Yearly Archives: 2022

Exercise MCQ

Question 1

Three angles of quadrilateral are 80^°, 95° and 112°. Its fourth angle is

(a) 78°

(b) 73°

(c) 85°

(d) 100°Solution 1

Question 2

The angles of a quadrilateral are in the ratio 3:4:5:6. The smallest of these angles is

(a) 45°

(b) 60°

(c) 36°

(d) 48°Solution 2

Question 3

In the given figure, ABCD is a parallelogram in which ∠BAD = 75° and ∠CBD = 60°. Then, ∠BDC =?

(a) 60°

(b) 75°

(c) 45°

(d) 50°

Solution 3

Question 4

ABCD is a rhombus such that ∠ACB = 50°. Then, ∠ADB = ?

(a) 40° 

(b) 25° 

(c) 65°  

(d) 130° Solution 4

Correct option: (a)

ABCD is a rhombus.

⇒ AD ∥ BC and AC is the transversal.

⇒ ∠DAC = ∠ACB  (alternate angles)

⇒ ∠DAC = 50° 

In ΔAOD, by angle sum property,

∠AOD + ∠DAO + ∠ADO = 180° 

⇒ 90° + ∠50° + ∠ADO = 180° 

⇒ ∠ADO = 40° 

⇒ ∠ADB = 40° Question 5

In which of the following figures are the diagonals equal?

1. Parallelogram
2. Rhombus
3. Trapezium
4. Rectangle

Solution 5

Question 6

If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a

a. trapezium

b. parallelogram

c. rectangle

d. rhombusSolution 6

Question 7

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is

1. 10 cm
2. 12 cm
3. 9cm
4. 8cm

Solution 7

Question 8

The length of each side of a rhombus is 10 cm and one of its diagonals is of length 16 cm. The length of the other diagonal is

Solution 8

Question 9

A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is

(a) 55° 

(b) 70° 

(c) 45° 

(d) 50° Solution 9

Correct option: (b)

∠DAO + ∠OAB = ∠DAB

⇒ ∠DAO + 35° = 90° 

⇒ ∠DAO = 55° 

ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.

OA = OD

⇒ ∠ODA = ∠DAO (angles opposte to equal sides are equal)

⇒ ∠ODA = 55° 

In DODA, by angle sum property,

∠ODA + ∠DAO + ∠AOD = 180° 

⇒ 55° + ∠55° + ∠AOD = 180° 

⇒ ∠AOD = 70° Question 10

If ABCD is a parallelogram with two adjacent angles ∠A = ∠B, then the parallelogram is a

1. rhombus
2. trapezium
3. rectangle
4. none of these

Solution 10

Question 11

In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A and ∠B respectively, ∠C = 70° and ∠D = 30°. Then, ∠AOB =?

1. 40°
2. 50°
3. 80°
4. 100°

Solution 11

Question 12

The bisectors of any adjacent angles of a parallelogram intersect at

1. 30°
2. 45°
3. 60°
4. 90°

Solution 12

Question 13

The bisectors of the angles of a parallelogram enclose a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 13

Correct option: (c)

The bisectors of the angles of a parallelogram enclose a rectangle.Question 14

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S then PQRS is a

(a) rectangle

(b) parallelogram

(c) rhombus

(d) quadrilateral whose opposite angles are supplementarySolution 14

Correct option: (d)

In ΔAPB, by angle sum property,

∠APB + ∠PAB + ∠PBA = 180° 

In ΔCRD, by angle sum property,

∠CRD + ∠RDC + ∠RCD = 180° 

Now, ∠SPQ + ∠SRQ = ∠APB + ∠CRD

= 360° – 180° 

= 180° 

Now, ∠PSR + ∠PQR = 360° – (∠SPQ + ∠SRQ)

= 360° – 180° 

= 180° 

Hence, PQRS is a quadrilateral whose opposite angles are supplementary. Question 15

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 15

Question 16

The figure formed by joining the mid-points of the adjacent sides of a square is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 16

Question 17

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 17

Question 18

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 18

Question 19

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 19

Question 20

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if

(a) ABCD is a parallelogram

(b) ABCD is a rectangle

(c) diagonals of ABCD are equal

(d) diagonals of ABCD are perpendicular to each otherSolution 20

Correct option: (d)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ ∥ RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM ∥ NO and PN ∥ MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle if AC ⊥ BD. Question 21

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if

(a) ABCD is a parallelogram

(b) ABCD is a rhombus

(c) diagonals of ABCD are equal

(d) diagonals of ABCD are perpendicular to each otherSolution 21

Correct option: (c)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ ∥ RS and QR ∥ SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus if diagonals of ABCD are equal. Question 22

The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if

(a) ABCD is a rhombus

(b) diagonals of ABCD are equal

(c) diagonals of ABCD are perpendicular

(d) diagonals of ABCD are equal and perpendicularSolution 22

Correct option: (d)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular. Question 23

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

1. 108°
2. 54°
3. 72°
4. 81°

Solution 23

Question 24

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angles of the parallelogram is

1. 68°
2. 102°
3. 112°
4. 136°

Solution 24

Question 25

If ∠A, ∠B, ∠C and ∠D of a quadrilateral ABCD taken in order, are in the ratio 3:7:6:4, then ABCD is a

1. rhombus
2. kite
3. trapezium
4. parallelogram

Solution 25

Question 26

Which of the following is not true for a parallelogram?

1. Opposite sides are equal.
2. Opposite angles are equal.
3. Opposite angles are bisected by the diagonals.
4. Diagonals bisect each other.

Solution 26

Question 27

If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a

1. square
2. rhombus
3. rectangle
4. kite

Solution 27

Question 28

In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD =?

1. 45°
2. 55°
3. 60°
4. 75°

Solution 28

Question 29

If area of a ‖gm with side ɑ and b is A and that of a rectangle with side ɑ and b is B, then

(a) A > B

(b) A = B

(c) A < B

(d) A ≥ BSolution 29

Question 30

In the given figure, ABCD is a ‖gm and E is the mid-point at BC, Also, DE and AB when produced meet at F. Then,

Solution 30

Question 31

P is any point on the side BC of a ΔABC. P is joined to A. If D and E are the midpoints of the sides AB and AC respectively and M and N are the midpoints of BP and CP respectively then quadrilateral DENM is

(a) a trapezium

(b) a parallelogram

(c) a rectangle

(d) a rhombusSolution 31

Correct option: (b)

In ΔABC, D and E are the mid-points of sides AB and AC respectively.

Hence, DENM is a parallelogram.Question 32

The parallel sides of a trapezium are ɑ and b respectively. The line joining the mid-points of its non-parallel sides will be

Solution 32

Question 33

In a trapezium ABCD, if E and F be the mid-points of the diagonals AC and BD respectively. Then, EF =?

Solution 33

Question 34

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB=?

1. 45°
2. 60°
3. 90°
4. 30°

Solution 34

Question 35

In the given figures, ABCD is a rhombus. Then

(a) AC² + BD² = AB²

(b) AC² + BD² = 2AB²

(c) AC² + BD² = 4AB²

(d) 2(AC² + BD²)=3AB²

Solution 35

Question 36

In a trapezium ABCD, if AB ‖ CD, then (AC² + BD²) =?

(a) BC² + AD² + 2BC. AD

(b) AB² +CD² + 2AB.CD

(c) AB² + CD² + 2AD. BC

(d) BC² + AD² + 2AB.CD

Solution 36

Question 37

Two parallelogram stand on equal bases and between the same parallels. The ratio of their area is

1. 1:2
2. 2:1
3. 1:3
4. 1:1

Solution 37

Question 38

In the given figure, AD is a median of ΔABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF =?

Solution 38

Question 39

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30°and ∠AOB = 70°. Then, ∠DBC =?

1. 40°
2. 35°
3. 45°
4. 50°

Solution 39

Question 40

Three statement are given below:

1. In a ‖gm, the angle bisectors of two adjacent angles enclose a right angle.
2. The angle bisectors of a ‖gm form a rectangle.
3. The triangle formed by joining the mid-point of the sides of an isosceles triangle is not necessarily an isosceles.

Which is true?

