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Exercise 8.1

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer

Let x be the common ratio between the angles.
Sum of the interior angles of the quadrilateral = 360°
Now,
3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°
Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer

Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.
∠A = ∠B (by CPCT)
also,
∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Thus ABCD is a rectangle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

 Answer

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

Answer

Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show,
 AC = BD, AO = OC and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
∠ABC = ∠BAD = 90°
AC = AD (Given)
Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
also, ∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer

Given,
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.
To prove,
Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.
Thus, AB = CD by CPCT. — (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB || CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.
Thus, AD = CD by CPCT. — (ii)
also,
AD = BC and AD = CD
⇒ AD = BC = CD = AB — (ii)
also,  ∠ADC = ∠BCD  by CPCT.
and ∠ADC + ∠BCD = 180° (co-interior angles)
⇒ 2∠ADC = 180°
⇒ ∠ADC = 90° — (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Answer

(i)
In ΔADC and ΔCBA,
AD = CB (Opposite sides of a ||gm)
DC = BA (Opposite sides of a ||gm)
AC = CA (Common)
Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.
Thus,
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
Thus, AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved)
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a ||gm)
Thus, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer

Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
Similarly, we can prove that diagonal AC bisects ∠A.

Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.

8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.

Answer

(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.

(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
∠CBD = ∠ADB
⇒ ∠CDB = ∠ADB
Thus, BD bisects ∠D

Page No: 147

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram



Answer

(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||gm)
Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a ||gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii)  and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ

Answer

(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.

Answer

12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD

Answer

Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
⇒ ∠D = ∠C

(iii) In ΔABC and ΔBAD,
AB = AB (Common)
∠DBA = ∠CBA
AD = BC (Given)
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

Answer

Given,
AC = AD and AB bisects ∠A
To prove,
ΔABC ≅ ΔABD
Proof,
In ΔABC and ΔABD,
AB = AB (Common)
AC = AD (Given)
∠CAB = ∠DAB (AB is bisector)
Therefore, ΔABC ≅ ΔABD by SAS congruence condition.
BC and BD are of equal length.

Page No: 119

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

Answer

Given,
AD = BC and ∠DAB = ∠CBA

(i) In ΔABD and ΔBAC,
AB = BA (Common)
∠DAB = ∠CBA (Given)
AD = BC (Given)
Therefore, ΔABD ≅ ΔBAC by SAS congruence condition.
(ii) Since, ΔABD ≅ ΔBAC
Therefore BD = AC by CPCT
(iii) Since, ΔABD ≅ ΔBAC
Therefore ∠ABD = ∠BAC by CPCT

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Answer

Given,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In ΔAOD and ΔBOC,
∠A = ∠B (Perpendicular)
∠AOD = ∠BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, ΔAOD ≅ ΔBOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

Answer

Given,
l || m and p || q
To prove,
ΔABC ≅ ΔCDA
Proof,
In ΔABC and ΔCDA,
∠BCA = ∠DAC (Alternate interior angles)
AC = CA (Common)
∠BAC = ∠DCA (Alternate interior angles)
Therefore, ΔABC ≅ ΔCDA by ASA congruence condition.

5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

 Answer

Given,
l is the bisector of an angle ∠A.
BP and BQ are perpendiculars.

(i) In ΔAPB and ΔAQB,
∠P = ∠Q (Right angles)
∠BAP = ∠BAQ (l is bisector)
AB = AB (Common)
Therefore, ΔAPB ≅ ΔAQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.

Page No: 120

6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Answer

Given,
AC = AE, AB = AD and ∠BAD = ∠EAC
To show,
BC = DE
Proof,
∠BAD = ∠EAC (Adding ∠DAC both sides)
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
In ΔABC and ΔADE,
AC = AE (Given)
∠BAC = ∠EAD
AB = AD (Given)
Therefore, ΔABC ≅ ΔADE by SAS congruence condition.
BC = DE by CPCT.

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

Answer

Given,
P is mid-point of AB.
∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
In ΔDAP ≅ ΔEBP,
∠DPA = ∠EPB
AP = BP (P is mid-point of AB)
∠BAD = ∠ABE (Given)
Therefore, ΔDAP ≅ ΔEBP by ASA congruence condition.
(ii) AD = BE by CPCT.

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB

Answer

Given,
∠C = 90°, M is the mid-point of AB and DM = CM

(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, ΔAMC ≅ ΔBMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒ 90° + ∠B = 180°
⇒ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC (byy CPCT, already proved)
Therefore, ΔDBC ≅ ΔACB by SAS congruence condition.

(iv)  DC = AB (ΔDBC ≅ ΔACB)
⇒ DM = CM = AM = BM (M is mid-point)
⇒ DM + CM = AM + BM
⇒ CM + CM = AB
⇒ CM = 1/2AB

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC                     (ii) AO bisects ∠A

Answer

Given,
AB = AC, the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,
∴ ∠B = ∠C
⇒ 1/2∠B = 1/2∠C
⇒ ∠OBC = ∠OCB (Angle bisectors.)
⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,
AB = AC (Given)
AO = AO (Common)
OB = OC (Proved above)
Therefore, ΔAOB ≅ ΔAOC by SSS congruence condition.
∠BAO = ∠CAO (by CPCT)
Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.

