Question

the ratio of the sides of a triangular garden is 2:3:4 if the perimeter of the garden is 108 m then find its area ​

Answer 1

Given,

The ratio of the sides of a triangular garden is $2:3:4$ if the perimeter of the garden is $108 m$

Let the sides of the garden be '$x$'

The perimeter of triangle $=2x+3x+4x=108$

$9x=108\\x=12m$

Sides of the triangles is $2\times12=24$

$3\times12=36$, $4\times12=48$

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{a+b+c}{2}$

Here, $a=24,b=36,c=48$

$s=\frac{24+36+48}{2} \\s=\frac{108}{2}\\s=54$

Area $=\sqrt{54(54-24)(54-36)(54-48)}$

Area $=\sqrt{54(30)(18)(6)}$

Area $=\sqrt{174960}$

Area $=418.28$

Therefore, the area of a triangle is $418.28$.

Answer 2

Given,

The sides of triangle are given the ratio $2:3:4$

This ratio can be written as $2x,3x,4x$

The perimeter of the garden

$P=108m$

$2x+3x+4x=108$

$9x=108$

$x=12cm$

So, the sides of triangle becomes

$2x=2(12)=24\\3x=3(12)=36\\4x=4(12)=48$

Area of a triangle formula

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

Here $s$ is semi perimeter of a triangle

$s=\frac{a+b+c}{2}$

$=\frac{24+36+48}{2}$

$=\frac{108}{2} \\=54$

$s=54m$

Now,

$A=\sqrt[]{54(54-24)(54-36)(54-48)}$

$A=\sqrt[]{54(30)(18)(6)} \\A=\sqrt[]{174960} \\A=418.28$

Therefore, the area of triangle is $418.28m$