The lengths of the sides of a triangle are in the ratio $3:4:5$ and its perimeter is $144\ cm$. Find the area of the triangle and the height corresponding to the longest side.

Given:

The lengths of the sides of a triangle are in the ratio $3:4:5$ and its perimeter is $144\ cm$.

To do:

We have to find the area of the triangle and the height corresponding to the longest side.

Solution:

Let the sides of the triangle be $a=3x, b=4x$ and $c=5x$.

Perimeter $=144 \mathrm{~cm}$

This implies,

$3x+4x+5x=144\ cm$

$12x=144\ cm$

$x=\frac{144}{12}=12\ cm$

Therefore,

$a=3x=3(12)=36\ cm$

$b=4x=4(12)=48\ cm$

$c=5x=5(12)=60\ cm$

$s=\frac{\text { Perimeter }}{2}$

$=\frac{144}{2}$

$=72$

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

$=\sqrt{72(72-36)(72-48)(72-60)}$

$=\sqrt{72 \times 36 \times 24 \times 12}$

$=\sqrt{12 \times 3 \times 2 \times 12 \times 3 \times 12 \times 2 \times 12}$

$=12 \times 12 \times 3 \times 2$

$=864 \mathrm{~cm}^{2}$

Length of the altitude on the longest side $60 \mathrm{~cm}=\frac{\text { Area } \times 2}{\text { Base }}$

$=\frac{864 \times 2}{60}$

$=\frac{1728}{60}$

$=\frac{288}{10}$

$=28.8 \mathrm{~cm}$

The area of the triangle is $864\ cm^2$ and the height corresponding to the longest side is $28.8\ cm$.