Answer

Given,
AD is the perpendicular bisector of BC
To show,
AB = AC
Proof,
In ΔADB and ΔADC,
AD = AD (Common)
∠ADB = ∠ADC
BD = CD (AD is the perpendicular bisector)
Therefore, ΔADB ≅ ΔADC by SAS congruence condition.
AB = AC (by CPCT)

Page No: 124

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Answer

Given,
BE and CF are altitudes.
AC = AB
To show,
BE = CF
Proof,
In ΔAEB and ΔAFC,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.
Thus, BE = CF by CPCT.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer

Given,
BE = CF

(i) In ΔABE and ΔACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
BE = CF (Given)
Therefore, ΔABE ≅ ΔACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

Answer

Given,
ABC and DBC are two isosceles triangles.
To show,
∠ABD = ∠ACD
Proof,
In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

Answer

Given,
AB = AC and AD = AB
To show,
∠BCD is a right angle.
Proof,
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
AD = AB
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
Similarly in ΔADC,
∠CAD = 180° – 2∠ACD — (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii)
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°

7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Answer

Given,
∠A = 90° and AB = AC
A/q,
AB = AC
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90°
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

Answer

Let ABC be an equilateral triangle.
BC = AC = AB (Length of all sides is same)
⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)
Also,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
Therefore, ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

Exercise 7.4

1. Show that in a right angled triangle, the hypotenuse is the longest side.

Answer

ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° and ∠B is 90°.
Since, B is the largest angle of the triangle, the side opposite to it must be the largest.
So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.

2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Answer

Given,
∠PBC < ∠QCB
Now,
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° – ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° – ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.

3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Answer

Given,
∠B < ∠A and ∠C < ∠D
Now,
AO <  BO — (i) (Side opposite to the smaller angle is smaller)
OD < OC —(ii) (Side opposite to the smaller angle is smaller)
Adding (i) and (ii)
AO + OD < BO + OC
⇒ AD < BC

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
Show that ∠A > ∠C and ∠B > ∠D.

Answer


In ΔABD,
AB < AD < BD
∴ ∠ADB < ∠ABD — (i) (Angle opposite to longer side is larger.)
Now,
In ΔBCD,
BC < DC < BD
∴ ∠BDC < ∠CBD — (ii)
Adding (i) and (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
⇒ ∠ADC < ∠ABC
⇒ ∠B > ∠D
Similarly,
In ΔABC,
∠ACB < ∠BAC — (iii) (Angle opposite to longer side is larger.)
Now,
In ΔADC,
∠DCA < ∠DAC — (iv)
Adding (iii) and (iv) we get,
∠ACB + ∠DCA < ∠BAC + ∠DAC
⇒ ∠BCD < ∠BAD
⇒ ∠A > ∠C

5. In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Answer

Given,
PR > PQ and PS bisects ∠QPR
To prove,
∠PSR > ∠PSQ
Proof,
∠PQR > ∠PRQ — (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS — (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR + ∠QPS — (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)
Adding (i) and (ii)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]

Page No: 133

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer

Let l is a line segment and B is a point lying o it. We drew a line AB perpendicular to l. Let C be any other point on l.
To prove,
AB < AC
Proof,
In ΔABC,
∠B = 90°
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90°
∴ ∠C must be acute angle. or ∠C < ∠B
⇒ AB < AC (Side opposite to the larger angle is larger.)

https://youtu.be/8zyFiuaWVec

Exercise 6.1

1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer
Given,
∠AOC + ∠BOE = 70° and ∠BOD = 40°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°

Page No: 97

2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer

Given,
∠POY = 90° and a : b = 2 : 3
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°

3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer

Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° – ∠PQR — (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° – ∠PRQ
⇒ ∠PRQ = 180° – ∠PQR — (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° – ∠PQR
Therefore,  ∠PQS = ∠PRT

4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Answer

Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) +  (w + z) = 360°
⇒ (x + y) +  (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.

5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Answer

Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS — (i)
∠POS = ∠POR – ∠ROS = 90° – ∠ROS — (ii)
Subtracting (ii) from (i)
∠QOS – ∠POS = 90° + ∠ROS – (90° – ∠ROS)
⇒ ∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS
⇒ ∠QOS – ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.

6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Answer

Given,
∠XYZ = 64°
YQ bisects ∠ZYP

∠XYZ +∠ZYP = 180° (Linear Pair)
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠ZYP = 2∠ZYQ
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 58° = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + 58°
⇒ ∠XYQ = 122°
also,
reflex ∠QYP = 180° + ∠XYQ
∠QYP = 180° + 122°
⇒ ∠QYP = 302°

https://youtu.be/jum4L0ChTLw

Exercise 6.2

1. In Fig. 6.28, find the values of x and y and then show that AB || CD.

 

Answer

x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.

