## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.5

Question 1.

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

Solution:

In ΔPQR, ∠P = 90°

PQ = 10 cm, PR = 24 cm

Using Pythagoras Theorem,

QR^{2} = PQ^{2} + PR^{2} = 10^{2} + 24^{2} = 100 + 576 = 676 = (26)^{2}

QR = 26 cm

Question 2.

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

Solution:

In ΔABC, ∠C = 90°

AB = 25 cm and AC = 7 cm

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

⇒ (25)^{2} = (7)^{2} + BC^{2}

⇒ 625 = 49 + BC^{2}

⇒ BC^{2} = 625 – 49 = 576 = (24)^{2}

⇒ BC = 24 cm

Question 3.

Find the value of x in each of the following figures. All measurements are in centimeters.

Solution:

(i) In ΔABC, ∠C = 90°

AB = 29, BC = 21, AC = x

AB^{2} = AC^{2} + BC^{2} (Using Pythagoras theorem)

⇒ (29)^{2} = (21)^{2} + x^{2}

⇒ 841 = 441 + x^{2}

⇒ x^{2} = 841 – 441 = 400 = (20)^{2}

x = 20

(ii) In ΔABC, AD ⊥ BC

AB = 37, AC = 37

In right ΔABD

AB^{2} = AD^{2} + BD^{2} (Pythagoras Theorem)

⇒ (37)^{2} = (12)^{2} + BD^{2}

⇒ 1369 = 144 + BD^{2}

⇒ BD^{2} = 1369 – 144

⇒ BD^{2} = 1225 = (35)^{2}

⇒ BD = 35 cm

But AD bisects BC at D

BC = 2 × BD

⇒ x = 2 × 35 = 70 cm

(iii) In right ΔABE, ∠B = 90°

AE^{2} = AB^{2} + BE^{2} (Pythagoras Theorem)

⇒ (13)^{2} = 12^{2} + BE^{2}

⇒ 169 = 144 + BE^{2}

⇒ BE^{2} = 169 – 144 = 25 = (5)^{2}

⇒ BE = 5 cm

Similarly in right ΔCDE, ∠D = 90°

CE^{2} = CD^{2} + ED^{2}

⇒ 10^{2} = 6^{2} + ED^{2}

⇒ 100 = 36 + ED^{2}

⇒ ED^{2} = 100 – 36 = 64 = (8)^{2}

⇒ ED = 8 cm

Now, BD = x = BE + ED = 5 + 8 = 13 cm

Question 4.

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

(i) 4 cm, 5 cm, 7 cm

(ii) 1.5 cm, 2 cm, 2.5 cm

(iii) 7 cm, 5.6 cm, 4.2 cm

Solution:

(i) Sides are : 4 cm, 5 cm, 7 cm

(Longest side)^{2} = (7)^{2} = 49

Sum of squares of other two sides

= 4^{2} + 5^{2}

= 16 + 25

= 41

49 ≠ 41

These are not the sides of the right triangle.

(ii) Sides are : 1.5 cm, 2 cm, 2.5 cm

(Longest side)^{2} = (2.5)^{2} = 6.25

Sum of the squares of the other two sides

= 1.5^{2} + 2^{2}

= 2.25 + 4

= 6.25

6.25 = 6.25

There are the sides of a right triangle

and right angle is opposite to the side 2.5 cm

(iii) Sides are : 7 cm, 5.6 cm, 4.2 cm

(Longest side)^{2} = 7^{2} = 49

Sum of the squares of the other two sides

= (5.6)^{2} + (4.2)^{2}

= 31.36 + 17.64

= 49

49 = 49

These are the sides of a right triangle

Right angle is opposite to the side 5.6 cm

Question 5.

A 15 m long ladder reaches a window 12 m high from the ground on placing it against a wall. How far is the foot of the ladder from the wall?

Solution:

Let length of ladder AB = 15 m

and height of wind AC = 12 m

BC is the distance from wall to the foot of ladder

In right ΔABC, ∠C = 90°

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

⇒ (15)^{2} = 12^{2} + BC^{2}

⇒ 225 = 144 + BC^{2}

⇒ BC^{2} = 225 – 144 = 81 = (9)^{2}

⇒ BC = 9 m

Distance of the foot of the ladder and the wall = 9 m

Question 6.

Find the area and the perimeter of the rectangle whose length is 15 cm and the length of one diagonal is 17 cm.

Solution:

Length of a rectangle = 15 cm

and length of its one diagonal = 17 cm

In right ΔABC

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

⇒ 17^{2} = 15^{2} + BC^{2}

⇒ 289 = 225 + BC^{2}

⇒ BC^{2} = 289 – 225 = 64 = (8)^{2}

Breadth = 8 cm

Area = Length × Breadth = 15 × 8 = 120 cm^{2}

and perimeter = 2(Length + Breadth)

= 2(15 + 8)

= 2 × 23

= 46 cm

Question 7.

If the diagonals of a rhombus measure 10 cm and 24 cm, find its perimeter.

Solution:

Length of diagonals of a rhombus are 10 cm and 24 cm

The diagonals of a rhombus bisect each other at right angles.

O is the mid-point of AC and BD

AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm

and BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm

Now in right ΔAOB

AB^{2} = AO^{2} + BO^{2} (Pythagoras Theorem)

⇒ AB^{2} = 12^{2} + 5^{2}

⇒ AB^{2} = 144 + 25 = 169 = (13)^{2}

⇒ AB = 13 cm

Perimeter of rhombus = 4 × Side = 4 × 13 = 52 cm

Question 8.

The side of a rhombus is 5 cm. If the length of one diagonal of the rhombus is 8 cm, then find the length of the other diagonal.

Solution:

Side of a rhombus = 5 cm

Length of one diagonal (AC) = 8 cm

The diagonal of a rhombus bisect

each other at right angles.

AO = OC = 4 cm, BO = OD

In right ΔAOB,

AB^{2} = AO^{2} + BO^{2}

5^{2} = 4^{2} + BO^{2}

25 = 16 + BO^{2}

BO^{2} = 25 – 16 = 9 = (3)^{2}

BO = 3 cm

and diagonal BD = 2 × 3 = 6 cm