Question

the perimeter of a triangular field is 840 M and its sides are in the ratio 6 ratio 7 ratio 8 find the area of the field

Answer 1

Given that perimeter = 840
sides are 6x,7x,8x

therefore

6x+7x+8x= 840
21x= 840
x= 840/21
x= 40

then it's sides are 6×40=240
7×40=280
8×40=320

S = A+B+C/2
= 280+240+320/2
=840/2
= 420

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

$= \sqrt{420(420 - 240)(420 - 280)(420 - 320)}$

$= \sqrt{420 \times 180 \times 140 \times 100}$
$= 10 \sqrt{2 \times 2 \times 5 \times 3 \times 7 \times 2 \times2 \times 5 \times 7 \times 2 \times 2 \times 3 \times 3 \times 5}$
$= 10\sqrt{ {2}^{6} \times {5}^{3} \times {3}^{3} \times {7}^{2} }$
$= 10 \times {2}^{3} \times 5 \times 3 \times 7 \sqrt{5 \times 3}$
$= 8400 \sqrt{15}$
area = 8400root 15 m^2.


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Answer 2

step by step explanation