Page No: 104

2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Answer

Given,
AB || CD and CD || EF
y : z = 3 : 7
Now,
x + y = 180° (Angles on the same side of transversal.)
also,
∠O = z (Corresponding angles)
and, y + ∠O = 180° (Linear pair)
⇒ y + z = 180°
A/q,
y = 3w and z = 7w
3w + 7w = 180°
⇒ 10 w = 180°
⇒ w = 18°
∴ y = 3×18° = 54°
and, z = 7×18° = 126°
Now,
x + y = 180°
⇒ x + 54° = 180°
⇒ x = 126°

3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer

Given,
AB || CD
EF ⊥ CD
∠GED = 126°
A/q,
∠FED = 90° (EF ⊥ CD)
Now,
∠AGE = ∠GED (Since, AB || CD and GE is transversal. Alternate interior angles.)
∴ ∠AGE = 126°
Also, ∠GEF = ∠GED – ∠FED
⇒ ∠GEF = 126° – 90°
⇒ ∠GEF = 36°
Now,
∠FGE +∠AGE = 180° (Linear pair)
⇒ ∠FGE = 180° – 126°
⇒ ∠FGE = 54°

4. In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]

 

∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 70°
Also,
∠RST + ∠SRY = 180° (Angles on the same side of transversal.)
⇒ 130° + ∠SRY = 180°
⇒ ∠SRY = 50°
Now,
∠QRX +∠SRY + ∠QRS = 180°
⇒ 70° + 50° + ∠QRS = 180°
⇒ ∠QRS = 60°

5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

 Answer

Given,
AB || CD, ∠APQ = 50° and ∠PRD = 127°
A/q,
x = 50° (Alternate interior angles.)
∠PRD + ∠RPB = 180° (Angles on the same side of transversal.)
⇒ 127° + ∠RPB = 180°
⇒ ∠RPB = 53°
Now,
y + 50° + ∠RPB = 180° (AB is a straight line.)
⇒ y + 50° + 53° = 180°
⇒ y + 103° = 180°
⇒ y = 77°

6. In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

 

 As PQ || RS
So, BE || CF

From (i) and (ii),

⇒ ∠ABC = ∠DCB

https://youtu.be/rKakJ64X8Nc

Exercise 6.3

1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer

Given,
∠SPR = 135° and ∠PQT = 110°
A/q,
∠SPR +∠QPR = 180° (SQ is a straight line.)
⇒ 135° +∠QPR = 180°
⇒ ∠QPR = 45°
also,

∠PQT +∠PQR = 180° (TR is a straight line.)
⇒ 110° +∠PQR = 180°
⇒ ∠PQR = 70°
Now,
∠PQR +∠QPR + ∠PRQ = 180° (Sum of the interior angles of the triangle.)
⇒ 70° + 45° + ∠PRQ = 180°
⇒ 115° + ∠PRQ = 180°
⇒ ∠PRQ = 65°

2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Answer

Given,
∠X = 62°, ∠XYZ = 54°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
A/q,
∠X +∠XYZ + ∠XZY = 180° (Sum of the interior angles of the triangle.)
⇒ 62° + 54° + ∠XZY = 180°
⇒ 116° + ∠XZY = 180°
⇒ ∠XZY = 64°
Now,
∠OZY = 1/2∠XZY (ZO is the bisector.)
⇒ ∠OZY = 32°
also,
∠OYZ = 1/2∠XYZ (YO is the bisector.)
⇒ ∠OYZ = 27°
Now,
∠OZY +∠OYZ + ∠O = 180° (Sum of the interior angles of the triangle.)
⇒ 32° + 27° + ∠O = 180°
⇒ 59° + ∠O = 180°
⇒ ∠O = 121°

3. In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer

Given,
AB || DE, ∠BAC = 35° and ∠ CDE = 53°
A/q,
∠BAC = ∠CED (Alternate interior angles.)
∴ ∠CED = 35°
Now,
∠DCE +∠CED + ∠CDE = 180° (Sum of the interior angles of the triangle.)
⇒ ∠DCE + 35° + 53° = 180°
⇒ ∠DCE + 88° = 180°
⇒ ∠DCE = 92°

4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer

Given,
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
A/q,
∠PRT +∠RPT + ∠PTR = 180° (Sum of the interior angles of the triangle.)
⇒ 40° + 95° + ∠PTR = 180°
⇒ 40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 45°
∠PTR = ∠STQ = 45° (Vertically opposite angles.)
Now,
∠TSQ +∠PTR + ∠SQT = 180° (Sum of the interior angles of the triangle.)
⇒ 75° + 45° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 60°

Page No: 108

5. In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Answer

Given,
PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
A/q,
x +∠SQR = ∠QRT (Alternate angles  as QR is transveersal.)
⇒ x + 28° = 65°
⇒ x = 37°
also,
∠QSR = x
⇒ ∠QSR = 37°
also,
∠QRS +∠QRT = 180° (Linea pair)
⇒ ∠QRS + 65° = 180°
⇒ ∠QRS = 115°
Now,
∠P + ∠Q+ ∠R +∠S = 360° (Sum of the angles in a quadrilateral.)
⇒ 90° + 65° + 115° + ∠S = 360°
⇒ 270° + y + ∠QSR = 360°
⇒ 270° + y + 37° = 360°
⇒ 307° + y = 360°
⇒ y = 53°

6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2∠QPR.