1. I only
2. II only
3. I and II
4. II and III

Solution 40

Question 41

Three statements are given below:

 I. In a rectangle ABCD, the diagonal AC bisects ∠A as well as ∠C.

 II. In a square ABCD, the diagonal AC bisects ∠A as well as ∠C

 III. In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C

Which is true?

1. I only
2. II and III
3. I and III
4. I and II

Solution 41

Question 42

In a quadrilateral PQRS, opposite angles are equal. If SR = 2 cm and PR = 5 cm then determine PQ. Solution 42

Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.

⇒ PQ = SR (opposite sides of parallelogram are equal)

⇒ PQ = 2 cmQuestion 43

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer.Solution 43

The given statement is false.

Diagonals of a parallelogram bisect each other. Question 44

What special name can be given to a quadrilateral PQRS if ∠P + ∠S = 118°?Solution 44

In quadrilateral PQRS, ∠P and ∠S are adjacent angles.

Since the sum of adjacent angles ≠ 180°, PQRS is not a parallelogram.

Hence, PQRS is a trapezium. Question 45

All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.Solution 45

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are acute, the sum will be less than 360°. Question 46

All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.Solution 46

The given statement is true.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are right angles,

Sum of all angles of a quadrilateral = 4 × 90° = 360° Question 47

All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.Solution 47

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are obtuse, the sum will be more than 360°. Question 48

Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.Solution 48

We know that the sum of all the four angles of a quadrilateral is 360°.

Here,

70° + 115° + 60° + 120° = 365° ≠ 360° 

Hence, we cannot form a quadrilateral with given angles. Question 49

What special name can be given to a quadrilateral whose all angles are equal?Solution 49

A quadrilateral whose all angles are equal is a rectangle. Question 50

If D and E are respectively the midpoints of the sides AB and BC of ΔABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6 cm then determine the length of DE.Solution 50

D and E are respectively the midpoints of the sides AB and BC of ΔABC.

Thus, by mid-point theorem, we have

Question 51

In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If ∠Q = 56°, determine ∠R.Solution 51

Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.

Now, adjacent angles of parallelogram are supplementary.

⇒ ∠Q + ∠R = 180° 

⇒ 56° + ∠R = 180° 

⇒ ∠R = 124° Question 52

In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not?

Solution 52

AFDE is a parallelogram

⇒ AF = ED …(i)

BDEF is a parallelogram.

⇒ FB = ED …(ii)

From (i) and (ii),

AF = FB Question 53

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together.

Is quadrilateral ABCD a ‖gm?

1. Diagonal AC and BD bisect each other.
2. Diagonal AC and BD are equal.

The correct answer is : (a)/ (b)/ (c)/ (d).Solution 53

Correct option: (a)

If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.

So, I gives the answer.

If the diagonals are equal, then the quad. ABCD is a parallelogram.

So, II gives the answer.Question 54

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is quadrilateral ABCD a rhombus?

1. Quad. ABCD is a ‖gm.
2. Diagonals AC and BD are perpendicular to each other.

The correct answer is: (a) / (b)/ (c)/ (d).Solution 54

Correct option: (c)

If the quad. ABCD is a ‖gm, it could be a rectangle or square or rhombus.

So, statement I is not sufficient to answer the question.

If the diagonals AC and BD are perpendicular to each other, then the ‖gm could be a square or rhombus.

So, statement II is not sufficient to answer the question.

However, if the statements are combined, then the quad. ABCD is a rhombus.Question 55

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is ‖gm ABCD a square?

1. Diagonals of ‖gm ABCD are equal.
2. Diagonals of ‖gm ABCD intersect at right angles.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 55

Correct option: (c)

If the diagonals of a ‖gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.

If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.

However, if both the statements are combined, then ‖gm ABCD will be a square.Question 56

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is quad. ABCD a parallelogram?

1. Its opposite sides are equal.
2. Its opposite angles are equal.

The correct answer is: (a)/ (b)/ (c)/ (d)Solution 56

Correct option: (b)

If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.

If the opposite angles are equal, then the quad. ABCD is a parallelogram.Question 57

 Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angle is 100°.	The sum of all the angle of a quadrilateral is 360°.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 57

Question 58

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.Then, PQRS is a parallelogram.	The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 58

The Reason (R) is true and is the correct explanation for the Assertion (A).Question 59

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C.	The diagonals of a rhombus bisect each other at right angles.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 59

Question 60

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
Every parallelogram is a rectangle.	The angle bisectors of a parallelogram form a rectangle.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 60

Question 61

Each question consists of two statement, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The diagonals of a ‖gm bisect each other.	If the diagonals of a ‖gm are equal and intersect at right angles, then the parallelogram is a square.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 61

Question 62

Column I	Column II
(a) Angle bisectors of a parallelogram form a	(p) parallelogram
(b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a	(q) rectangle
(c) The quadrilateral formed by joining the mid-points of the pairs of adjacent side of a rectangle is a	(r) square
(d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is	(s) rhombus

The correct answer is:

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….Solution 62

Question 63

Column I	Column II
(a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7cm. If P and Q are the mid-points of AD and BC respectively, then PQ =	(p) equal
(b) In the given figure, PQRS is a ‖gm whose diagonal intersect at O. If PR = 13 cm, then OR=	(q) at right angle
(c) The diagonals of a square are	(r) 8.5 cm
(d) The diagonals of a rhombus bisect each other	(s) 6.5 cm

The correct answer is:

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….Solution 63

Exercise Ex. 10B

Question 1

In the adjoining figure, ABCD is a parallelogram in which =72⁰. Calculate _,and .

Solution 1

Question 2

In the adjoining figure , ABCD is a parallelogram in which

 and . Calculate _.

Solution 2

Question 3

In the adjoining figure, M is the midpoint of side BC of parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.

Solution 3

ABCD is a parallelogram.

Hence, AD || BC.

⇒ ∠DAM = ∠AMB (alternate angles)

⇒ ∠BAM = ∠AMB (since ∠BAM = ∠DAM)

⇒ BM = AB (sides opposite to equal angles are equal)

But, AB = CD (opposite sides of a parallelogram)

⇒ BM = AB = CD ….(i)

Question 4

In a adjoining figure, ABCD is a parallelogram in which =60^o. If the parallelogram in which and meet DC at P, prove that (i) PB=90^o, (ii) AD=DP and PB=PC=BC, (iii) DC=2AD.

Solution 4

Question 5

_{In the adjoining figure, ABCD is a parallelogram in which}

_Calculate 

Solution 5

Question 6

_{In a ||gm ABCD , if , find the value of x and the measure of each angle of the parallelogram.}Solution 6

Question 7

_{If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram .}Solution 7

Question 8

_{Find the measure of each angle of parallelogram , if one of its angles is  less than twice the smallest angle.}Solution 8

Question 9

_{ABCD is a parallelogram in which AB=9.5 cm and its parameter is 30 cm. Find the length of each side of the parallelogram.}Solution 9

Question 10

_{In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.}

Solution 10

Question 11

_{The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively . Find the length of each side of the rhombus.}Solution 11

Question 12

_{Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.}Solution 12

Question 13

_{In each of the figures given below, ABCD is a rectangle . Find the values of x and y in each case.}

Solution 13

Question 14

In a rhombus ABCD, the altitude from D to the side AB bisect AB. Find the angle of the rhombusSolution 14

Let the altitude from D to the side AB bisect AB at point P.

Join BD.

In ΔAMD and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMD = ∠BMD (Each 90°)

MD = MD (common)

∴ ΔAMD ≅ ΔBMD (by SAS congruence criterion)

⇒ AD = BD (c.p.c.t.)

But, AD = AB (sides of a rhombus)

⇒ AD = AB = BD

⇒ ΔADB is an equilateral triangle.

⇒ ∠A = 60° 

⇒ ∠C = ∠A = 60° (opposite angles are equal)

⇒ ∠B = 180° – ∠A = 180° – 60° = 120° 

⇒ ∠D = ∠B = 120° 

Hence, in rhombus ABCD, ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.Question 15

_{In the adjoining figure , ABCD is square . A line segment CX cuts AB at X and the diagonal BD at O, such that . Find the value of x.}

*Back answer incorrectSolution 15

Question 16

In a rhombus ABCD show that diagonal AC bisect ∠A as well as ∠C and diagonal BD bisect ∠B as well as ∠D.Solution 16

In ΔABC and ΔADC,

AB = AD (sides of a rhombus are equal)

BC = CD (sides of a rhombus are equal)

AC = AC (common)

∴ ΔABC ≅ ΔADC (by SSS congruence criterion)

⇒ ∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)

⇒ AC bisects ∠A as well as ∠C.

