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 + a2

a2 = 225 – 144

a2 = 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 ,

AC2 = BC2 + AB2

AC2 = (5 m)2 + (12 m)2

AC2 = 25 m2+ 144 m2

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) PQ2 + QR2= RP2
(ii) PQ2 + RP2= QR2
(iii) RP2 + QR2 = PQ2

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 + x2

1681 = 1600 + x2

x2 = 1681 – 1600

x2 = 81

x = 9 cm

= 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 ,

OA2 + OB2 = AB2

82 + 152 = AB2

64 + 225 = AB2

289 = AB2

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

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