Question

The sides of triangle are in the ratio 10:17:21 and its perimeter is 96cm.find the area of the ∆.​

Answer 1

Let the sides be 10x, 17x and 21x.

$10x + 17x + 21x = 96$

$48x = 96$

$x = \frac{96}{48}$

$= 2$

$2s = 96$

$s = \frac{96}{2} = 48$

Sides of the triangle are

$10x = 10 \times 2 = 20$

$17x = 17 \times 2 = 34$

$21x = 21 \times 2 = 42$

Area

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{48(48 - 20)(48 - 34)(48 - 42)}$

$= \sqrt{48 \times 28 \times 14 \times 4}$

$= 112 \sqrt{3}$

Therefore, the area is

$112 \sqrt{3}$

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