# EXERCISE 6.5

QUESTION 1

**PQR is a triangle right angled at P . If PQ = 10 cm and PR = 24 cm , find QR**

Sol :

By applying Pythagoras theorem in △PQR,

(PQ)^{2} + (PR)^{2} = (RQ)^{2}

(10)^{2} + (24)^{2} = (RQ)^{2}

100 + 576 = (QR)^{2}

676 = (QR)^{2}

QR = 26 cm

QUESTION 2

**ABC is a triangle right angled at C. If AB = 25 cm and AC = 7 cm, find BC.**

By appling Pythagoras theorem in △ABC ,

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

(BC)^{2} = (AB)^{2} – (AC)^{2}

(BC)^{2} = (25)^{2} – (7)^{2}

(BC)^{2 }= 625 – 49

(BC)^{2 }= 576

BC = 24 cm

QUESTION 3

**A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a . Find the distance of the foot of the ladder from the wall .**

Sol :

By applying Pythagoras theorem,

(15)^{2} = (12)^{2} + a

225 = 144 + a^{2}

a^{2} = 225 – 144

a^{2} = 81 m

a = 9 m

Therefore , the distance of the foot of the ladder from the wall is 9 cm

QUESTION 4

**Which of the following can be the sides of a right triangle?**

**(i) 2.5 cm, 6.5 cm, 6 cm**

**(ii) 2 cm, 2 cm, 5 cm**

**(iii) 1.5 cm, 2 cm, 2.5 cm**

**In the case of the right-angled triangles , identify the right angles .**

Sol :

**(i)**

2.5 cm, 6.5 cm, 6 cm

(2.5)^{2} = 6.25

(6.5)^{2} = 42.25

(6)^{2} = 36

It can be observed that ,

36 + 6.25 = 42.25

(6)^{2} + (2.5)^{2} = (6.5)^{2}

The square of the length of one side is the sum of the squares of the lengths of the remaining two sides . Hence , these are the sides of a right- angled triangle . Right angle will be in front of the side of 6.5 cm measure .

(ii) 2 cm , 2 cm , 5 cm

(2)^{2} = 4

(2)^{2} = 4

(5)^{2} = 25

Here , (2)^{2}+ (2)^{2} ≠ (5)^{2}

The square of the length of one side is not equal to the sum of the squares of the remaining two sides . Hence , these sides are not of a right-angled triangle .

(iii) 1.5 cm , 2 cm , 2.5 cm

(1.5)^{2} + (2)^{2} = (2.5)^{2}

The square of the length of one side is the sum of the squares of the lengths of the remaining two sides . Hence , these are the sides of a right-angled triangle .

QUESTION 5

A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Sol :

In the given figure , BC represents the unbroken part of the tree . Point C represents the point where the tree broke and CA represents the broken part of the tree . Triangle ABC , thus formed , is right-angled at B .

Applying Pythagoras theorem in △ABC ,

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

AC^{2} = (5 m)^{2} + (12 m)^{2}

AC^{2} = 25 m^{2}+ 144 m^{2}

AC = 13 m

Thus , original height of the tree = AC + CB = 13 m + 5 m = 18 m

QUESTION 6

Angles Q and R of a △PQR are 25° and 65°.

Write which of the following is true :

(i) PQ^{2} + QR^{2}= RP2

(ii) PQ^{2} + RP^{2}= QR^{2}

(iii) RP^{2} + QR^{2} = PQ^{2}

Sol :

The sum of the measures of all interior angles of a triangle is 180°

∠PQR + ∠PRQ + ∠QPR = 180°

25° + 65° + ∠QPR = 180°

90° + ∠QPR = 180°

∠QPR = 180° – 90°

∠QPR = 90°

Therefore , △PQR is right-angled at point P

Hence , (PR)^{2}+ (PQ)^{2} = (QR)^{2}

Thus , (ii) is true

QUESTION 7

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Sol :

In a rectangle , all interior angles of 90° measure . Therefore , Pythagoras theorem can be applied here .

(41)^{2} = (40)^{2} + x^{2}

1681 = 1600 + x^{2}

x^{2} = 1681 – 1600

x^{2} = 81

x = 9 cm

Perimeter = 2(Length + Breadth)

= 2(x + 40)

= 2(9 + 40)

= 98 cm

QUESTION 8

**The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.**

Sol:

Let ABCD be a rhombus (all sides are of equal length) and its diagonals, AC and BD, are

intersecting each other at point 0. Diagonals in a rhombus bisect each other at 90°. It can be

observed that

By applying Pythagoras theorem in △AOB ,

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

8^{2} + 15^{2} = AB^{2}

64 + 225 = AB^{2}

289 = AB^{2}

AB = 17

Therefore , the length of the sides of rhombus is 17 cm

Perimeter of rhombus = 4 × side of the rhombus

= 4 × 17

= 68 cm