Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 2 Pythagoras Theorem.

## Practice Set 2.1 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem

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 = \(\sqrt { MQ\times QP }\) [Taking square root of both sides]

= \(\sqrt { 9\times 4 } \)

= 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 = \(\frac { 100 }{ 8 } \)

= 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 = \(\frac { 1 }{ 2 } \) RP [Side opposite to 30°]

∴6 = \(\frac { 1 }{ 2 } \) RP

∴ RP = 6 × 2 = 12 units

Also, PS = \(\frac{\sqrt{3}}{2}\) RP [Side opposite to 60°]

= \(\frac{\sqrt{3}}{2}\) × 12

= \(6 \sqrt{3}\) units

∴ RP = 12 units, PS = 6 \(\sqrt { 3 }\) 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 = \(\frac { 90° }{ 2 } \) [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} = \(\frac { 100 }{ 2 } \)

∴x^{2} = 50

∴ x = \(\sqrt { 50 }\) [Taking square root of both sides]

= \(=\sqrt{25 \times 2}=5 \sqrt{2}\)

∴side of square is 5\(\sqrt { 2 }\) cm.

= 4 × 5 \(\sqrt { 2 }\)

∴ Perimeter of square = 20 \(\sqrt { 2 }\) 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 = \(\frac { 144 }{ 8 } \)

∴ EG = 18 units

ii. In ∆FGD, ∠FGD = 90° [Given]

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

= 122 + 82 = 144 + 64

= 208

∴ FD = \(\sqrt { 208 }\) [Taking square root of both sides]

∴ FD = 4 \(\sqrt { 13 }\) units

iii. In ∆EGF, ∠EGF = 90° [Given]

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

= 182 + 122 = 324 + 144

= 468

∴ EF = \(\sqrt { 468 }\) [Taking square root of both sides]

∴ EF = 6 \(\sqrt { 13 }\) 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 = \(\sqrt { 1369 }\) [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 = \(\frac { 1 }{ 2 } \) 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.