Class 10: Maths Chapter 2 solutions. Complete Class 10 Maths Chapter 2 Notes.

Contents

## Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 2- Pythagoras Theorem

Maharashtra Board 10th Maths Chapter 2, Class 10 Maths Chapter 2 solutions

#### Practice Set 2.1

**Question 1.Identify, with reason, which of the following are Pythagorean triplets.**

i. (3,5,4)

ii. (4,9,12)

iii. (5,12,13)

iv. (24,70,74)

v. (10,24,27)

vi. (11,60,61)

**Solution:**

i. Here, 5

^{2}= 25

3

^{2}+ 4

^{2}= 9 + 16 = 25

∴ 5

^{2}= 3

^{2}+ 4

^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3,5,4) is a Pythagorean triplet.

ii. Here, 12^{2} = 144

4^{2} + 9^{2}= 16 + 81 =97

∴ 12^{2} ≠ 4^{2} + 92

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (4,9,12) is not a Pythagorean triplet.

iii. Here, 13^{2} = 169

5^{2} + 12^{2} = 25 + 144 = 169

∴ 13^{2} = 5^{2} + 12^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (5,12,13) is a Pythagorean triplet.

iv. Here, 74^{2} = 5476

24^{2} + 70^{2} = 576 + 4900 = 5476

∴ 74^{2} = 24^{2} + 70^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (24, 70,74) is a Pythagorean triplet.

v. Here, 27^{2} = 729

10^{2} + 24^{2} = 100 + 576 = 676

∴ 27^{2} ≠ 10^{2} + 24^{2}

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (10,24,27) is not a Pythagorean triplet.

vi. Here, 61^{2} = 3721

11^{2} + 60^{2} = 121 + 3600 = 3721

∴ 61^{2} = 11^{2} + 60^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (11,60,61) is a Pythagorean triplet.

**Question 2.In the adjoining figure, ∠MNP = 90°, seg NQ ⊥ seg MP,MQ = 9, QP = 4, find NQ.Solution:**

In ∆MNP, ∠MNP = 90° and [Given]

seg NQ ⊥ seg MP

NQ

^{2}= MQ × QP [Theorem of geometric mean]

∴ NQ = MQ×QP−−−−−−−−−√ [Taking square root of both sides]

= 9×4−−−−√

= 3 × 2

∴NQ = 6 units

**Question 3.In the adjoining figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q – M – R, PM = 10, QM = 8, find QR.Solution:**

In ∆PQR, ∠QPR = 90° and [Given]

seg PM ⊥ seg QR

∴ PM

^{2}= OM × MR [Theorem of geometric mean]

∴ 10

^{2}= 8 × MR

∴ MR = 1008

= 12.5

Now, QR = QM + MR [Q – M – R]

= 8 + 12.5

∴ QR = 20.5 units

**Question 4.See adjoining figure. Find RP and PS using the information given in ∆PSR.Solution:**

In ∆PSR, ∠S = 90°, ∠P = 30° [Given]

∴ ∠R = 60° [Remaining angle of a triangle]

∴ ∆PSR is a 30° – 60° – 90° triangle.

RS = 12 RP [Side opposite to 30°]

∴6 = 12 RP

∴ RP = 6 × 2 = 12 units

Also, PS = 3√2 RP [Side opposite to 60°]

= 3√2 × 12

= 63–√ units

∴ RP = 12 units, PS = 6 3–√ units

**Question 5.For finding AB and BC with the help of information given in the adjoining figure, complete the following activity.Solution:**

AB = BC [Given]

∴ ∠BAC = ∠BCA [Isosceles triangle theorem]

Let ∠BAC = ∠BCA = x (i)

In ∆ABC, ∠A + ∠B + ∠C = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ x + 90° + x = 180° [From (i)]

∴ 2x = 90°

∴ x = 90°2 [From (i)]

∴ x = 45°

**Question 6.Find the side and perimeter of a square whose diagonal is 10 cm.Solution:**

Let ꠸ABCD be the given square.

l(diagonal AC) = 10 cm

Let the side of the square be ‘x’ cm.

In ∆ABC,

∠B = 90° [Angle of a square]

∴ AC

^{2}= AB

^{2}+ BC

^{2}[Pythagoras theorem]

∴ 10

^{2}= x

^{2}+ x

^{2}

∴ 100 = 2x

^{2}

∴ x

^{2}= 1002

∴x

^{2}= 50

∴ x = 50−−√ [Taking square root of both sides]

= =25×2−−−−−√=52–√

∴side of square is 52–√ cm.