Similarly,

In ΔBAD and ΔBCD,

AB = BC (sides of a rhombus are equal)

AD = CD (sides of a rhombus are equal)

BD = BD (common)

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

⇒ BD bisects ∠B as well as ∠D.Question 17

In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM = CN. Show that AC and MN bisect each other.

Solution 17

In ΔAMO and ΔCNO

∠MAO = ∠NCO (AB ∥ CD, alternate angles)

AM = CN (given)

∠AOM = ∠CON (vertically opposite angles)

∴ ΔAMO ≅ ΔCNO (by ASA congruence criterion)

⇒ AO = CO and MO = NO (c.p.c.t.)

⇒ AC and MN bisect each other.Question 18

In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that  and , prove that AQCP is a parallelogram.

Solution 18

Question 19

In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O.A line segment EOF is drawn to meet AB at E and DC at F . Prove that OE=OF.

Solution 19

Question 20

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.Solution 20

∠DCM = ∠DCN + ∠MCN

⇒ 90° = ∠DCN + 60° 

⇒ ∠DCN = 30° 

In ΔDCN,

∠DNC + ∠DCN + ∠D = 180° 

⇒ 90° + 30° + ∠D = 180° 

⇒ ∠D = 60° 

⇒ ∠B = ∠D = 60° (opposite angles of parallelogram are equal)

⇒ ∠A = 180° – ∠B = 180° – 60° = 120° 

⇒ ∠C = ∠A = 120° 

Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.Question 21

ABCD is rectangle in which diagonal AC bisect ∠A as well as ∠C. Show that (i) ABCD is square, (ii) diagonal BD bisect ∠B as well as ∠D.Solution 21

(i) ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

⇒ ∠BAC = ∠DAC ….(i)

And ∠BCA = ∠DCA ….(ii)

Since every rectangle is a parallelogram, therefore

AB ∥ DC and AC is the transversal.

⇒ ∠BAC = ∠DCA (alternate angles)

⇒ ∠DAC = ∠DCA [From (i)]

Thus, in ΔADC,

AD = CD (opposite sides of equal angles are equal)

But, AD = BC and CD = AB (ABCD is a rectangle)

⇒ AB = BC = CD = AD

Hence, ABCD is a square.

(ii) In ΔBAD and ΔBCD,

AB = CD

AD = BC

BD = BD

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

Hence, diagonal BD bisects ∠B as well as ∠D.Question 22

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE= AB. Prove that ED bisects BC.

Solution 22

Question 23

In the adjoining figure , ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, prove that AF=2AB.

Solution 23

Question 24

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.Solution 24

l ∥ m and t is a transversal.

⇒ ∠APR = ∠PRD (alternate angles)

⇒ ∠SPR = ∠PRQ (PS and RQ are the bisectors of ∠APR and ∠PRD)

Thus, PR intersects PS and RQ at P and R respectively such that ∠SPR = ∠PRQ i.e., alternate angles are equal.

⇒ PS ∥ RQ

Similarly, we have SR ∥ PQ.

Hence, PQRS is a parallelogram.

Now, ∠BPR + ∠PRD = 180° (interior angles are supplementary)

⇒ 2∠QPR + 2∠QRP = 180° (PQ and RQ are the bisectors of ∠BPR and ∠PRD)

⇒ ∠QPR + ∠QRP = 90° 

In ΔPQR, by angle sum property,

∠PQR + ∠QPR + ∠QRP = 180° 

⇒ ∠PQR + 90° = 180° 

⇒ ∠PQR = 90° 

Since PQRS is a parallelogram,

∠PQR = ∠PSR

⇒ ∠PSR = 90° 

Now, ∠SPQ + ∠PQR = 180° (adjacent angles in a parallelogram are supplementary)

⇒ ∠SPQ + 90° = 180° 

⇒ ∠SPQ = 90° 

⇒ ∠SRQ = 90° 

Thus, all the interior angles of quadrilateral PQRS are right angles.

Hence, PQRS is a rectangle.Question 25

K, L, M and N are point on the sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM = DN. Prove that KLMN is a square.Solution 25

AK = BL = CM = DN (given)

⇒ BK = CL = DM = AN (i)(since ABCD is a square)

In ΔAKN and ΔBLK,

AK = BL (given)

∠A = ∠B (Each 90°)

AN = BK [From (i)]

∴ ΔAKN ≅ ΔBLK (by SAS congruence criterion)

⇒ ∠AKN = ∠BLK and ∠ANK = ∠BKL (c.p.c.t.)

But, ∠AKN + ∠ANK = 90° and ∠BLK + ∠BKL = 90° 

⇒ ∠AKN + ∠ANK + ∠BLK + ∠BKL = 90° + 90° 

⇒ 2∠AKN + 2∠BKL = 180° 

⇒ ∠AKN + ∠BKL = 90° 

⇒ ∠NKL = 90° 

Similarly, we have

∠KLM = ∠LMN = ∠MNK = 90° 

Hence, KLMN is a square.Question 26

A is given . if lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB forming  , as shown in the adjoining figure, show that 

Solution 26

Question 27

In the adjoining figure, is a triangle and through A,B, C lines are drawn , parallel respectively to BC, CA and AB, intersecting at P, Q and R. prove that the perimeter of  is double the perimeter of  .

Solution 27

Exercise Ex. 10A

Question 1

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.Solution 1

Let the measure of the fourth angle = x° 

For a quadrilateral, sum of four angles = 360° 

⇒ x° + 75° + 90° + 75° = 360° 

⇒ x° = 360° – 240° 

⇒ x° = 120° 

Hence, the measure of fourth angle is 120°. Question 2

The angle of the quadrilateral are in the ratio 2:4:5:7. Find the angles.Solution 2

Question 3

In the adjoining figure , ABCD is a trapezium in which AB || DC. If =55⁰ and = 70⁰, find  _and .

Solution 3

Since AB || DC

Question 4

In the adjoining figure , ABCD is a square andis an equilateral triangle . Prove that

(i)AE=BE, (ii) =15⁰

Solution 4

Given:

Question 5

In the adjoining figure , BMAC and DNAC. If BM=DN, prove that AC bisects BD.

Solution 5

Question 6

In given figure, ABCD is a quadrilateral in which AB=AD and BC= DC. Prove that (i) AC bisects  and , (ii) BE=DE,

(iii) 

Solution 6

Question 7

In the given figure , ABCD is a square and _PQR=90⁰. If PB=QC= DR, prove that (i) QB=RC, (ii) PQ=QR, (iii) QPR=45⁰.

Solution 7

Question 8

If O is a point with in a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD.Solution 8

Given: O is a point within a quadrilateral ABCD

Question 9

In the adjoining figure, ABCD is a quadrilateral and AC is one of the diagonals .Prove that:

(i)AB+BC+CD+DA> 2AC

(ii)AB+BC+CD>DA

(iii)AB+BC+CD+DA>AC+BD

Solution 9

Given: ABCD is a quadrilateral and AC is one of its diagonals.

Question 10

Prove that the sum of all the angles of a quadrilateral is 360^0.Solution 10

Given: ABCD is a quadrilateral.

Exercise Ex. 10C

Question 1

P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that

(i) PQ ∥ AC and PQ = 

(ii) PQ ∥ SR

(iii) PQRS is a parallelogram.

Solution 1

(i) In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

(ii) In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii), we have

PQ = SR and PQ ∥ SR

(iii) Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram. Question 2

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse.Solution 2

Let ΔABC be an isosceles right triangle, right-angled at B.

⇒ AB = BC

Let PBSR be a square inscribed in ΔABC with common ∠B.

⇒ PB = BS = SR = RP

Now, AB – PB = BC – BS

⇒ AP = CS ….(i)

In ΔAPR and ΔCSR

AP = CS  [From (i)

∠APR = ∠CSR (Each 90°)

PR = SR (sides of a square)

∴ ΔAPR ≅ ΔCSR (by SAS congruence criterion)

⇒ AR = CR (c.p.c.t.)

