Class 7 The Triangle and its Properties Ex 6.5

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

Solution:-

Let us draw a rough sketch of a right-angled triangle.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Image 37

By the rule of Pythagoras’ Theorem,

QR2 = PQ2 + PR2

QR2 = 102 + 242

QR2 = 100 + 576

QR2 = 676

QR = √676

QR = 26 cm

Hence, the length of the hypotenuse QR = 26 cm

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

Solution:-

Let us draw a rough sketch of the right-angled triangle.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Image 38

By the rule of Pythagoras’ Theorem,

AB2 = AC2 + BC2

252 = 72 + BC2

625 = 49 + BC2

BC2 = 625 – 49

BC2 = 576

BC = √576

BC = 24 cm

Hence, the length of the BC = 24 cm

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.

Solution:-

By the rule of Pythagoras’ Theorem,

152 = 122 + a2

225 = 144 + a2

a2 = 225 – 144

a2 = 81

a = √81

a = 9 m

Hence, the length of a = 9 m

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, 2cm, 2.5 cm

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

Solution:-

(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm

Let us assume the largest value is the hypotenuse side, i.e., b = 6.5 cm.

Then, by Pythagoras’ theorem,

b2 = a2 + c2

6.52 = 2.52 + 62

42.25 = 6.25 + 36

42.25 = 42.25

The sum of squares of two sides of the triangle is equal to the square of the third side,

∴ the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side, 6.5 cm.

(ii) Let a = 2 cm, b = 2 cm, c = 5 cm

Let us assume the largest value is the hypotenuse side, i.e. c = 5 cm.

Then, by Pythagoras’ theorem,

c2 = a2 + b2

52 = 22 + 22

25 = 4 + 4

25 ≠ 8

The sum of squares of two sides of the triangle is not equal to the square of the third side,

∴ the given triangle is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

Let us assume the largest value is the hypotenuse side, i.e., b = 2.5 cm.

Then, by Pythagoras’ theorem,

b2 = a2 + c2

2.52 = 1.52 + 22

6.25 = 2.25 + 4

6.25 = 6.25

The sum of squares of two sides of the triangle is equal to the square of the third side,

∴ the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side 2.5 cm.

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.

Solution:-

Let ABC is the triangle and B be the point where the tree is broken at the height of 5 m from the ground.

Treetop touches the ground at a distance of AC = 12 m from the base of the tree,

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Image 40

By observing the figure, we came to conclude that a right-angle triangle is formed at A.

From the rule of Pythagoras’ theorem,

BC2 = AB2 + AC2

BC2 = 52 + 122

BC2 = 25 + 144

BC2 = 169

BC = √169

BC = 13 m

Then, the original height of the tree = AB + BC

= 5 + 13

= 18 m

6. Angles Q and R of a ΔPQR are 25o and 65o.

Write which of the following is true:

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Image 41

Solution:-

Given that ∠Q = 25o, ∠R = 65o

Then, ∠P =?

We know that sum of the three interior angles of a triangle is equal to 180o.

∠PQR + ∠QRP + ∠RPQ = 180o

25o + 65o + ∠RPQ = 180o

90o + ∠RPQ = 180o

∠RPQ = 180 – 90

∠RPQ = 90o

Also, we know that the side opposite to the right angle is the hypotenuse.

∴ QR2 = PQ2 + PR2

Hence, (ii) is true.

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

Solution:-

Let ABCD be the rectangular plot.

Then, AB = 40 cm and AC = 41 cm , BC =?

According to Pythagoras’ theorem,

From the right angled \(\triangle\) ABC,

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

\(41^2 = 402 + \text{BC}^2\)

\(\text{BC}^2 = 41^2 – 40^2\)

\(\text{BC}^2 = 1681 – 1600\)

\(\text{BC}^2 = 81\)

\(\text{BC} = \sqrt{81}\)

\(\text{BC} = 9\) cm

Hence, the perimeter of the rectangle plot = 2 (length + breadth)

length = \(40\) cm, breadth = 9\) cm

\(= 2(40 + 9)\)

\(= 2 × 49\)

\(= 98\) cm

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

Solution:-

Let PQRS be a rhombus, all sides of the rhombus have equal length, and its diagonal PR and SQ are intersecting each other at point O. Diagonals in the rhombus bisect each other at 90o.

\(\text{PO} = ({\text{PR}\over 2})\)

\(= \dfrac{16}{2}\)

= 8 cm\(\text{SO} = ({\text{SQ}\over 2})\)

=\( 30\over2\)

= 15 cm

In \(\triangle \text{POS}\) and apply the Pythagoras theorem,

\(\text{PS}^2 = \text{PO}^2 + \text{SO}^2\)

\(\text{PS}^2 = 82 + 152\)

\(\text{PS}^2 = 64 + 225\)

\(\text{PS}^2 = 289\)

\(\text{PS} = \sqrt{289}\)

\(\text{PS} = 17\) cm

Hence, the length of the side of the rhombus is 17 cm

The perimeter of the rhombus = 4 × side of the rhombus

= 4 × 17

= 68 cm

∴ The perimeter of the rhombus is 68 cm.