Answer

Given,
Bisectors of ∠PQR and ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR (Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒ ∠QTR = ∠TRS – ∠TQR — (i)
also,
∠SRP = ∠QPR + ∠PQR
⇒ 2∠TRS = ∠QPR + 2∠TQR
⇒ ∠QPR =  2∠TRS – 2∠TQR
⇒ 1/2∠QPR =  ∠TRS – ∠TQR — (ii)
Equating (i) and (ii)
∠QTR – ∠TQR = 1/2∠QPR
Hence proved.

https://youtu.be/UBrZiKewNmQ

Exercise 5.1

1. Which of the following statements are true and which are false? Give reasons for your
answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In Fig. 5.9, if AB = PQ and PQ = XY, then AB = XY.

Answer

(i) False. There can be infinite line drawn passing through a single point.

(ii) False. Only one line can be drawn which passes through two distinct points.

(iii) True. A terminated line can be produced indefinitely on both the sides.
In geometry, a line can be extended in both direction. A line means infinite long length.

(iv) True. If two circles are equal, then their radii are equal.
By superposition, we will find that the centre and circumference of the both circles coincide. Hence, their radius must be equal.

(v) True. By Euclid’s first axiom things which are equal to the same thing, are equal to one another.

2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines                 (ii) perpendicular lines                (iii) line segment  
(iv) radius of a circle         (v) square

Answer

Yes, other terms need to be defined first which are:
Plane: A plane is flat surface on which geometric figures are drawn.
Point: A point is a dot drawn on a plane surface and is dimensionless.
Line: A line is collection of points which can extends in both direction and has only length not breadth.

(i) Parallel lines: When two or more never intersect each other in a plane and perpendicular distance between them is always constant then they are said to be parallel lines.

(ii) Perpendicular lines: When two lines intersect each other at right angle in a plane then they are said to be perpendicular to each other.

(iii) Line segment: A line segment is a part of a line with two end points and cannot be extended further.

(iv) Radius of circle: The fixed distance between the centre and the circumference of the circle is called the radius of the circle.

(v) Square: A square is a quadrilateral in which all the four sides are equal and each internal angle is a right angle.

3. Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.

Answer

Undefined terms in the postulates:
→ Many points lie in a plane. But here it is not given about the position of the point C whether it lies on the line segment joining AB or not.
→ Also, there is no information about the plane whether the points are in same plane or not.
Yes, these postulates are consistent when we deal with these two situation:
(i) Point C is lying in between and on the line segment joining A and B.
(ii) Point C not lies on the line segment joining A and B.

No, they don’t follow from Euclid’s postulates. They follow the axioms.

Page No: 86

4. If a point C lies between two points A and B such that AC = BC, then prove that AC = 1/2 AB. Explain by drawing the figure.

Answer

Here, AC = BC
Now, adding AC both sides.
AC + AC = BC + AC
also, BC +AC  = AB (as it coincides with line segment AB)
∴ 2 AC = AB (If equals are added to equals, the wholes are equal.)
⇒ AC = 1/2 AB.

5. In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.

Answer
 

Let A and B be the line segment and points P and Q be two different mid points of AB.
Now,
∴ P and Q are midpoints of AB.
Therefore AP=PB and also AQ = QB.
also, PB + AP = AB (as it coincides with line segment AB)
Similarly, QB + AQ = AB.
Now,
AP + AP = PB + AP (If equals are added to equals, the wholes are equal.)
⇒ 2 AP = AB — (i)
Similarly, 
2 AQ = AB — (ii)
From (i) and (ii)
2 AP  = 2 AQ (Things which are equal to the same thing are equal to one another.)
⇒ AP = AQ (Things which are double of the same things are equal to one another.)
Thus, P and Q are the same points. This contradicts the fact that P and Q are two different mid points of AB. Thus, it is proved hat every line segment has one and only one mid-point.

6. In Fig. 5.10, if AC = BD, then prove that AB = CD.

 
Answer

Given, AC = BD
From the figure,
AC = AB + BC
BD = BC + CD   
⇒ AB + BC = BC + CD
According to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.
Subtracting BC both sides,
AB + BC – BC = BC + CD – BC
AB = CD

7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)

Answer

Axiom 5 : The whole is always greater than the part.
Take an example of a cake. When it is whole it will measures 2 pound but when we took out a part from it and measures its weigh it will came out lower than the previous one. So, the fifth axiom of Euclid is true for all the universal things. That is why it is considered a ‘universal truth’.

https://youtu.be/VuFtU6PVj9U

Exercise 5.2

1. How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?

Answer

The fifth postulates is about parallel lines.
When two or more never intersect each other in a plane and perpendicular distance between them is always constant then they are said to be parallel lines.
Two facts of the postulates:
(i) If P doesn’t lie on l then we can draw a line through P which will be parallel to the line l.
(ii) There will be only one line can be drawn through P which is parallel to the line l.

2. Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.

