Haryana State Board HBSE 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.5

Question 1.

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

Solution:

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

⇒ (QR)^{2} =(PR)^{2} – (PQ)^{2}

= (24)^{2}-(10)^{2}

= 576 – 100 = 476

∴ QR = \(\sqrt{476}\)

= \(\sqrt{2 \times 2 \times 7 \times 17}=\sqrt[2]{119}\)

Question 2.

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

Solution:

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

(25)^{2} = 7^{2} + (BC)^{2}

625-49 = (BC)^{2}

576 = (BC)^{2}

\(\sqrt{576}\) =, BC

24 = BC

∴ 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 (Fig. 6.56). Find the distance of the foot of the ladder from the wall.

Solution:

Let ? = x

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

225 -144 = x^{2}

81 = x^{2}

\(\sqrt{81}\) = x

9 = x

∴ x = 9

The distance of the foot of the ladder from the wall = 9mm.

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 right angled triangles, identify the right angles.

Solution:

In a right triangle

h^{2} = p^{2} + b^{2}

(iii) (1.5 cm, 2 cm, 2.5 cm) is the right angle

h^{2} = p^{2} + b^{2}

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

6.25 = 2.25 + 4

6.25 = 6.25

This is satisfied.

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.

Solution:

Let the original height of tree = x

∴ x^{2} = (5)^{2} + (12)^{2}

= 25 + 144 = 169

∴ x = \(\sqrt{169}\)

∴ The original height of the tree = 13 m.

Question 6.

Angle Q and R of a APQR are 25° and 65°. Write which of the following is true:

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

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

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

Solution:

∠QPR = 180° – (25° + 65°)

= 180°-90°

∴ ∠QPR = 90

By Pythagoras Property,

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

i.e. h^{2} = p^{2} + b^{2}

hence, QR^{2} = (PQ)^{2} + (RP)^{2} i.e. (ii) is true.

Question 7.

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

Solultion:

From Fig 6.88

(BC) = (BD) – (CD)

= (41) – (40)

= 1681 – 1600 = 81

∴ BC = \(\) = 9 cm.

Thus the perimeter of the rectangle

= 2 ( l x b) = 2(40 + 9)

= 2 x 49 = 98 cm.

Question 8.

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

Solution:

Area of Rhombus

= \(\frac{1}{2}\) x d_{1} x d_{2} Sq.units.

(where d_{1} and d_{2} are diagonals of rhombus) Side of Rhombus (a) = \(\frac{1}{2} \sqrt{\left(d_{1}\right)^{2}+\left(d_{2}\right)^{2}}\)

Perimeter of Rhombus = 4 x a

Here,

d_{1} = 16 cm

d_{2} = 30 cm

∴ Side of Rhombus (a)

= \(\frac{1}{2} \times \sqrt{(16)^{2}+(30)^{2}}=\frac{1}{2} \times \sqrt{256 \times 900}\)

= \(\frac{1}{2}\) x \(\sqrt{(16)^{2}+(30)^{2}}\) = \(\frac{1}{2}\) x 34 =17 cm z z

∴ Perimeter of Rhombuse

= 4 x a = 4 x 17= 68

∴ Perimeter of Rhombus = 68 cm.