# Heron’s Formula - Class 9 - Mathematics

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## Extra Questions - Heron’s Formula | NCERT | Mathematics | Class 9

Cube root of the perimeter of a triangular field is $7$ $\mathrm{m}$ and its sides are in the ratio $14:15:20$. Find the area of the triangle.

The sides of the triangular field are given in the ratio **14:15:20**, and we can denote the sides as $14x$, $15x$, and $20x$ respectively.

Given that the cube root of the perimeter is $7$ $\mathrm{m}$, the perimeter itself is: $$ (7)^3 = 343 \text{ m} $$

The sum of the sides, which is the perimeter, is represented by: $$ 14x + 15x + 20x = 343 $$ Adding the sides, we get: $$ 49x = 343 $$ From the above, solving for $x$ yields: $$ x = \frac{343}{49} = 7 $$

Thus, the sides of the triangle are:

First side $(a) = 14x = 14 \times 7 = 98 \text{ m}$

Second side $(b) = 15x = 15 \times 7 = 105 \text{ m}$

Third side $(c) = 20x = 20 \times 7 = 140 \text{ m}$

To find the area of the triangle, we use **Heron's Formula**, where:
$$
s = \frac{a + b + c}{2}
$$
Plugging the sides into the formula gives us the semi-perimeter $s$:
$$
s = \frac{98 + 105 + 140}{2} = 171.5 \text{ m}
$$

The area $A$ of the triangle using Heron's formula is: $$ A = \sqrt{s(s-a)(s-b)(s-c)} $$ Substituting the values we get: $$ A = \sqrt{171.5(171.5-98)(171.5-105)(171.5-140)} $$ Calculating further: $$ A = \sqrt{171.5 \times 73.5 \times 66.5 \times 31.5} \approx \sqrt{2016283} \text{ m}^2 $$

Thus, the **area of the triangle** is approximately:
$$
A = 2016283 \text{ m}^2
$$
This calculation provides the required area based on the perimeter and given side ratios.

If 8, 15, and $c$ form a Pythagorean triplet, then the value of $c$ is

A) 14

B) 18

C) 17

D) $\mathbf{16}$

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