (x² + y²) – xy (a² + b²)

= abx² + aby² – a²xy – b²xy

= abx² – a²xy + aby² – b²xy

= ax (bx – ay) + by(ay – bx)

= (bx – ay) (ax – by)Question 25

Factorise:

a² + ab (b + 1) + b³Solution 25

a² + ab (b + 1) + b³

= a² + ab² + ab + b³

= a² + ab + ab² + b³

= a (a + b) + b² (a + b)

= (a + b) (a + b²)Question 26

Factorise:

a³ + ab (1 – 2a) – 2b²Solution 26

a³ + ab (1 – 2a) – 2b²

= a³ + ab – 2a²b – 2b²

= a (a² + b) – 2b (a² + b)

= (a² + b) (a – 2b)Question 27

Factorise:

2a² + bc – 2ab – acSolution 27

2a² + bc – 2ab – ac

= 2a² – 2ab – ac + bc

= 2a (a – b) – c (a – b)

= (a – b) (2a – c)Question 28

Factorise:

(ax + by)² + (bx – ay)²Solution 28

(ax + by)² + (bx – ay)²

= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy

= a²x² + b²y² + b²x² + a²y²

= a²x² + b²x² + b²y² + a²y²

= x² (a² + b²) + y²(a² + b²)

= (a² + b²) (x² + y²)Question 29

Factorise:

a (a + b – c) – bcSolution 29

a (a + b – c) – bc

= a² + ab – ac – bc

= a(a + b) – c (a + b)

= (a – c) (a + b)Question 30

Factorise:

a(a – 2b – c) + 2bcSolution 30

a(a – 2b – c) + 2bc

= a² – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)Question 31

Factorise:

a²x² + (ax² + 1)x + aSolution 31

a²x² + (ax² + 1)x + a

= a²x² + ax³ + x + a

= ax² (a + x) + 1 (x + a)

= (ax² + 1) (a + x)Question 32

Factorise:

ab (x² + 1) + x (a² + b²)Solution 32

ab (x² + 1) + x (a² + b²)

= abx² + ab + a²x + b²x

= abx² + a²x + ab + b²x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)Question 33

Factorise:

x² – (a + b) x+ abSolution 33

x² – (a + b) x+ ab

= x² – ax – bx + ab

= x (x – a) – b(x – a)

= (x – a) (x – b)Question 34

Factorise:

Solution 34