= 4 × 5 2–√

∴ Perimeter of square = 20 2–√ cm

**Question 7.In the adjoining figure, ∠DFE = 90°, FG ⊥ ED. If GD = 8, FG = 12, find**

i. EG

ii. FD, and

iii. EF

**Solution:**

i. In ∆DEF, ∠DFE = 90° and FG ⊥ ED [Given]

∴ FG

^{2}= GD × EG [Theorem of geometric mean]

∴ 122 = 8 × EG .

∴ EG = 1448

∴ EG = 18 units

ii. In ∆FGD, ∠FGD = 90° [Given]

∴ FD^{2} = FG^{2} + GD^{2} [Pythagoras theorem]

= 122 + 82 = 144 + 64

= 208

∴ FD = 208−−−√ [Taking square root of both sides]

∴ FD = 4 13−−√ units

iii. In ∆EGF, ∠EGF = 90° [Given]

∴ EF^{2} = EG^{2} + FG^{2} [Pythagoras theorem]

= 182 + 122 = 324 + 144

= 468

∴ EF = 468−−−√ [Taking square root of both sides]

∴ EF = 6 13−−√ units

**Question 8.Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.Solution:**

Let ꠸ABCD be the given rectangle.

AB = 12 cm, BC 35 cm

In ∆ABC, ∠B = 90° [Angle of a rectangle]

∴ AC

^{2}= AB

^{2}+ BC

^{2}[Pythagoras theorem]

= 122 + 352

= 144 + 1225

= 1369

∴ AC = 1369−−−−√ [Taking square root of both sides]

= 37 cm

∴ The diagonal of the rectangle is 37 cm.

**Question 9.In the adjoining figure, M is the midpoint of QR. ∠PRQ = 90°.Prove that, PQ**

^{2}= 4 PM

^{2}– 3 PR

^{2}.Solution:

Proof:

In ∆PQR, ∠PRQ = 90° [Given]

PQ

^{2}= PR

^{2}+ QR

^{2}(i) [Pythagoras theorem]

RM = 12 QR [M is the midpoint of QR]

∴ 2RM = QR (ii)

∴ PQ

^{2}= PR

^{2}+ (2RM)

^{2}[From (i) and (ii)]

∴ PQ

^{2}= PR

^{2}+ 4RM

^{2}(iii)

Now, in ∆PRM, ∠PRM = 90° [Given]

∴ PM

^{2}= PR

^{2}+ RM

^{2}[Pythagoras theorem]

∴ RM

^{2}= PM

^{2}– PR

^{2}(iv)

∴ PQ

^{2}= PR

^{2}+ 4 (PM

^{2}– PR

^{2}) [From (iii) and (iv)]

∴ PQ

^{2}= PR

^{2}+ 4 PM

^{2}– 4 PR

^{2}

∴ PQ

^{2}= 4 PM

^{2}– 3 PR

^{2}

**Question 10.Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4.2 m. Find the width of the street.Solution:**

Let AC and CE represent the ladder of length 5.8 m, and A and E represent windows of the buildings on the opposite sides of the street. BD is the width of the street.

AB = 4 m and ED = 4.2 m

In ∆ABC, ∠B = 90° [Given]

AC

^{2}= AB

^{2}+ BC

^{2}[Pythagoras theorem]

∴ 5.8

^{2}= 4

^{2}+ BC

^{2}

∴ 5.8

^{2}– 4

^{2}= BC

^{2}

∴ (5.8 – 4) (5.8 + 4) = BC

^{2}

∴ 1.8 × 9.8 = BC

^{2}

CE

^{2}= CD

^{2}+ DE

^{2}[Pythagoras theorem]

∴ 5.8

^{2}= CD

^{2}+ 4.22

∴ 5.8

^{2}– 4.2

^{2}= CD

^{2}

∴ (5.8 – 4.2) (5.8 + 4.2) = CD

^{2}

∴ 1.6 × 10 = CD

^{2}

∴ CD

^{2}= 16

∴ CD = 4m (ii) [Taking square root of both sides]

Now, BD = BC + CD [B – C – D]

= 4.2 + 4 [From (i) and (ii)]

= 8.2 m

∴ The width of the street is 8.2 metres.

