Class 9 Maths Question Answers - Herons Formula

Q1. What is the area of an equilateral triangle whose side is 2 cm?

Solution: Area of an equilateral triangle

Q2. Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm.

Q3. If the area of an equilateral is √3/4 cm² then find the side of the triangle.

Q4.  If the perimeter of an equilateral triangle is 180 cm. Then its area will be ? 

Solution. Given, Perimeter = 180 cm

3a = 180 (Equilateral triangle)

a = 60 cm

Semi-perimeter = 180/2 = 90 cm

Now as per Heron’s formula,

= √s (s-a)(s-b)(s-c)

In the case of an equilateral triangle, a = b = c = 60 cm

Substituting these values in the Heron’s formula, we get the area of the triangle as:

A = √[90(90 – 60)(90 – 60)(90 – 60)]

= √(90× 30 × 30 × 30)

A = 900√3 cm²

Q5. The sides of a triangle are 122 m, 22 m and 120 m respectively. The area of the triangle is ? 

Solution: Given,

a = 122 m

b = 22 m

c = 120 m

Semi-perimeter, s = (122 + 22 + 120)/2 = 132 m

Using heron’s formula:

= √s (s-a)(s-b)(s-c)

= √[132(132 – 122)(132 – 22)(132 – 120)]

= √(132 × 10 × 110 × 12)

= 1320 sq.m

Q6.  The sides of a triangle are in the ratio 12: 17: 25 and its perimeter is 540 cm. The area is:

Solution: The ratio of the sides is 12: 17: 25

Perimeter = 540 cm

Let the sides of the triangle be 12x, 17x and 25x.

Hence,

12x + 17x + 25x = 540 cm

54x = 540 cm

x = 10

Therefore,

a = 12x = 12 × 10 = 120

b = 17x = 17 × 10 = 170

c = 25x = 25 × 10 = 250

Semi-perimeter, s = 540/2 = 270 cm

Using Heron’s formula:

$𝐴=\; \surd s\; (s-a)(s-b)(s-c)\surd s\; (s-a)(s-b)(s-c)\surd s\; (s-a)(s-b)(s-c)\surd s\; (s-a)(s-b)(s-c)$

= √[270(270 – 120)(270 – 170)(270 – 250)]

= √(270 × 150 × 100 × 20)

= 9000 sq.cm

Q7. Find the area of an equilateral triangle having side length equal to √3/4 cm (using Heron’s formula)

Solution: Here, a = b = c = √3/4

Semiperimeter = (a + b + c)/2 = 3a/2 = 3√3/8 cm

Using Heron’s formula,

$𝐴=\; \surd s\; (s-a)(s-b)(s-c)\surd s\; (s-a)(s-b)(s-c)\surd s\; (s-a)(s-b)(s-c)$

= √[(3√3/8) (3√3/8 – √3/4)(3√3/8 – √3/4)(3√3/8 – √3/4)]

= 3√3/64 sq.cm

Q8. The base of a right triangle is 8 cm and the hypotenuse is 10 cm. Its area will be

Solution: Given: Base = 8 cm and Hypotenuse = 10 cm

Hence, height = √[(10² – 8²) = √36 = 6 cm

Therefore, area = (½)×b×h = (½)×8×6 = 24 cm².

Q9. The area of an isosceles triangle having a base 2 cm and the length of one of the equal sides 4 cm, is

Solution: Given that a = 2 cm, b= c = 4 cm 

s = (2 + 4 + 4)/2 = 10/2 = 5 cm

By using Heron’s formula, we get:

A =√[5(5 – 2)(5 – 4)(5 – 4)] = √[(5)(3)(1)(1)] = √15 cm².

Q10. The perimeter of an equilateral triangle is 60 m. The area is

Solution:

Given: Perimeter of an equilateral triangle = 60 m

3a = 60 m (As the perimeter of an equilateral triangle is 3a units)

a = 20 cm.

We know that area of equilateral triangle = (√3/4)a² square units

A = (√3/4)20²

A = (√3/4)(400) = 100√3 m².