Answer

Yes, Euclid’s fifth postulate imply the existence of parallel lines.
If the sum of the interior angles will be equal to sum of the two right angles then two lines will not meet each other on either sides and therefore they will be parallel to each other.

m and n will be parallel if
∠1 + ∠3 = 180°
Or  ∠3 + ∠4 = 180°

https://youtu.be/MxNJ_K6ybLs

Exercise 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be x and that of a pen to be y).

Answer

Let the cost of pen be y and the cost of notebook be x.
A/q,
Cost  of a notebook = twice the pen = 2y.
∴2y = x
⇒ x – 2y = 0
This is a linear equation in two variables to represent this statement.

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

(i) 2x + 3y = 9.35               (ii) x – y/5 – 10 = 0                  (iii) -2x + 3y = 6                (iv) x = 3y

(v) 2x = -5y                        (vi) 3x + 2 = 0                        (vii) y – 2 = 0                     (viii) 5 = 2x

Answer

(i) 2x + 3y = 9.35
⇒ 2x + 3y – 9.35 = 0
On comparing this equation with ax + by + c = 0, we get
a = 2x, b = 3 and c = -9.35

(ii) x – y/5 – 10 = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -1/5 and c = -10


(iii) -2x + 3y = 6
⇒ -2x + 3y – 6 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 3 and c = -6

(iv) x = 3y
⇒ x – 3y = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -3 and c = 0

(v) 2x = -5y
⇒ 2x + 5y = 0
On comparing this equation with ax + by + c = 0, we get
a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 3, b = 0 and c = 2

(vii) y – 2 = 0
⇒ 0x + y – 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 0, b = 1 and c = -2

(viii) 5 = 2x
⇒ -2x + 0y + 5 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 0 and c = 5

https://youtu.be/_31qm2RCxmw

Exercise 4.2

1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,               (ii) only two solutions,             (iii) infinitely many solutions

Answer

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

2. Write four solutions for each of the following equations:

    (i) 2x + y = 7             (ii) πx + y = 9                (iii) x = 4y

Answer

(i) 2x + y = 7
⇒ y = 7 – 2x
→ Put x = 0,
y = 7 – 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 – 2 × 1 ⇒ y = 5
(1, 5) is the solution.
→ Now, put x = 2
y = 7 – 2 × 2 ⇒ y = 3
(2, 3) is the solution.
→ Now, put x = -1
y = 7 – 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9
⇒ y = 9 – πx
→ Put x = 0,
y = 9 – π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 – π×1 ⇒ y = 9-π
(1, 9-π) is the solution.
→ Now, put x = 2
y = 9 – π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution.
→ Now, put x = -1
y = 9 – π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution.
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution.
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

3. Check which of the following are solutions of the equation x – 2y = 4 and which are not:

    (i) (0, 2)              (ii) (2, 0)             (iii) (4, 0)            (iv) (√2, 4√2)              (v) (1, 1)

Answer

(i) Put x = 0 and y = 2 in the equation x – 2y = 4.
0 – 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x – 2y = 4.
2 – 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x – 2y = 4.
4 – 2×0 = 4
⇒ 4 = 4
∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x – 2y = 4.
√2 – 2×4√2 = 4 ⇒ √2 – 8√2 = 4 ⇒ √2(1 – 8) = 4
⇒ -7√2  ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x – 2y = 4.
1 – 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer

Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7

https://youtu.be/RzvMsJVBsvo

Exercise 4.3

1. Draw the graph of each of the following linear equations in two variables:
   (i) x + y = 4                  (ii) x – y = 2               (iii) y = 3x              (iv) 3 = 2x + y

Answer

(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0

x	0	4
y	4	0

(ii) x – y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0

x	0	2
y	-2	0

(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3

x	0	1
y	0	3

(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1

x	0	1
y	3	1

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer

Here, x = 2 and y =14.
Thus, x + y = 1
also, y = 7x ⇒ y – 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y – 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.

3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Answer

The point (3, 4) lies on the graph of the equation.
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7
⇒ a = 5/3

4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer

Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3

x	0	-3/5
y	3	0

5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
   For Fig. 4. 6                           For Fig. 4.7
   (i) y = x                                  (i) y = x + 2
   (ii) x + y = 0                           (ii) y = x – 2
   (iii) y = 2x                              (iii) y = –x + 2
   (iv) 2 + 3y = 7x                     (iv) x + 2y = 6

Answer

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.

Page No: 75

6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
                     (i) 2 units               (ii) 0 unit

Answer

Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit

x	2	0
y	10	0

7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Answer

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0

x	0	50	100
y	100	50	0

 

 

8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:

                                           F = (9/5)C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

 

Answer

 

(i) F = (9/5)C + 32

When C = 0 then F = 32

also, when C = -10 then F = 14

C	0	-10
F	32	14

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30  + 32
⇒ F = 54 + 32
⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C  + 32
⇒ (9/5)C = 95 – 32
⇒ C = 63 × 5/9
⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C  + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0  + 32
⇒ F = 32

(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F – 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.

https://youtu.be/wnegZVDCOzo

Exercise 4.4

1. Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
y = 3

 

2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Answer

(i) in one variable, it is represented as
x = -9/2

https://youtu.be/qezAU35eZm0

Exercise 3.1

1. How will you describe the position of a table lamp on your study table to another person?

Answer

To describe the position of a table lamp on the study table, we have two take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates.
Let the distance of point from X axis is x and from Y axis is y then the the position of the table lamp in terms of coordinates is (x,y).