**Question 1.Verify that (3,4,5), (5,12,13), (8,15,17), (24,25,7) are Pythagorean triplets. (Textbook pg. no. 30)Solution:**

i. Here, 5

^{2}= 25

3

^{2}+ 4

^{2}= 9 + 16 = 25

∴ 5

^{2}= 3

^{2}+ 4

^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ 3,4,5 is a Pythagorean triplet.

ii. Here, 13^{2} = 169

5^{2} + 12^{2} = 25 + 144 = 169

∴ 13^{2} = 5^{2} + 12^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ 5,12,13 is a Pythagorean triplet.

iii. Here, 17^{2} = 289

8^{2} + 15^{2} = 64 + 225 = 289

∴ 17^{2} = 8^{2} + 15^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ 8,15,17 is a Pythagorean triplet.

iv. Here, 25^{2} = 625

7^{2} + 24^{2 }= 49 + 576 = 625

∴ 25^{2} = 7^{2} + 24^{2}

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ 24,25, 7 is a Pythagorean triplet.

**Question 2.Assign different values to a and b and obtain 5 Pythagorean triplets. (Textbook pg. no. 31)Solution:**

i. Let a = 2, b = 1

a

^{2}+ b

^{2}= 2

^{2}+ 1

^{2}= 4 + 1 = 5

a

^{2}– b

^{2}= 2

^{2}– 1

^{2}= 4 – 1 = 3

2ab = 2 × 2 × 1 = 4

∴ (5, 3, 4) is a Pythagorean triplet.

ii. Let a = 4,b = 3

a^{2} + b^{2} = 4^{2} + 3^{2} = 16 + 9 = 25

a^{2} – b^{2} = 4^{2} – 3^{2} = 16 – 9 = 7

2ab = 2 × 4 × 3 = 24

∴ (25, 7, 24) is a Pythagorean triplet.

iii. Let a = 5, b = 2

a^{2} + b^{2} = 5^{2} + 2^{2} = 25 + 4 = 29

a^{2} – b^{2} = 5^{2} – 2^{2} = 25 – 4 = 21

2ab = 2 × 5 × 2 = 20

∴ (29, 21, 20) is a Pythagorean triplet.

iv. Let a = 4,b = 1

a^{2} + b^{2} = 4^{2} + 1^{2} = 16 + 1 = 17

a^{2} – b^{2} = 42 – 12 = 16 – 1 = 15

2ab = 2 × 4 × 1 = 8

∴ (17, 15, 8) is a Pythagorean triplet.

v. Let a = 9, b = 7

a^{2} + b^{2} = 92 + 72 = 81 + 49 = 130

a^{2} – b^{2} = 92 – 72 = 81 – 49 = 32

2ab = 2 × 9 × 7 = 126

∴ (130,32,126) is a Pythagorean triplet.

Note: Numbers in Pythagorean triplet can be written in any order.

#### Practice Set 2.2

**Question 1.In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.**

**Solution:**

In ∆PQR, point S is the midpoint of side QR. [Given]

∴ seg PS is the median.

∴ PQ

^{2}+ PR

^{2}= 2 PS

^{2}+ 2 SR

^{2}[Apollonius theorem]

∴ 11

^{2}+ 17

^{2}= 2 (13)

^{2}+ 2 SR

^{2}

∴ 121 + 289 = 2 (169)+ 2 SR

^{2}

∴ 410 = 338+ 2 SR

^{2}

∴ 2 SR

^{2}= 410 – 338

∴ 2 SR

^{2}= 72

∴ SR

^{2}= 722 = 36

∴ SR = 36−−√ [Taking square root of both sides]

= 6 units Now, QR = 2 SR [S is the midpoint of QR]

= 2 × 6

∴ QR = 12 units

**Question 2.In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.Solution:**

Let CD be the median drawn from the vertex C to side AB.

BD = 12 AB [D is the midpoint of AB]

= 12 × 10 = 5 units

In ∆ABC, seg CD is the median. [Given]

∴ AC

^{2}+ BC

^{2}= 2 CD

^{2}+ 2 BD

^{2}[Apollonius theorem]

∴ 7

^{2}+ 9

^{2}= 2 CD

^{2}+ 2 (5)

^{2}

∴ 49 + 81 = 2 CD

^{2}+ 2 (25)

∴ 130 = 2 CD

^{2}+ 50

∴ 2 CD

^{2}= 130 – 50

∴ 2 CD

^{2}= 80

∴ CD

^{2}= 802 = 40

∴ CD = 40−−√ [Taking square root of both sides]

= 2 10−−√ units

∴ The length of the median drawn from point C to side AB is 2 10−−√ units.