Thus, point R bisects the hypotenuse AC.Question 3

In the adjoining figure , ABCD is a ||gm in which E and F are the midpoints of AB and CD respectively . If GH is a line segment that cuts AD, EF, and BC at G, P and H respectively, prove that GP = PH.

Solution 3

Question 4

M and N are points on opposites sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. show that MN is bisected at O.Solution 4

In ΔAOM and ΔCON

∠MAO = ∠OCN  (Alternate angles)

AO = OC (Diagonals of a parallelogram bisect each other)

∠AOM = ∠CON  (Vertically opposite angles)

∴ ΔAOM ≅ ΔCON  (by ASA congruence criterion)

⇒ MO = NO (c.p.c.t.)

Thus, MN is bisected at point O. Question 5

In the adjoining figure, PQRS is a trapezium in which PQ ∥ SR and M is the midpoint of PS. A line segment MN ∥ PQ meets QR at N. Show that N is the midpoint of QR.

Solution 5

Construction: Join diagonal QS. Let QS intersect MN at point O.

PQ ∥ SR and MN ∥ PQ

⇒ PQ ∥ MN ∥ SR

By converse of mid-point theorem a line drawn, through the mid-point of any side of a triangle and parallel to another side bisects the third side. 

Now, in ΔSPQ

MO ∥ PQ and M is the mid-point of SP

So, this line will intersect QS at point O and O will be the mid-point of QS.

Also, MN ∥ SR

Thus, in ΔQRS, ON ∥ SR and O is the midpoint of line QS.

So, by using converse of mid-point theorem, N is the mid-point of QR.Question 6

In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.Solution 6

PM is the bisector of ∠P.

⇒ ∠QPM = ∠SPM ….(i)

PQRS is a parallelogram.

∴ PQ ∥ SR and PM is the transversal.

⇒ ∠QPM = ∠MS (ii)(alternate angles)

From (i) and (ii),

∠SPM = ∠PMS ….(iii)

⇒ MS = PS = 9 cm (sides opposite to equal angles are equal)

Now, ∠RMT = ∠PMS (iv)(vertically opposite angles)

Also, PS ∥ QT and PT is the transversal.

∠RTM = ∠SPM

⇒ ∠RTM = ∠RMT

⇒ RT = RM (sides opposite to equal angles are equal)

RM = SR – MS = 12 – 9 = 3 cm

⇒ RT = 3 cmQuestion 7

In the adjoining figure , ABCD is a trapezium in which AB|| DC and P, Q are the mid points of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ=QE, (ii) PR||AB, (iii) AR=RC.

Solution 7

Question 8

In the adjoining figure, AD is a medium of  _{and DE|| BA. Show that BE is also a median of .}

Solution 8

Question 9

_{In the adjoining figure , AD and BE are the medians of  and DF|| BE. Show that .}

Solution 9

Question 10

_{Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.}Solution 10

Question 11

_{In the adjoining figure, D,E,F are the midpoints of the sides BC, CA, and AB respectively, of . Show that .}

Solution 11

Question 12

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rectangle is a rhombus.Solution 12

Question 13

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rhombus is a rectangle.Solution 13

Question 14

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a square is a square.Solution 14

Question 15

Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.Solution 15

Question 16

The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.Solution 16

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP  [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus.Question 17

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.Solution 17

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ || RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle. Question 18

The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC ⊥ BD then prove that the quadrilateral formed is a square.Solution 18

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square.

Exercise MCQ

Question 1

Out of the following given figures which are on the same base but not between the same parables?

Solution 1

Question 2

In which of the following figures, you find polynomials on the same base and between the same parallels?

Solution 2

Question 3

The median of a triangle divides it into two

1. triangles of equal area
2. congruent triangles
3. isosceles triangle
4. right triangles

Solution 3

Question 4

The area of quadrilateral ABCD in the given figure is

1. 57 cm²
2. 108 cm²
3. 114 cm²
4. 195 cm²

Solution 4

Question 5

The area of trapezium ABCD in the given figure is

1. 62 cm²
2. 93 cm²
3. 124 cm²
4. 155 cm²

Solution 5

Question 6

In the given figure, ABCD is a ∥gm in which AB = CD = 5 cm and BD ⊥ DC such that BD = 6.8 cm, Then the area of ‖gm ABCD = ?

1. 17 cm²
2. 25 cm²
3. 34 cm²
4. 68 cm²

Solution 6

Question 7

In the given figure, ABCD is a ∥gm in which diagonals Ac and BD intersect at O. If ar(‖gm ABCD) is 52 cm², then the ar(ΔOAB)=?

1. 26 cm²
2. 18.5 cm²
3. 39 cm²
4. 13 cm²

Solution 7

Question 8

In the given figure, ABCD is a ∥gm in which DL ⊥ AB, If AB = 10 cm and DL = 4 cm, then the ar(‖gm ABCD) = ?

1. 40 cm²
2. 80 cm²
3. 20 cm²
4. 196 cm²

Solution 8

Question 9

The area of ∥gm ABCD is

(a) AB × BM

(b) BC × BN

(c) DC × DL

(d) AD × DL

Solution 9

Correct option: (c)

Area of ∥gm ABCD = Base × Height = DC × DL Question 10

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 1 : 1

(c) 2 : 1

(d) 3 : 1Solution 10

Correct option: (b)

Parallelograms on equal bases and between the same parallels are equal in area. Question 11

In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.

(a) true

(b) false

Solution 11

Correct option: (a)

ΔBMP and parallelogram ABPQ are on the same base BP and between the same parallels AQ and BP.

Parallelograms ABPQ and ABCD are on the same base AB and between the same parallels AB and PD.

Question 12

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to

(a) 

(b) 

(c) 

(d) ar(ΔABC)

Solution 12

Correct option: (a)

ΔABC is divided into four triangles of equal area.

A(parallelogram AFDE) = A(ΔAFE) + A(DFE)

Question 13

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is

1. 192 cm²
2. 96 cm²
3. 64 cm²
4. 80 cm²

Solution 13

Question 14

Two parallel sides of a trapezium are 12 cm and 8 cm long and the distance between them is 6.5 cm. The area of the trapezium is

1. 74 cm²
2. 32.5 cm²
3. 65 cm²
4. 130 cm²

Solution 14

Question 15

In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD)= ?

1. 24 cm²
2. 40 cm²
3. 55 cm²
4. 27.5 cm²

Solution 15

Question 16

In a quadrilateral ABCD, it is given that BD = 16 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?

1. 256 cm²
2. 128 cm²
3. 64 cm²
4. 96 cm²

Solution 16

Question 17

ABCD is a rhombus in which ∠C = 60°.

Then, AC : BD = ?

Solution 17

Question 18

In the given figure ABCD and ABFE are parallelograms such that ar(quad. EABC) = 17cm² and ar(‖gm ABCD) = 25 cm². Then, ar(ΔBCF) = ?

1. 4 cm²
2. 4.8 cm²
3. 6 cm²
4. 8 cm²

Solution 18

Question 19

ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC. Then, ar(ΔBDE) : ar(ΔABC) = ?

(a) 1 : 2

(b) 1 : 4

(c) 

(d) 3 : 4Solution 19

Question 20

In a ‖gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(‖gm ABCD) = 16 cm², then ar(‖gm APQD) = ?

1. 8 cm²
2. 12 cm²
3. 6 cm²
4. 9 cm²

Solution 20

Question 21

The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a

1. rectangle of area 24 cm²
2. square of area 24 cm²
3. trapezium of area 24 cm²
4. rhombus of area 24 cm²

Solution 21

Question 22

In ΔABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(ΔBED) = ?

Solution 22

Question 23

The vertex A of ΔABC is joined to a point D on BC. If E is the midpoint of AD, then ar(ΔBEC) = ?

Solution 23

Question 24

In ΔABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(ΔBOE) = ?

Solution 24

Question 25

If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is

1. 1 : 2
2. 1 : 3
3. 1 : 4
4. 3 : 4

Solution 25

Question 26

In the given figure ABCD is a trapezium in which AB ‖ DC such that AB = a cm and DC = b cm, If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?

