Chapter 6: Triangles

Solution: –

(c) 18 m

From the question it is given that,

The top of a broken tree touches the ground at a distance of 12 m from its base

The broken height of the tree = 5 m

By using Pythagoras theorem,

Hypotenuse² = Base² + Height²

Hypotenuse² = 12² + 5²

Hypotenuse² = 144 + 25

Hypotenuse² = 169

Hypotenuse = √169

Hypotenuse = 13

So, the total height of tree = 5 + 13 = 18 m

Solution: –

(b) a scalene triangle only

A scalene triangle is a triangle that has three unequal sides.

Solution:-

(b) 5 m

Consider PQ is the tree of height 7m and RS is the tree of height 4 m.

So, consider the triangle QRE, from the Pythagoras theorem,

QR² = QE² + ER²

QR² = 3² + 4²

QR² = 9 + 16

QR² = 25

QR = √25

QR = 5

Therefore, the distance between the top of the two trees is 5m.

Solution: –

(c) Obtuse angled triangle

We know that, sum of interior angles of triangle is equal to 180^o.

Let us assume the 3^rd angle be Q,

Then, 40^o + 40^o + Q = 180^o

80^o + Q = 180^o

Q = 180 – 80

Q = 100^o

An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles. Since a triangle’s angles must sum to 180^o.

Solution:-

(c) Equilateral

In an equilateral triangle, each angle has measure 60^o.

Solution:-

(d) 142 cm

Consider the rectangle PQRS,

Given, length of rectangle PQ = 60 cm, Diagonal of the rectangle = 61 cm.

To find out the height of the rectangle, consider the right angled triangle PQR.

From the Pythagoras theorem, PR² = PQ² + RQ²

61² = 60² + RQ²

3721 = 3600 + RQ²

RQ² = 3721 – 3600

RQ² = 121

RQ = √121

RQ = 11 cm

Then, the perimeter of the rectangle PQRS = 2 (Length + Breadth)

= 2 (60 + 11)

= 2 (71)

= 142 cm

Solution: –

(a) 40^o

Given, In ∆PQR, PQ = QR so it is an isosceles triangle.

Then, ∠P = ∠R

So, let us assume two angles be x

x + x + 100^o = 180^o

2x = 180^o – 100^o

2x = 80^o

x = 80^o/2

x = 40^o

Therefore, x = ∠P = ∠R = 40^o

Solution: –

(d) Difference of any two sides of a triangle is greater than the third side.

The difference of the lengths of any two sides of a triangle is always smaller than the length of the third side.

Solution:-

(b) 80^o

We know that, the exterior angle is equal to sum of opposite interior angles.

So, ∠ACD = ∠A + ∠B

As ΔACB is an isosceles triangle with AC = BC

Therefore, ∠A must be equal to ∠B

∠ACD = 40^o + 40^o

= 80^o

Solution: –

(c) 3 cm and 16 cm

From the question it is given that, the length of two sides of a triangle are 7 cm and 9 cm.

Let us assume the length of the third side of the triangle be ‘P’.

We Know that, the sum of the two sides of the triangle is greater than the third side.

So, 7 + 9 > P

16 > P

Now, difference between two sides = 9 – 7 = 2

Therefore, the third side is greater than 2 and smaller than 16.

i.e. 3 cm and 16 cm