**Question 3.In the adjoining figure, seg PS is the median of APQR and PT ⊥ QR. Prove that,**

i. PR

^{2}= PS

^{2}+ QR × ST + (QR2)

^{2}

ii. PQ

^{2}= PS

^{2}– QR × ST + (QR2)

^{2}

**Solution:**

i. QS = SR = 12 QR (i) [S is the midpoint of side QR]

∴ In ∆PSR, ∠PSR is an obtuse angle [Given]

and PT ⊥ SR [Given, Q-S-R]

∴ PR

^{2}= SR

^{2}+PS

^{2}+ 2 SR × ST (ii) [Application of Pythagoras theorem]

∴ PR

^{2}= (12 QR)

^{2}+ PS

^{2}+ 2 (12 QR) × ST [From (i) and (ii)]

∴ PR

^{2}= (QR2)

^{2}+ PS

^{2}+ QR × ST

∴ PR

^{2}= PS

^{2}+ QR × ST + (QR2)

^{2}

ii. In.∆PQS, ∠PSQ is an acute angle and [Given]

PT ⊥QS [Given, Q-S-R]

∴ PQ^{2} = QS^{2} + PS^{2} – 2 QS × ST (iii) [Application of Pythagoras theorem]

∴ PR^{2} = (12 QR)^{2} + PS^{2} – 2 (12 QR) × ST [From (i) and (iii)]

∴ PR^{2} = (QR2)^{2} + PS^{2} – QR × ST

∴ PR^{2} = PS^{2} – QR × ST + (QR2)^{2}

**Question 4.In ∆ABC, point M is the midpoint of side BC. If AB ^{2 }+ AC^{2} = 290 cm, AM = 8 cm, find BC.Solution:**

In ∆ABC, point M is the midpoint of side BC. [Given]

∴ seg AM is the median.

∴ AB

^{2}+ AC

^{2}= 2 AM

^{2}+ 2 MC

^{2}[Apollonius theorem]

∴ 290 = 2 (8)

^{2}+ 2 MC

^{2}

∴ 145 = 64 + MC

^{2}[Dividing both sides by 2]

∴ MC

^{2}= 145 – 64

∴ MC

^{2}= 81

∴ MC = 81−−√ [Taking square root of both sides]

MC = 9 cm

Now, BC = 2 MC [M is the midpoint of BC]

= 2 × 9

∴ BC = 18 cm

**Question 5.In the adjoining figure, point T is in the interior of rectangle PQRS. Prove that, TS ^{2} + TQ^{2} = TP^{2} + TR^{2}. (As shown in the figure, draw seg AB || side SR and A – T – B)**

**Given**: ꠸PQRS is a rectangle.

Point T is in the interior of ꠸PQRS.

To prove: TS

^{2}+ TQ

^{2}= TP

^{2}+ TR

^{2}

Construction: Draw seg AB || side SR such that A – T – B.

**Solution:**

Proof:

꠸PQRS is a rectangle. [Given]

∴ PS = QR (i) [Opposite sides of a rectangle]

In ꠸ASRB,

∠S = ∠R = 90° (ii) [Angles of rectangle PQRS]

side AB || side SR [Construction]

Also ∠A = ∠S = 90° [Interior angle theorem, from (ii)]

∠B = ∠R = 90°

∴ ∠A = ∠B = ∠S = ∠R = 90° (iii)

∴ ꠸ASRB is a rectangle.

∴ AS = BR (iv) [Opposite sides of a rectanglel

In ∆PTS, ∠PST is an acute angle

and seg AT ⊥ side PS [From (iii)]

∴ TP

^{2}= PS

^{2}+ TS

^{2}– 2 PS.AS (v) [Application of Pythagoras theorem]

In ∆TQR., ∠TRQ is an acute angle

and seg BT ⊥ side QR [From (iii)]

∴ TQ

^{2}= RQ

^{2}+ TR

^{2}– 2 RQ.BR (vi) [Application of pythagoras theorem]

TP

^{2}– TQ

^{2}= PS

^{2}+ TS

^{2}– 2PS.AS

-RQ

^{2}– TR

^{2}+ 2RQ.BR [Subtracting (vi) from (v)]

∴ TP

^{2}– TQ

^{2}= TS

^{2}– TR

^{2}+ PS

^{2}

– RQ

^{2}-2 PS.AS +2 RQ.BR

∴ TP

^{2}– TQ

^{2}= TS

^{2}– TR

^{2}+ PS

^{2}

– PS

^{2}– 2 PS.BR + 2PS.BR [From (i) and (iv)]