1. a : b
2. (a + 3b) : (3a + b)
3. (3a + b) : (a + 3b)
4. (2a + b) : (3a + b)

Solution 26

Question 27

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD is

1. a rectangle
2. a ‖gm
3. a rhombus
4. all of these

Solution 27

Question 28

In the given figure, a ‖gm ABCD and a rectangle ABEF are of equal area, Then,

1. perimeter of ABCD = perimeter of ABEF
2. perimeter of ABCD < perimeter of ABEF
3. perimeter of ABCD > perimeter of ABEF
4. 

Solution 28

Question 29

In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If AD = cm, then area of the rectangle is

1. 32 cm²
2. 40 cm²
3. 44 cm²
4. 48 cm²

Solution 29

Question 30

Which of the following is a false statement?

1. A median of a triangle divides it into two triangles of equal areas.
2. The diagonals of a ∥gm divide it into four triangles of equal areas.
3. In a ΔABC, if E is the midpoint of median AD, then ar(ΔBED) =

4. In a trap. ABCD, it is given that AB ‖ DC and the diagonals AC and BD intersect at O. Then, ar(ΔAOB) = ar(ΔCOD).

Solution 30

Question 31

Which of the following is a false statement?

1. If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm².
2. 
3. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
4. If the area of a ‖gm with one side 24 cm and corresponding height h cm is 192 cm², then h = 8 cm.

Solution 31

Question 32

Look at the statements given below:

1. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
2. In a ‖gm ABCD, it is given that AB = 10 cm. The altitudes DE on AB and BF on AD being 6 cm and 8 cm respectively, then AD = 7.5 cm.

3. 

Which is true?

1. I only
2. II only
3. I and II
4. II and III

Solution 32

Question 33

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In a trapezium ABCD we have AB ‖ DC and the diagonals AC and BD intersect at O.Then, ar(ΔAOD) = ar(ΔBOC)	Triangles on the same base and between the same parallels are equal in areas.

The correct answer is: (a) / (b) / (c) / (d).Solution 33

Question 34

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If ABCD is a rhombus whose one angle is 60°, then the ratio of the lengths of its diagonals is .	Median of a triangle divides it into two triangles of equal area.

The correct answer is: (a) / (b) / (c) / (d).Solution 34

Question 35

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The diagonals of a ‖gm divide it into four triangles of equal area.	A diagonal of a ‖gm divides it into two triangles of equal area.

The correct answer is: (a) / (b) / (c) / (d).Solution 35

Question 36

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm2.

The correct answer is: (a) / (b) / (c) / (d).Solution 36

Question 37

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In the given figure, ABCD is a ‖gm in which DE ⊥ AB and BE ⊥ AD. If AB = 16 cm, DE = 8cm and BF = 10cm, then AD is 12 cm.	Area of a ‖gm = base × height.

The correct answer is: (a) / (b) / (c) / (d).Solution 37

Exercise Ex. 11A

Question 2

In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of ||gm ABCD.

Solution 2

Question 3

In a parallelogram ABCD, it is being given that AB= 10 cm and the altitudes corresponding to the sides AB and AD are DL =6 cm and BM =8cm, respectively Find AD.

Solution 3

Question 5

Find the area of the trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.Solution 5

Question 6(ii)

Calculatethe area of trapezium PQRS , givenin Fig.(ii)

Solution 6(ii)

Question 6(i)

Calculate the area of quadrilateral ABCD givenin Fig (i)

Solution 6(i)

Question 8

BD is one of the diagonals of a quad. ABCD . If , show that 

Solution 8

Question 10

In the adjoining figure , ABCD is a quadrilateral in which diag. BD =14cm . If  such that AL=8 cm. and CM =6 cm, find the area of quadrilateral ABCD.

Solution 10

Question 13

In the adjoining figure , ABCD is a trapezium in which AB||DC and its diagonals AC and BD intersects at O. prove that 

Solution 13

Question 14

_{In the adjoining figure , DE||BC. Prove that} 

Solution 14

Question 15

Prove that the median divides a triangle into two triangles of equal area.Solution 15

Question 16

Show that the diagonal divides a parallelogram into two triangles of equal area.Solution 16

Question 20

_{In adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O. If BO=OD, prove that}

Solution 20

Question 21

The vertex A of  _{is joined to a point D on the side BC. The mid-point of AD is E. prove that}

Solution 21

Question 22

_{D is the mid-point of side BC of   and E is the midpoint of BD. If O is midpoint of AE, prove that} 

Solution 22

Question 24

_{In the adjoining figure , ABCD is a quadrilateral. A line through D , parallel to AC , meets BC produced in P. prove that} 

Solution 24

Question 25

In the adjoining figure ,  _{are on the same base BC with A and D on opposite sides of BC such that . Show that BC bisects AD.}

Solution 25

Question 28

_{P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of ||gm ABCD. Show that PQRS is a parallelogram and also that}

_{Ar(||gm PQRS)=x ar (||gm ABCD)}

Solution 28

Question 30

The base BC of   _{is divided at D such that .}

_{Prove that .}Solution 30

Question 32

_{The given figure shows the pentagon ABCDE. EG, drawn parallel to DA, meetsBA producedat G ,andCF, drawnparallel to DB, meets AB produced at F.}

_Showthat 

Solution 32

Question 34

_{In the adjoining figure , the point D divides the side BC of   in the ratio m:n. Prove that} 

Solution 34

Exercise Ex. 11

Question 1

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Fig (i)

Fig (ii)

Fig (iii)

Fig (iv)

Fig (v)

Fig (vi)

Solution 1

Following figures lie on the same base and between the same parallels:

Figure (i): No

Figure (ii): No

Figure (iii): Yes, common base – AB, parallel lines – AB and DE

Figure (iv): No

Figure (v): Yes, common base – BC, parallel lines – BC and AD

Figure (vi): Yes, common base – CD, parallel lines – CD and BPQuestion 4

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.Solution 4

Question 7

In the adjoining figure, ABCD is a trapezium in which AB ∥ DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

Solution 7

Question 9

M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm², find ar(Δ ABC).Solution 9

Construction: Join AC

Diagonal AC divides the parallelogram ABCD into two triangles of equal area.

⇒ A(ΔADC) = A(ΔABC) ….(i)

ΔADC and parallelogram ABCD are on the same base CD and between the same parallel lines DC and AM.

Since M is the mid-point of AB,

A(AMCD) = A(ΔADC) + A(ΔAMC)

Question 11

If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(ΔAPB) = ar(ΔBQC).Solution 11

Since ΔAPB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC, we have

Similarly, ΔBQC and parallelogram ABCD are on the same base BC and between the same parallels BC and AD, we have 

From (i) and (ii),

A(ΔAPB) = A(ΔBQC) Question 12

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that

Solution 12

(i) Parallelograms MNPQ and ABPQ are on the same base PQ and between the same parallels PQ and MB.

(ii) ΔATQ and parallelogram ABPQ are on the same base AQ and between the same parallels AQ and BP. 

Question 17

In the adjoining figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that

ar(Δ ABC) = ar(Δ ABD).

Solution 17

We know that median of a triangle divides it into two triangles of equal area.

Now, AO is the median of ΔACD.

⇒ A(ΔCOA) = A(ΔDOA) ….(i)

And, BO is the median of ΔBCD.

⇒ A(ΔCOB) = A(ΔDOB) ….(ii)

Adding (i) and (ii), we get

A(ΔCOA) + A(ΔCOB) = A(ΔDOA) + A(ΔDOB)

⇒ A(ΔABC) = A(ΔABD)Question 18

D and E are points on sides AB and AC respectively of Δ ABC such that ar(ΔBCD) = ar(ΔBCE). Prove that DE ∥ BC.Solution 18

Since ΔBCD and ΔBCE are equal in area and have a same base BC.

Therefore,

Altitude from D of ΔBCD = Altitude from E of ΔBCE

⇒ ΔBCD and ΔBCE are between the same parallel lines.

⇒ DE ∥ BC Question 19

P is any point on the diagonal AC of parallelogram ABCD. Prove that ar(ΔADP) = ar(ΔABP).Solution 19

Construction: Join BD.

