Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.3

1. Find the length of the unknown side in each of following figures

Question (i).

Answer:

Take a = 3 cm, b = 4 cm and unknown side = c

By Pythagoras Theorem

c^{2} = a^{2} + b^{2}

c^{2} = (3)^{2} + (4)^{2}

c^{2} = 9 + 16

c^{2} = 25

∴ c = \(\sqrt{25}\)

c = 5

Thus, the length of unknown side = 5 cm.

Question (ii).

Answer:

Take a = 15 cm, b = 20 cm

By Pythagoras Theorem

c^{2} = a^{2} + b^{2}

∴ c^{2} = (15)^{2} + (20)^{2}

c^{2} = 225 + 400

c^{2} = 625

∴ c = \(\sqrt{625}\)

c = 25

Thus, the length of unknown side = 5 cm

2. Which of the following can be the sides of a right triangle ?

(i) 4 cm, 5 cm, 7 cm

(ii) 1.5 cm, 2 cm, 2.5 cm

(iii) 2 cm, 2 cm, 5 cm

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

Solutions:

(i) Let in ΔABC, the longest side is AB = 7 cm

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

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

= 16 + 25 = 41

(BC)^{2} + (AC)^{2} = 41

Also AB^{2} = (7)^{2} = 49

Since AB^{2} ≠ (BC)^{2} + (AC)^{2}

∴ The triangles with the given sides is not a right triangle.

(ii) Let in ΔABC, the longest side is AB = 2.5 cm

(AB)^{2} = (2.5)^{2} = 6.25 ….(1)

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

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

= 2.25 + 4

= 6.25

∴ (BC)^{2} + (AC)^{2} = 6.25 ….(2)

From (1) and (2)

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

Therefore, the given triangle is a right triangle.

The angle opposite to the longest side is right angle.

(iii) Let in ΔABC, the longest side is AB = 5 cm

(AB)^{2} = (5)^{2}

(AB)^{2} = 25 ….(1)

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

(BC)^{2} + (AC)^{2} = 4 + 4

(BC)^{2} + (AC)^{2} = 8 …..(2)

From (1) and (2)

(AB)^{2} ≠ (BC)^{2} + (AC)^{2}

Therefore the triangle whose sides are 5 cm, 2 cm and 2 cm is not a right triangle.

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

Solution:

Let ABCD be a rectangle with length AB = 15 cm and diagonal AC = 17 cm.

In ΔABC, ∠B = 90° (Each angle of a rectangle)

By Pythagoras Theorem,

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

(17)^{2} = (15)^{2} + (BC)^{2}

289 = 225 + (BC)^{2}

(BC)^{2} = 289 – 225 = 64

BC = 8 cm

Area of rectangle ABCD

= AB × BC

= 15 cm × 8 cm

= 120 cm^{2}

Perimeter of rectangle ABCD = 2(AB + BC)

= 2(15 cm + 8 cm)

= 2(23 cm) = 46 cm^{2}

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

Solution:

Let AB be the ladder and BC be the distance of the foot of the ladder from the wall then AB = 15 m and AC = 12 m

By Pythagoras Theorem,

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

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

225 = BC^{2} + 144

BC^{2} = 225 – 144

BC^{2} = 81

BC =9

Hence the distance of the foot of the ladder from the wall is 9 m.

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

Solution:

Let ABCD be a rhombus with side AB = 5 cm and diagonal AC = 8 cm

Let diagonal AC and BD bisect each other at O.

Then OA = OC = \(\frac {8}{2}\) cm = 4 cm

The diagonals of a rhombus bisect each other at right angle

∴ In right angled ΔAOB,

AO = 4 cm, AB = 5 cm

By Pythagoras Theorem,

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

(4)^{2} + OB^{2} = (5)^{2}

16 + OB^{2} = 25

OB^{2} = 25 – 16 = 9

OB = 3 cm

Diagonal BD = 2 × OB = 2 × 3cm = 6cm

Therefore other diagonal of rhombus = 6 cm.