∴ TP

^{2}– TQ

^{2}= TS

^{2}– TR

^{2}

∴ TS

^{2}+ TQ

^{2}= TP

^{2}+ TR

^{2}

**Question 1.In ∆ABC, ∠C is an acute angle, seg AD Iseg BC. Prove that: AB ^{2} = BC^{2} + A^{2} – 2 BC × DC. (Textbook pg. no. 44)Given: ∠C is an acute angle, seg AD ⊥ seg BC.To prove: AB^{2} = BC^{2} + AC^{2} – 2BC × DCSolution:**

Proof:

∴ LetAB = c, AC = b, AD = p,

∴ BC = a, DC = x

BD + DC = BC [B – D – C]

∴ BD = BC – DC

∴ BD = a – x

In ∆ABD, ∠D = 90° [Given]

AB

^{2}= BD

^{2}+ AD

^{2}[Pythagoras theorem]

∴ c

^{2}= (a – x)

^{2}+ [P

^{2}] (i)

∴ c

^{2}= a

^{2}– 2ax + x

^{2}+ [P

^{2}]

In ∆ADC, ∠D = 90° [Given]

AC

^{2}= AD

^{2}+ CD

^{2}[Pythagoras theorem]

∴ b

^{2}= p

^{2}+ [X

^{2}]

∴ p

^{2}= b

^{2}– [X

^{2}] (ii)

∴ c

^{2}= a

^{2}– 2ax + x

^{2}+ b

^{2}– x

^{2}[Substituting (ii) in (i)]

∴ c

^{2}= a

^{2}+ b

^{2}– 2ax

∴ AB

^{2}= BC

^{2}+ AC

^{2}– 2 BC × DC

**Question 2.In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. Prove that: AB ^{2} = BC^{2} + AC^{2} + 2 BC × CD. (Textbook pg. no. 40 and 4.1)Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC.To prove: AB^{2} = BC^{2} + AC^{2} + 2BC × CDSolution:**

Proof:

Let AD = p, AC = b, AB = c,

BC = a, DC = x

BD = BC + DC [B – C – D]

∴ BD = a + x

In ∆ADB, ∠D = 90° [Given]

AB

^{2}= BD

^{2}+ AD

^{2}[Pythagoras theorem]

∴ c

^{2}= (a + x)

^{2}+ p

^{2}(i)

∴ c

^{2}= a

^{2}+ 2ax + x

^{2}+ p

^{2}

Also, in ∆ADC, ∠D = 90° [Given]

AC

^{2}= CD

^{2}+ AD

^{2}[Pythagoras theorem]

∴ b

^{2}= x

^{2}+ p

^{2}

∴ p

^{2}= b

^{2}– x

^{2 }(ii)

∴ c

^{2}= a

^{2}+ 2ax + x

^{2}+ b

^{2}– x

^{2}[Substituting (ii) in (i)]

∴ c

^{2}= a

^{2}+ b

^{2}+ 2ax

∴ AB

^{2}= BC

^{2}+ AC

^{2}+ 2 BC × CD

**Question 3.In ∆ABC, if M is the midpoint of side BC and seg AM ⊥seg BC, then prove thatAB**

^{2}+ AC

^{2}= 2 AM

^{2}+ 2 BM

^{2}. (Textbook pg, no. 41)Given: In ∆ABC, M is the midpoint of side BC and seg AM ⊥ seg BC.To prove: AB

^{2}+ AC

^{2}= 2 AM

^{2}+ 2 BM

^{2}Solution:

Proof:

In ∆AMB, ∠M = 90° [segAM ⊥ segBC]

∴ AB2 = AM2 + BM2 (i) [Pythagoras theorem]

Also, in ∆AMC, ∠M = 90° [seg AM ⊥ seg BC]

∴ AC2 = AM2 + MC2 (ii) [Pythagoras theorem]

∴ AB

^{2}+ AC

^{2 }= AM

^{2}+ BM

^{2}+ AM

^{2}+ MC

^{2}[Adding (i) and (ii)]

∴ AB

^{2}+ AC

^{2}= 2 AM

^{2}+ BM

^{2}+ BM

^{2}[∵ BM = MC (M is the midpoint of BC)]

∴ AB

^{2}+ AC

^{2}= 2 AM

^{2}+ 2 BM

^{2}

**Download PDF**

Maharashtra Board Solutions for Class 10-Maths (Part 2): Chapter 2- Pythagoras Theorem

**Chapterwise Maharashtra Board Solutions Class 10 Maths (Part 2) :**

- Chapter 1- Similarity
- Chapter 2- Pythagoras Theorem
- Chapter 3- Circle
- Chapter 4- Geometric Constructions
- Chapter 5- Co-ordinate Geometry
- Chapter 6- Trigonometry
- Chapter 7- Mensuration

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