 

https://youtu.be/usRYLKnTT_E

(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?

(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii)The coordinates of the point L.
(viii) The coordinates of the point M.

 

(i) The coordinates of B is (-5, 2).

(ii) The coordinates of C is (5, -5).

(iii) The point identified by the coordinates (-3, -5) is E.

(iv) The point identified by the coordinates (2, -4) is G.

(v) Abscissa means x coordinate of point D. So, abscissa of the point D is 6.

(vi) Ordinate means y coordinate of point H. So, ordinate of point H is -3.

(vii) The coordinates of the point L is (0, 5).

(viii) The coordinates of the point M is (- 3, 0).

https://youtu.be/ILVmMPtFim4

Exercise 3.3

1. In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.

Answer

 

(-2, 4) → Second quadrant
(3, -1) → Fourth quadrant
(-1, 0) → Second quadrant
(1, 2) → First quadrant
(-3, -5) → Third quadrant

2. Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

x	-2	-1	0	1	3
y	8	7	-1.25	3	-1

Answer

Points (x,y) on the plane. 1unit = 1 cm

https://youtu.be/Zsql4HVImSk

1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(iii) 3√t + t√2

There is only one variable y with whole number power so this polynomial in one variable.

There is only one variable t but in 3√t power of t is 1/2 which is not a whole number so 3√t + t√2 is not a polynomial.

There is only one variable y but 2/y = 2y-1 so the power is not a whole number so y + 2/y is not a polynomial.

There are three variable x, y and t and there powers are whole number so this polynomial in three variable.

2. Write the coefficients of x² in each of the following:
(i) 2 + x² + x
(ii) 2 – x² + x³

(iv) √2x – 1

Answer

(i) coefficients of x² = 1
(ii) coefficients of x² = -1
(iii) coefficients of x² = π/2
(iv) coefficients of x² = 0

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

(iii) y + y² +4
► Quadratic Polynomial

https://youtu.be/aZ-LJird–k

(i) p(y) = y² – y + 1

(i) p(y) = y² – y + 1
p(0) = (0)² – (0) + 1 = 1
p(1) = (1)² – (1) + 1 = 1
p(2) = (2)² – (2) + 1 = 3

(ii) p(t) = 2 + t + 2t² – t³
p(0) = 2 + 0 + 2 (0)² – (0)³ = 2
p(1) = 2 + (1) + 2(1)² – (1)3
= 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x³
p(0) = (0)³ = 0
p(1) = (1)³ = 1
p(2) = (2)³ = 8

(iii) p(x) = x² – 1, x = 1, -1

(iv) p(x) = (x + 1) (x – 2), x = -1, 2
(v) p(x) = x² , x = 0

(viii) p(x) = 2x + 1, x = 1/2

Answer

(i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0.
At, p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0
Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1.

(ii) If x = 4/5 is a zero of polynomial p(x) = 5x – π then p(4/5) should be 0.
At, p(4/5) = 5(4/5) – π = 4 – π
Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x – π.

(iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x² – 1, then p(1) and p(-1) should be 0.
At, p(1) = (1)² – 1 = 0 and
At, p(-1) = (-1)² – 1 = 0
Hence, x = 1 and -1 are zeroes of the polynomial  p(x) = x² – 1.

(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x – 2), then p( – 1) and (2)should be 0.
At, p(-1) = (-1 + 1) (-1 – 2) = 0 (-3) = 0, and
At, p(2) = (2 + 1) (2 – 2) = 3 (0) = 0
Therefore, x = -1 and x = 2 are zeroes of the polynomial p(x) = (x +1) (x – 2).

(v) If x = 0 is a zero of polynomial p(x) = x², then p(0) should be zero.
Here, p(0) = (0)² = 0
Hence, x = 0 is a zero of the polynomial p(x) = x².



(viii)  If x = 1/2 is a zero of polynomial p(x) = 2x + 1 then p(1/2) should be 0.
At, p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of given polynomial p(x) = 2x + 1.

4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 

https://youtu.be/sIqZwHl8qw8

Exercises 2.3

1. Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – 1/2
(iii) x
(iv) x + π
(v) 5 + 2x

Answer

(i) x + 1
By long division,

Therefore, the remainder is 0.

(ii) x – 1/2
By long division,


Therefore, the remainder is 27/8.

(iii) x

Therefore, the remainder is 1.

(iv) x + π

Therefore, the remainder is [1 – 3π + 3π² – π³].

(v) 5 + 2x

Therefore, the remainder is -27/8.

2. Find the remainder when x³ – ax² + 6x – a is divided by x – a.

Answer

By Long Division,

Therefore, remainder obtained is 5a when x³ – ax² + 6x – a is divided by x – a.