Let the diagonals AC and BD intersect at point O.

Diagonals of a parallelogram bisect each other.

Hence, O is the mid-point of both AC and BD.

We know that the median of a triangle divides it into two triangles of equal area.

In ΔABD, OA is the median.

⇒ A(ΔAOD) = A(ΔAOB) ….(i)

In ΔBPD, OP is the median.

⇒ A(ΔOPD) = A(ΔOPB) ….(ii)

Adding (i) and (ii), we get

A(ΔAOD) + A(ΔOPD) = A(ΔAOB) + A(ΔOPB)

⇒ A(ΔADP) = A(ΔABP)Question 23

In a trapezium ABCD, AB ∥ DC and M is the midpoint of BC. Through M, a line PQ ∥ AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD).

Solution 23

In ΔMCQ and ΔMPB,

∠QCM = ∠PBM (alternate angles)

CM = BM (M is the mid-point of BC) 

∠CMQ = ∠PMB (vertically opposite angles)

∴ ΔMCQ ≅ ΔMPB

⇒ A(ΔMCQ) = A(ΔMPB)

Now,

A(ABCD) = A(APQD) + A(DMPB) – A(ΔMCQ)

⇒ A(ABCD) = A(APQD)Question 26

ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure. AP intersect CD at M. If ar(DMB) = 7 cm², find the area of parallelogram ABCD.

Solution 26

In ΔADM and ΔPCM,

∠ADM = ∠PCM (alternate angles)

AD = CP (AD = BC = CP) 

∠AMD = ∠PMC (vertically opposite angles)

∴ ΔADM ≅ ΔPCM

⇒ A(ΔADM) = A(ΔPCM)

And, DM = CM (c.p.c.t.)

⇒ BM is the median of ΔBDC.

⇒ A(ΔDMB) = A(ΔCMB) 

⇒ A(ΔBDC) = 2 × A(ΔDMB) = 2 × 7 = 14 cm²

Now,

A(parallelogram ABCD) = 2 × A(ΔBDC) = 2 × 14 = 28 cm² Question 27

In a parallelogram ABCD, any point E is taken on the side BC. AE and DC when produced meet at a point M. Prove that

ar(ΔADM) – ar(ABMC)Solution 27

Construction: Join AC and BM

Let h be the distance between AB and CD.

Question 29

In a triangle ABC, the medians BE and CF intersect at G. Prove that ar(ΔBCG) = ar(AFGE).Solution 29

Construction: Join EF

Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side,

FE ∥  BC

Clearly, ΔBEF and ΔCEF are on the same base EF and between the same parallel lines.

∴ A(ΔBEF) = A(ΔCEF)

⇒ A(ΔBEF) – A(ΔGEF) = A(ΔCEF) – A(ΔGEF)

⇒ A(ΔBFG) = A(ΔCEG) …(i)

We know that a median of a triangle divides it into two triangles of equal area.

⇒ A(ΔBEC) = A(ΔABE)

⇒ A(ΔBGC) + A(ΔCEG) = A(quad. AFGE) + A(ΔBFG)

⇒ A(ΔBGC) + A(ΔBFG) = A(quad. AFGE) + A(ΔBFG) [Using (i)]

⇒ A(A(ΔBGC) = A(quad. AFGE)Question 31

Solution 31

ΔDBC and ΔEBC are on the same base and between the same parallels.

⇒ A(ΔDBC) = A(ΔEBC) ….(i)

BE is the median of ΔABC.

Question 33

In the adjoining figure, CE ∥ AD and CF ∥ BA. Prove that ar(ΔCBG) = ar(ΔAFG).

Solution 33

ΔBCF and ΔACF are on the same base CF and between the same parallel lines CF and BA.

∴ A(ΔBCF) = A(ΔACF)

⇒ A(ΔBCF) – A(ΔCGF) = A(ΔACF) – A(ΔCGF)

⇒ A(ΔCBG) = A(ΔAFG)Question 35

In a trapezium ABCD, AB ∥ DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel sides, AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).

Solution 35

Construction: Join DB. Let DB cut MN at point Y.

M and N are the mid-points of AD and BC respectively.

⇒ MN ∥ AB ∥ CD

In ΔADB, M is the mid-point of AD and MY ∥ AB.

∴ Y is the mid-point of DB.

Similarly, in ΔBDC,

Now, MN = MY + YN

Construction: Draw DQ ⊥ AB. Let DQ cut MN at point P.

Then, P is the mid-point of DQ.

i.e. DP = PQ = h (say)

Question 36

ABCD is a trapezium in which AB ∥ DC, AB = 16 cm and DC = 24 cm. If E and F are respectively the midpoints of AD and BC, prove that Solution 36

Construction: Join AC. Let AC cut EF at point Y.

E and F are the mid-points of AD and BC respectively.

⇒ EF ∥ AB ∥ CD

In ΔADC, E is the mid-point of AD and EY ∥ CD.

∴ Y is the mid-point of AC.

Similarly, in ΔABC,

Now, EF = EY + YF

Construction: Draw AQ ⊥ DC. Let AQ cut EF at point P.

Then, P is the mid-point of AQ.

i.e. AP = PQ = h (say)

Question 37

In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of ΔABC. If PQ ∥ BC and CDP and BEQ are straight lines then prove that ar(ΔABQ) = ar(ΔACP).

Solution 37

Since D and E are the mid-points of AB and AC respectively,

DE ∥ BC ∥ PQ

In ΔACP, AP ∥ DE and E is the mid-point of AC.

⇒ D is the mid-point of PC (converse of mid-point theorem)

In ΔABQ, AQ ∥ DE and D is the mid-point of AB.

⇒ E is the mid-point of BQ (converse of mid-point theorem)

From (i) and (ii),

AP = AQ

Now, ΔACP and ΔABQ are on the equal bases AP and AQ and between the same parallels BC and PQ.

⇒ A(ΔACP) = A(ΔABQ)Question 38

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(ΔRSC) = ar(ΔPQB).

Solution 38

In ΔRSC and ΔPQB,

∠CRS = ∠BPQ (RC ∥ PB, corresponding angles)

∠RSC = ∠PQB (RC ∥ PB, corresponding angles)

SC = QB (opposite sides of a parallelogram BQSC)

∴ ΔRSC ≅ ΔPQB (by AAS congruence criterion)

⇒ A(ΔRSC) = A(ΔPQB)

Exercise MCQ

Question 1

The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is

1. 11.5 cm
2. 12 cm
3. 
4. 23 cm

Solution 1

Correct option: (b)

Question 2

A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is

1. 25 cm
2. 12.5 cm
3. 30 cm
4. 9 cm

Solution 2

Correct option: (c)

Question 3

In the given figure, BOC is a diameter of a circle and AB = AC. Then ∠ABC =?

1. 30°
2. 45°
3. 60°
4. 90°

Solution 3

Question 4

In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB =?

1. 30°
2. 15°
3. 60°
4. 90°

Solution 4

Question 5

In the given figure, O is the centre of a circle. If ∠OAB = 40° and C is a point on the circle, then ∠ACB =?

1. 40°
2. 50°
3. 80°
4. 100°

Solution 5

Question 6

In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then, the distance of CD from AB is

1. 8 cm
2. 15 cm
3. 18 cm
4. 6 cm

Solution 6

Question 7

AB and CD are two equal chords of a circle with centre O such that ∠AOB = 80°, then ∠COD =?

1. 100°
2. 80°
3. 120°
4. 40°

Solution 7

Question 8

In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, then radius of the circle is

1. 6 cm
2. 9 cm
3. 7.5 cm
4. 8 cm

Solution 8

Question 9

In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is

1. 10 cm
2. 12 cm
3. 6 cm
4. 8 cm

Solution 9

Question 10

In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB ‖ CD. If AB = 10 cm, then CD =?

1. 5 cm
2. 12.5 cm
3. 15 cm
4. 10 cm

Solution 10

Question 11

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠ACD = 25°, then ∠AOD =?

1. 50°
2. 75°
3. 90°
4. 100°

Solution 11

Question 12

In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If OD ⟘ AB such that OD = 6 cm, then AC =?

