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**Question 1: The sides of certain triangles are given below. Determine which of them are right triangles.**

**(i) 9 cm, 16 cm, 18 cm**

**(ii) 1 cm, 24 cm, 25 cm**

**(iii) 1.4 cm, 4.8 cm, 5 cm**

**(iv) 1.6 cm, 3.8 cm, 4 cm**

**(v) (a â€“ 1) cm, 2âˆša cm, (a + 1) cm**

**Solution:**

A given triangle to be right-angled, if it satisfies Pythagorean Theorem. That is, the sum of the squares of the two smaller sides must be equal to the square of the largest side.

(i) 9 cm, 16 cm, 18 cm

Longest side = 18

Now (18)Â² = 324

and (9)Â² + (16)Â² = 81 + 256 = 337

324 â‰ 337

It is not a right triangle.

(ii) 1 cm, 24 cm, 25 cm

Longest side = 25 cm

(25)Â² = 625

and (7)Â² x (24)Â² = 49 + 576 = 625

625 = 625

It is a right triangle.

(iii) 1.4 cm, 4.8 cm, 5 cm

Longest side = 5 cm

(5)Â² = 25

and (1.4)Â² + (4.8)Â² = 1.96 + 23.04 = 25.00 = 25

25 = 25

It is a right triangle.

(iv) 1.6 cm, 3.8 cm, 4 cm

Longest side = 4 cm

(4 )Â² = 16

and (1.6)Â² + (3.8)Â² = 2.56 + 14.44 = 17.00 = 17

16 â‰ 17

It is not a right triangle.

(v) (a- 1) cm, 2âˆša cm, (a + 1) cm

Longest side = (a + 1) cm

(a + 1)Â² = aÂ² + 2a + 1

and (a â€“ 1)Â² + (2 âˆša )Â² = aÂ² â€“ 2a + 1 + 4a = aÂ² + 2a + 1

aÂ² + 2a + 1 = aÂ² + 2a + 1

It is a right triangle.

**Question 2: A man goes 80 m due east and then 150 m due north. How far is he from the starting point?**

**Solution:**

A man goes 80 m from O to east side and reaches A, then he goes 150 m due north from A and reaches B.

Draw a figure based on given instructions:

From right âˆ†OAB,

By Pythagoras Theorem:

OB^{2} = OA^{2}+ AB^{2}

= (80)Â² + (150)Â²

= 6400 + 22500

= 28900

or OB = âˆš28900 = 170

Man is 170 m away from the starting point.

**Question 3: A man goes 10 m due south and then 24 m due west. How far is he from the starting point?**

**Solution:**

A man goes 10 m due south from O and reaches A and then 24 m due west from A and reaches B.

Draw a figure based on given instructions:

From right âˆ†OAB,

By Pythagoras Theorem:

OB^{2} = OA^{2}+ AB^{2}

= (10)Â² + (24)Â²

= 676

or OB = 26

Man is 26 m away from the starting point.

**Question 4: A 13-m-long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.**

**Solution:**

Height of the window = 12 m

Length of a ladder = 13 m

In the figures,

Let AB is ladder, A is window of building AC

By Pythagoras Theorem:

AB^{2} = AC^{2} + BC^{2}

(13)Â² = (12)Â² + xÂ²

169 = 144 + xÂ²

xÂ² = 169 â€“ 144 = 25

or x = 5

Distance between foot of ladder and building = 5 m.

**Question 5: A ladder is placed in such a way that its foot is at a distance of 15 m from a wall and its top reaches a window 20 m above the ground. Find the length of the ladder.**

**Solution:**

Height of window AC = 20 m

Let length of ladder AB = x m

Distance between the foot of the ladder and the building (BC) = 15 m

In the figure:

By Pythagoras Theorem:

AB^{2} = AC^{2} + BC^{2}

xÂ² = 20Â² + 15Â²

= 400 + 225

= 625

or x = 25

Length of ladder is 25 m

**Question 6: Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.**

**Solution:**

Height of first pole AB = 9 m and

Height of second pole CD = 14 m

Let distance between their tops CA = x m

From A, draw AE || BD meeting CD at E.

Then EA = DB = 12 m CE = CD â€“ ED = CD â€“ AB = 14-9 = 5 m

In right âˆ†AEC,

ACÂ² = AEÂ² + CEÂ²

= 122 + 52

= 144 + 25

= 169

or AC = 13

Distance between pole’s tops is 13 m

**Question 7: A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?**

**Solution:**

Length of wire = AC = 24 m

Height of the pole = AB = 18 m

Let Distance between the base of the pole and other end of the wire = BC = x m

In right âˆ†ABC,

By Pythagoras Theorem:

AC^{2} = AB^{2} + BC^{2}

(24)Â² = (18)Â² + xÂ²

576 = 324 + xÂ²

xÂ² = 576 â€“ 324 = 252

or x = 6v7

BC is 6v7m

**Question 8: In the given figure, O is a point inside a âˆ†PQR such that âˆ POR = 90Â°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that âˆ†PQR is right-angled.**

**Solution:**

In âˆ†PQR, O is a point in it such that

OP = 6 cm, OR = 8 cm and âˆ POR = 90Â°

PQ = 24 cm, QR = 26 cm

Now,

In âˆ†POR, âˆ O = 90Â°

PRÂ² = POÂ² + ORÂ²

= (6)Â² + (8)Â²

= 36 + 64

= 100

PR = 10

Greatest side QR is 26 cm

QRÂ² = (26)Â² = 676

and PQÂ² + PRÂ² = (24)Â² + (10)Â²

= 576 + 100

= 676

Which implies, 676 = 676

QRÂ² = PQÂ² + PRÂ²

âˆ†PQR is a right angled triangle and right angle at P.

**Question 9: âˆ†ABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.**

**Solution:**

In isosceles âˆ†ABC, AB = AC = 13 cm

Consider AL is altitude from A to BC and AL = 5 cm

Now, in right âˆ†ALB

AB^{2} = AL^{2} + BL^{2}

(13)Â² = (5)Â² + BLÂ²

169 = 25 + BLÂ²

BLÂ² = 169 â€“ 25 = 144

or BL = 12

Since L is midpoint of BC, then

BC = 2 x BC = 2 x 12 = 24

BC is 24 cm

**Question 10: Find the length of altitude AD of an isosceles âˆ†ABC in which AB = AC = 2a units and BC = a units.**

**Solution:**

In an isosceles âˆ†ABC in which AB = AC = 2a units, BC = a units

AD is the altitude. Therefore, D is the midpoint of BC

=> BD = a/2

We have two right triangles: Î”ADB and Î”ADC

By Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2}

(2a)^{2} = (a/2)^{2} + AD^{2}

## R S Aggarwal Solutions for Class 10 Maths Chapter 7 Triangles Exercise 7D

Class 10 Maths Chapter 7 Triangles Exercise 7D is based on the topics:

-Pythagoras Theorem

-Some important results based upon Pythagoras Theorem