6. A right triangle is isosceles. If the square of the hypotenuse is 50 m, what is length of each of its sides ?

Solution:

Let ΔABC is a right isosceles triangle in which (AC)^{2} = 50 m and AB = AC

∴ By Pythagoras Theorem

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

∴ AB^{2} + AB^{2} = AC^{2}

2AB^{2} = 50

AB^{2} = 25

AB = 5

Therefore length of each equal side = 5m.

7. ΔABC is a triangle right angled at C if AC = 8 cm and BC = 6 cm, find AB.

Solution:

In right angled triangle ABC right angle at C

AC = 8 cm and BC = 6 cm

By Pythagoras Theorem

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

AB^{2} = (8)^{2} + (6)^{2}

AB^{2} = 100

AB = 10 cm.

8. State whether the following triplets are Pythagorean or not.

Question (i).

(5, 7, 12)

Solution:

Let a = 5, b = 1, c = 12

∴ c^{2} = (12)^{2} = 144

a^{2} + b^{2} = (5)^{2} + (7)^{2}

= 25 + 49 = 74

∴ a^{2} + b^{2} ≠ c^{2}

∴ (5, 7, 12) is not a pythagorean triplet.

Question (ii).

(3, 4, 5)

Solution:

Let a = 3, b = 4, c = 5

∴ a^{2} + b^{2} = (3)^{2} + (4)^{2}

= 9 + 16 = 25

c^{2} = (5)^{2} = 25

∴ c^{2} = a^{2} + b^{2}

∴ (3, 4, 5) is a pythagorean triplet

Question (iii).

(8, 9, 10)

Solution:

Let a = 8, b = 9, c = 10

∴ a^{2} + b^{2 }= (8)^{2} + (9)^{2}

= 64 + 81 = 145

c^{2} = (10)^{2} = 100

c^{2} ≠ a^{2} + b^{2}

Therefore (8, 9, 10) is not a pythagorean triplet.

Question (iv).

(5, 12, 13)

Solution:

Let a = 5, b = 12, c = 13

∴ a^{2} + b^{2} = (5)^{2} + (12)^{2}

= 25 + 144 = 169

c^{2} = (13)^{2} = 169

a^{2} + b^{2} ≠ c^{2}

Therefore (5, 12, 13) is a pythagorean triplet.

9. Multiple Choice Questions :

Question (i).

In a ΔABC, if ∠A = 40° and ∠B = 55° then ∠C is

(a) 75°

(b) 80°

(c) 95°

(d) 85°

Answer:

(d) 85°

Question (ii).

If the angles of a triangle are 35°, 35° and 110°, then it is

(a) an isosceles triangle

(b) an equilateral triangle

(c) a scalene triangle

(d) right angled triangle

Answer:

(a) an isosceles triangle

Question (iii).

A triangle can have two

(a) right angles

(b) obtuse angles

(c) acute angles

(d) straight angles

Answer:

(c) acute angles

Question (iv).

A triangle whose angles measure 35°, 55° and 90° is

(a) acute angled

(b) right angled

(c) obtuse angled

(d) isosceles

Answer:

(b) right angled

Question (v).

A triangle is not possible whose angles measure

(a) 40°, 65°, 75°

(b) 50°, 56°, 74°

(c) 72°, 63°, 45°

(d) 67°, 42°, 81°

Answer:

(d) 67°, 42°, 81°

Question (vi).

A triangle is not possible with sides of lengths (in cm)

(a) 6, 4, 10

(b) 5, 3, 7

(c) 7, 8, 9

(d) 3.6, 5.4, 8

Answer:

(a) 6, 4, 10

Question (vii).

In a right angled triangle, the length of two legs are 6 cm and 8 cm. The length of the hypotenuse is

(a) 14 cm

(b) 10 cm

(c) 11 cm

(d) 12 cm

Answer:

(b) 10 cm