1. 9 cm
2. 12 cm
3. 15 cm
4. 7.5 cm

Solution 12

Question 13

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is

1. 3 cm
2. 
3. 
4. 6 cm

Solution 13

Question 14

 The angle in a semicircle measures

1. 45°
2. 60°
3. 90°
4. 36°

Solution 14

Question 15

Angles in the same segment of a circle area are

1. equal
2. complementary
3. supplementary
4. none of these

Solution 15

Question 16

In the given figures, ⧍ABC and ⧍DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°. Then, ∠BCD =?

1. 50°
2. 60°
3. 70°
4. 80°

Solution 16

Question 17

In the given figure, BOC is a diameter of a circle with centre O. If ∠BCA = 30°, then ∠CDA =?

1. 30°
2. 45°
3. 60°
4. 50°

Solution 17

Question 18

In the given figure, O is the centre of a circle. If ∠OAC = 50°, then ∠ODB =?

1. 40°
2. 50°
3. 60°
4. 75°

Solution 18

Question 19

In the given figure, O is the centre of a circle in which ∠OBA = 20° and ∠OCA = 30°. Then, ∠BOC =?

1. 50°
2. 90°
3. 100°
4. 130°

Solution 19

Question 20

In the given figure, O is the centre of a circle. If ∠AOB = 100° and ∠AOC = 90°, then ∠BAC =?

1. 85°
2. 80°
3. 95°
4. 75°

Solution 20

Question 21

In the given figure, O is the centre of a circle. Then, ∠OAB =?

1. 50°
2. 60°
3. 55°
4. 65°

Solution 21

Question 22

In the given figure, O is the centre of a circle and ∠AOC = 120°. Then, ∠BDC =?

1. 60°
2. 45°
3. 30°
4. 15°

Solution 22

Question 23

In the given figure, O is the centre of a circle and ∠OAB = 50°. Then, ∠CDA =?

1. 40°
2. 50°
3. 75°
4. 25°

Solution 23

Question 24

In the given figure, AB and CD are two intersecting chords of a circle. If ∠CAB = 40° and ∠BCD = 80°, then ∠CBD =?

1. 80°
2. 60°
3. 50°
4. 70°

Solution 24

Question 25

In the given figures, O is the centre of a circle and chords AC and BD intersect at E. If ∠AEB = 110° and ∠CBE = 30°, then ∠ADB =?

1. 70°
2. 60°
3. 80°
4. 90°

Solution 25

Question 26

In the given figure, O is the centre of a circle in which ∠OAB =20° and ∠OCB = 50°. Then, ∠AOC =?

1. 50°
2. 70°
3. 20°
4. 60°

Solution 26

Question 27

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If ∠ADC = 120°, then ∠BAC =?

1. 60°
2. 30°
3. 20°
4. 45°

Solution 27

Question 28

In the given figure ABCD is a cyclic quadrilateral in which AB ‖ DC and ∠BAD = 100°. Then ∠ABC =?

1. 80°
2. 100°
3. 50°
4. 40°

Solution 28

Question 29

In the given figure, O is the centre of a circle and ∠AOC =130°. Then, ∠ABC =?

1. 50°
2. 65°
3. 115°
4. 130°

Solution 29

Question 30

In the given figure, AOB is a diameter of a circle and CD AB. If ∠BAD = 30°, then ∠CAD =?

1. 30°
2. 60°
3. 45°
4. 50°

Solution 30

Question 31

In the given figure, O is the centre of a circle in which ∠AOC =100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD =?

1. 50°
2. 40°
3. 25°
4. 80°

Solution 31

Question 32

In the given figure, O is the centre of a circle and ∠OAB = 50°. Then, ∠BOD=?

1. 130°
2. 50°
3. 100°
4. 80°

Solution 32

Question 33

In the given figures, ABCD is a cyclic quadrilateral in which BC = CD and ∠CBD = 35°. Then, ∠BAD =?

1. 65°
2. 70°
3. 110°
4. 90°

Solution 33

Question 34

In the given figure, equilateral ⧍ ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then ∠BDC =?

1. 90°
2. 60°
3. 120°
4. 150°

Solution 34

Question 35

In the give figure, side Ab and AD of quad. ABCD are produced to E and F respectively. If ∠CBE = 100°, then ∠CDF =?

1. 100°
2. 80°
3. 130°
4. 90°

Solution 35

Question 36

In the given figure, O is the centre of a circle and ∠AOB = 140°. The, ∠ACB =?

1. 70°
2. 80°
3. 110°
4. 40°

Solution 36

Question 37

In the given figure, O is the centre of a circle and ∠AOB = 130°. Then, ∠ACB =?

1. 50°
2. 65°
3. 115°
4. 155°

Solution 37

Question 38

In the given figure, ABCD and ABEF are two cyclic quadrilaterals. If ∠BCD = 110°, then ∠BEF =?

1. 55°
2. 70°
3. 90°
4. 110°

Solution 38

Question 39

 In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ∠ADC = 95° and ∠ECF =20°. Then, ∠BAD =?

1. 95°
2. 85°
3. 105°
4. 75°

Solution 39

Question 40

Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD =?

1. 10.5 cm
2. 9.5 cm
3. 8.5 cm
4. 7.5 cm

Solution 40

Question 41

In the given figure, A and B are the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at points P and Q respectively. If AB = 4 cm, then the length of common chord PQ is

1. 3 cm
2. 6 cm
3. 7.5 cm
4. 9 cm

Solution 41

Question 42

In the given figure, ∠AOB = 90° and ∠ABC = 30°. Then ∠CAO =?

1. 30°
2. 45°
3. 60°
4. 90°

Solution 42

Ex. 12C

Question 1

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that  _.

_Find 

Solution 1

Question 2

_{In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If} 

Solution 2

Question 3

_{In the given figure , O is the centre of the circle and arc ABC subtends an angle of at the centre . If AB is extended to P, find .}

Solution 3

Question 4

_{In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced . If}  

Solution 4

Question 5

_{In the given figure, BD=DC and} 

Solution 5

Question 6

_{In the given figure, O is the centre of the given circle and measure of arc ABC is  Determine .}

Solution 6

Question 7

_{In the given figure,  is equilateral. Find} 

Solution 7

Question 8

_{In the adjoining figure, ABCD is a cyclic quadrilateral in which .}

Solution 8

Question 9

_{In the given figure , O is the centre of a circle and Find the values of x and y.}

Solution 9

Question 10

_{In the given figure, O is the centre of the circle and . Calculate the vales of x and y.}

Solution 10

Question 11

_{In the given figure , sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If ,  find the value of x.}

Solution 11

Question 12

_{In the given figure, AB is a diameter of a circle with centre O and DO||CB. If  , calculate}

_{Also , show that  is an equilateral triangle.}

Solution 12

Question 13Two chord AB and CD of a circle intersects each other at P outside the circle. If AB=6cm, BP=2cm and PD=2.5 cm, Find CD.

Solution 13

Question 14

_{In the given figure , O is the centre of a circle. If  , calculate}

Solution 14

Question 15

_{In the given figure,  is an isosceles triangle in which AB=AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.}

Solution 15

Question 16

_{In the given figure, AB and CD are two parallel chords of a circle . If BDE and ACE are straight lines , intersecting at E, prove that  is isosceles.}

Solution 16

Question 17

_{In the given figure,  Find the values of x and y.}

Solution 17

Question 18

_{In the given figure , ABCD is a quadrilateral in which AD=BC and .  Show that the pints A, B, C, D lie on a circle.}

Solution 18

Question 19

_{Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.}Solution 19

Question 20

_{Prove that the circles described with the four sides of a rhombus . as diameter , pass through the point of intersection of its diagonals.}Solution 20

Question 21

_{ABCD is a rectangle . Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.}Solution 21

Question 22

_{Give a geometrical construction for finding the fourth point lying on a circle passing through three given points , without finding the centre of the circle. Justify the construction.}Solution 22

Question 23

_{In a cyclic quadrilateral ABCD, if , show that the smaller of the two is} 
Solution 23

Question 24

_{The diagonal s of a cyclic quadrilateral are at right angles . Prove that perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.}Solution 24

Question 25

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.Solution 25

AB is the common hypotenuse of ΔACB and ΔADB.