3. Check whether 7 + 3x is a factor of 3x³ + 7x.

Answer

We have to divide 3x³ + 7x by 7 + 3x. If remainder comes out to be 0 then 7 + 3x will be a factor of 3x³ + 7x.
By Long Division,

As remainder is not zero so 7 + 3x is not a factor of 3x³ + 7x.

https://youtu.be/IXQ8DZKnyI0

Exercise 2.4

1. Determine which of the following polynomials has (x + 1) a factor:
(i) x³ + x² + x + 1

As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial

(iii)If (x + 1) is a factor of polynomial p(x) = x⁴ + 3x³ + 3x² + x + 1, p(- 1) must be 0. 
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial.

(iv) If (x + 1) is a factor of polynomial

p(-1) =  (-1)³ –  (-1)² –  (2 + √2) (-1) + √2
= -1 – 1 + 2 + √2 + √2
=2√2
As, p(-1) ≠ 0
Therefore,, x + 1 is not a factor of this polynomial.

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
(iii) p(x) = x³ – 4 x² + x + 6, g(x) = x – 3

Answer

(i) If g(x) = x + 1 is a factor of given polynomial p(x), p(- 1) must be zero.
p(x) = 2x³ + x² – 2x – 1
p(- 1) = 2(-1)³ + (-1)² – 2(-1) – 1
= 2(- 1) + 1 + 2 – 1 = 0
Hence, g(x) = x + 1 is a factor of given polynomial.

(ii) If g(x) = x + 2 is a factor of given polynomial p(x), p(- 2) must be 0.
p(x) = x³ +3x² + 3x + 1
p(-2) = (-2)³ + 3(- 2)² + 3(- 2) + 1
= -8 + 12 – 6 + 1
= -1

(iii) If g(x) = x – 3 is a factor of given polynomial p(x), p(3) must be 0.
p(x) = x³ – 4x² + x + 6
p(3) = (3)³ – 4(3)² + 3 + 6
= 27 – 36 + 9 = 0
Therefore,, g(x) = x – 3 is a factor of given polynomial.

Page No: 44

3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:(i) p(x) = x² + x + k
(ii) p(x) = 2x² + kx +  √2
(iii) p(x) = kx² – √2x + 1
(iv) p(x) = kx² – 3x + k

Answer

(i) If x – 1 is a factor of polynomial p(x) = x² + x + k, then

(ii) If x – 1 is a factor of polynomial p(x) = 2x² + kx +  √2, then
p(1) = 0
⇒ 2(1)² + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 – √2 = -(2 + √2)

(iii) If x – 1 is a factor of polynomial p(x) = kx² – √2x + 1, then
p(1) = 0
⇒ k(1)² – √2(1) + 1 = 0
⇒ k – √2 + 1 = 0
⇒ k = √2 – 1

(iv) If x – 1 is a factor of polynomial p(x) = kx² – 3x + k, then
p(1) = 0
⇒ k(1)² + 3(1) + k = 0
⇒ k – 3 + k = 0
⇒ 2k – 3 = 0
⇒ k = 3/2

4. Factorise:
(i) 12x² + 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4 

Answer

(i) 12x² + 7x + 1
= 12x² – 4x – 3x+ 1                   
= 4x (3x – 1) – 1 (3x – 1)
= (3x – 1) (4x – 1)

(ii) 2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
=  (x + 3) (2x + 1) 

(iii) 6x² + 5x – 6
= 6x² + 9x – 4x – 6

(iv) 3x² – x – 4
= 3x² – 4x + 3x – 4 
= x (3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)

5. Factorise:
(i) x³ – 2x² – x + 2

(iv) 2y³ + y² – 2y – 1

Answer

(i) Let p(x) = x³ – 2x² – x + 2
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = x³ – 2x² – x + 2
p(-1) = (-1)³ – 2(-1)² – (-1) + 2 = -1 -2 + 1 + 2 = 0
Therefore, (x+1) is the factor of  p(x)

 

Now, Dividend = Divisor × Quotient + Remainder

(x+1) (x² – 3x + 2)

(ii) Let p(x) = x³ – 3x² – 9x – 5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x³ – 2x² – x + 2
p(5) = (5)³ – 3(5)² – 9(5) – 5 = 125 – 75 – 45 – 5 = 0
Therefore, (x-5) is the factor of  p(x)

(x-5) (x² + 2x + 1)

= (x-5) (x+1) (x+1)

(iii) Let p(x) = x³ + 13x² + 32x + 20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) =  x³ + 13x² + 32x + 20
p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0
Therefore, (x+1) is the factor of  p(x)

 

Now, Dividend = Divisor × Quotient + Remainder

(x+1) (x² + 12x + 20)

= (x+1) (x² + 2x + 10x + 20)

= (x-5) {x(x+2) +10(x+2)}

= (x-5) (x+2) (x+10)

(iv) Let p(y) = 2y³ + y² – 2y – 1
Factors of ab = 2× (-1) = -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) =  2y³ + y² – 2y – 1
p(1) = 2(1)³ + (1)² – 2(1) – 1 = 2 +1 – 2 – 1 = 0
Therefore, (y-1) is the factor of  p(y)

 

 Now, Dividend = Divisor × Quotient + Remainder

(y-1) (2y² + 3y + 1)

= (y-1) (2y² + 2y + y + 1)

= (y-1) {2y(y+1) +1(y+1)}

= (y-1) (2y+1) (y+1)

Part-1 ( Q1 to Q12)

https://youtu.be/eUwEQgj6dgU

Part-2 ( Q13 to Q16)

https://youtu.be/1qa2EdkkBGE

https://youtu.be/JRB8Nhg_OMY

Exercise 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Yes. Zero is a rational number as it can be represented as 0/1 or 0/2.