⇒ ∠ACB = 90° and ∠BDC = 90° 

⇒ ∠ACB + ∠BDC = 180° 

⇒ The opposite angles of quadrilateral ACBD are supplementary.

Thus, ACBD is a cyclic quadrilateral.

This means that a circle passes through the points A, C, B and D.

⇒ ∠BAC = ∠BDC (angles in the same segment) Question 26

ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that Solution 26

Construction: Take a point E on the circle. Join BE, DE and BD.

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

⇒ ∠BAD = 2∠BED

Now, EBCD is a cyclic quadrilateral.

⇒ ∠BED + ∠BCD = 180° 

⇒ ∠BCD = 180° – ∠BED

In ΔBCD, by angle sum property

∠CBD + ∠CDB + ∠BCD = 180° 

Ex. 12A

Question 1

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of a chord from the centre of the circle.Solution 1

Question 2

Find the length of the chord which is at the distance of 3 cm from the centre of a circle of radius 5 cm.Solution 2

Question 3

A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find the radius of the circle.Solution 3

Question 4

In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6cm respectively. Calculate the distance between the chords if they are

(i) On the same side of the centre

(ii) On the opposite side of the centre.Solution 4

Question 5

Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of a radius 17 cm. Find the distance between the chords.Solution 5

Question 6

In the given figure , the diameter CD of a circle with centre O is perpendicular to chord AB. If AB=12 cm and CE=3cm, calculate the radius of the circle.

Solution 6

Question 7

In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE=ED=8 cm and EB=4 cm. Find the radius of the circle.

Solution 7

Question 8

In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC||DO and AC=2xOD.

Solution 8

Question 9

In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects _{. Prove that AB=CD.}

Solution 9

Question 10

_{Prove that the diameter of the circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.}Solution 10

Question 11

_{Prove that two different circles cannot intersects other at more than two points.}Solution 11

Question 12

_{Two circle of the radii 10 cm and 8 cm intersects each other , and the length of the common chord is 12 cm. Find the distance between their centres.}

Solution 12

Question 13

_{Two equal circle intersects in P and Q. A straight line through P meets the circles in A and B. Prove that QA= QB.}

Solution 13

Question 14

_{If a diameter of the circle bisects each of the two chords of a circle then prove that the chords are parallel.}Solution 14

Question 15

_{In the adjoining figure , two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q . Find the length of PQ.}

Solution 15

Question 16

_{In the given figure , AB is a chord of a circle with centre O and AB is produced to C such that BC=OB. Also CO is a joined and produced to meet the circle in D. If , prove that x=3y.}

Solution 16

Question 17

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre then prove that 4q² = p² + 3r².Solution 17

Let O be the centre of a circle with radius r.

⇒ OB = OC = r

Let AC = x

Then, AB = 2x

Let OM ⊥ AB

⇒ OM = p 

Let ON ⊥ AC

⇒ ON = q

In ΔOMB, by Pythagoras theorem,

OB² = OM² + BM²

In ΔONC, by Pythagoras theorem,

OC² = ON² + CN²

Question 18

_{In the adjoining figure , O is the centre of a circle . If AB and AC are chordsof a circlesuch thatAB=AC, , prove that PB=QC.}

Solution 18

Question 19

_{In the adjoining figure, BC is a diameter of a circle with circle with centre O. If AB and CD are two chords such that AB|| CD, prove that AB=CD.}

Solution 19

Question 20

_{An equilateral triangle of side 9 cm is inscribed in a circle . Find the radius of the circle.}Solution 20

Question 21

Solution 21

Question 22

_{In the adjoining figure , OPQR is a square. A circle drawn with centre O cuts the square in X and Y. prove that QX =QY.}

Solution 22

Question 23

Two circle with centres O and O’ intersect at two points A and B. A line PQ is drawn parallel to OO’ through A or B, intersecting the circles at P and Q. Prove that PQ = 2OO’.Solution 23

Draw OM ⊥ PQ and O’N ⊥ PQ

⇒ OM ⊥ AP

⇒ AM = PM (perpendicular from the centre of a circle bisects the chord)

⇒ AP = 2AM ….(i) 

And, O’N ⊥ PQ

⇒ O’N ⊥ AQ

⇒ AN = QN (perpendicular from the centre of a circle bisects the chord)

⇒ AQ = 2AN ….(ii)

Now,

PQ = AP + PQ

⇒ PQ = 2AM + 2AN

⇒ PQ = 2(AM + AN)

⇒ PQ = 2MN

⇒ PQ = 2OO’  (since MNO’O is a rectangle)

Ex. 12B

Question 1

(i) In figure (1) , O is the centreof the circle . If _{(ii) In figure(2), A, B and C are three points on the circlewithcentre O such that  .}

Solution 1

Question 2

_{In the given figure, O is the centre of the circle and  .}

_{Calculate the value of   .}

Solution 2

Question 3

_{In the given figure , O is the centre of the circle .If   , find the value of} 

Solution 3

Question 4

_{In the given figure , O is the centre of the circle. If} 

Solution 4

Question 5

_{In the given figure, O is the centre of the circle .If   find .}

Solution 5

Question 6

In the given figure , _{, calculate} 

Solution 6

Question 7

In the adjoining figure , DE is a chord parallel to diameter AC of the circle with centre O. If _{, calculate .}

Solution 7

Question 8

_{In the adjoining figure, O is the centre of the circle. Chord CD is parallel to diameter AB. If   calculate} 

Solution 8

Question 9

_{In the given figure, AB and CD are straight lines through the centre O of a circle. If  , find} 

Solution 9

Question 10

_{In the given figure , O is the centre of a circle,  , find .}

Solution 10

Question 11

_{In the adjoining figure , chords AC and BD of a circle with centre O, intersect at right angles at E. if  , calculate .}

Solution 11

Question 12

_{In the given figure , O is the centre of a circle in which .  Find}  

Solution 12

Question 13

In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y.

Solution 13

Given, ∠AOD = 90° and ∠OEC = 90° 

⇒ ∠AOD = ∠OEC

But ∠AOD and ∠OEC are corresponding angles.

⇒ OD || BC and OC is the transversal. 

∴ ∠DOC = ∠OCE (alternate angles)

⇒ ∠DOC = 30° (since ∠OCE = 30°)

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

⇒ ∠DOC = 2∠DBC

Now, ∠ABE = ∠ABC = ∠ABD + ∠DBC = 45° + 15° = 60° 

In ΔABE,

∠BAE + ∠AEB + ∠ABE = 180° 

⇒ x + 90° + 60° = 180° 

⇒ x + 150° = 180° 

⇒ x = 30° Question 14

In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB

Solution 14

Construction: Join AC

Given, BD = OD

Now, OD = OB (radii of same circle)

⇒ BD = OD = OB

⇒ ΔODB is an equilateral triangle.

⇒ ∠ODB = 60° 

We know that the altitude of an equilateral triangle bisects the vertical angle.

Now, ∠CAB = ∠BDC (angles in the same segment)

⇒ ∠CAB = ∠BDE = 30° Question 15

_{In the given figure, PQ is a diameter of a circle with centre O. If} 

Solution 15

Question 16

In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line.

Solution 16

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

⇒ ∠APB = 2∠ACB

Now, ACD is a straight line.

⇒ ∠ACB + ∠DCB = 180° 

⇒ 75° + ∠DCB = 180° 

⇒ ∠DCB = 105° 

Again,

Question 17

_{In the given figure , . Show that BC is equal to the radius of the circumcircle of  whose centre is O.}

Solution 17

Question 18

In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E. Prove that

Solution 18

Join AC and BC

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

⇒ ∠AOC = 2∠ABC ….(i)

Similarly, ∠BOD = 2∠BCD ….(ii)

Adding (i) and (ii),

∠AOC + ∠BOD = 2∠ABC + 2∠BCD

⇒ ∠AOC + ∠BOD = 2(∠ABC + ∠BCD)

⇒ ∠AOC + ∠BOD = 2(∠EBC + ∠BCE)

⇒ ∠AOC + ∠BOD = 2(180° – ∠CEB)

⇒ ∠AOC + ∠BOD = 2(180° – [180° – ∠AEC])

⇒ ∠AOC + ∠BOD = 2∠AEC

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