2. Find six rational numbers between 3 and 4.

Answer

There are infinite rational numbers in between 3 and 4.
3 and 4 can be represented as 24/8 and 32/8 respectively.

https://youtu.be/WwvqdJvSWvg

Exercise 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

Answer

No, the square roots of all positive integers are not irrational. For example √4 = 2.

3. Show how √5 can be represented on the number line.

Answer

Step 1: Let AB be a line of length 2 unit on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.

https://youtu.be/15ZUeooo9x0

Exercise 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) 36/100
= 0.36 (Terminating)

(ii) 1/11
0.09090909… = 0.9 (Non terminating repeating)

(iii)
= 33/8 = 4.125 (Terminating)

(iv) 3/13
= 0.230769230769… = 0.230769 (Non terminating repeating)

(v) 2/11
= 0.181818181818… = 0.18 (Non terminating repeating)

(vi) 329/400
= 0.8225 (Terminating)

2. You know that 1/7 = 0.142857.Can you predict what the decimal expansion of 2/7, 3/7, 4/7, 5/7, 6/7 are without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of 1/7 carefully.]

Answer

Yes. We can be done this by:


3. Express the following in the form p/q where p and q are integers and q ≠ 0.

1000x = 1.001001…
1000x = 1 + x
999x = 1

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Answer

1/17 = 0.0588235294117647
There are 16 digits in the repeating block of the decimal expansion of 1/17.
Division Check:

6. Look at several examples of rational numbers in the form p/q (q ≠ 0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer

We observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:
1/2 = 0.5, denominator q = 2¹
7/8 = 0.875, denominator q = 2³
4/5 = 0.8, denominator q = 5¹

We can observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

0.7207200720007200007200000…

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

0.73073007300073000073…

9. Classify the following numbers as rational or irrational:

(ii) √225

Since the this number is non-terminating recurring, therefore, it is a rational number.

Since the number is non-terminating non-repeating, therefore, it is an irrational number.

https://youtu.be/qla_5QAJgQ0

1. Visualise 3.765 on the number line using successive magnification.

https://youtu.be/YbI8J1SUzjs

(iii) 2√7/7√7

(iv) 1/√2

(v) 2π

 

Answer

 

(i) 2 – √5 = 2 – 2.2360679… = – 0.2360679…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) – √23 = 3 + √23 – √23 = 3 = 3/1
Since the number is rational number as it can represented in p/q form.

(iii) 2√7/7√7 = 2/7
Since the number is rational number as it can represented in p/q form.

(iv) 1/√2 = √2/2 = 0.7071067811…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

2. Simplify each of the following expressions:

 

(i) (3 + √3) (2 + √2)

(ii) (3 + √3) (3 – √3)

(iii) (√5 + √2)²
(iv) (√5 – √2) (√5 + √2)

 

Answer

 

(i) (3 + √3) (2 + √2)

⇒ 3 × 2 + 2 + √3 + 3√2+ √3 ×√2

⇒ 6 + 2√3 +3√2 + √6

 

(ii) (3 + √3) (3 – √3) [∵ (a + b) (a – b) = a² – b²]

⇒ 3² – (√3)²
⇒ 9 – 3

⇒ 6

 

(iii) (√5 + √2)² [∵ (a + b)² = a² + b² + 2ab]
⇒ (√5)² + (√2)² + 2 ×√5 × √2
⇒ 5 + 2 + 2 × √5× 2 

⇒ 7 +2√10

(iv) (√5 – √2) (√5 + √2) [∵ (a + b) (a – b) = a² – b²]

⇒ (√5)² – (√2)² 

⇒ 5 – 2

⇒ 3

 

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

 

Answer

 

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realise that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

 

4. Represent √9.3 on the number line.

 

Answer

Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 unit.
Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.
Step 3: Draw a semi circle with radius OC and centre O.
Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D. Also, Join OD.
Step 5: Now, OBD is a right angled triangle.
Here, OD = 10.3/2 (radius of semi circle), OC = 10.3/2, BC = 1
OB = OC – BC = (10.3/2) – 1 = 8.3/2
Using Pythagoras theorem,
OD² = BD² + OB²
⇒ (10.3/2)² = BD2 + (8.3/2)²
⇒ BD² = (10.3/2)² – (8.3/2)²
⇒ BD² = (10.3/2 – 8.3/2) (10.3/2 + 8.3/2)
⇒ BD² = 9.3
⇒ BD² =  √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

5. Rationalise the denominators of the following:

(i) 1/√7

(ii) 1/√7-√6

(iii) 1/√5+√2

(iv) 1/√7-2

 

Answer

https://youtu.be/Zh_78OJEG38

Answer



2. Find:

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3. Simplify:

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