Question

The sides of a triangle are in the ratio $3:4:5$ and its area is $54c{m}^2$. Find the length of the sides of the triangle.

Answer 1

Given:

The sides of a triangle are in the ratio $3:4:5$

So, let the sides of the triangle be $3k,4k,5k$

Here,

$(5k{)}^2=25k^2$

$(3k{)}^2+(4k{)}^2=9k^2+16k^2=25k^2$

$\Rightarrow (5k{)}^2=(3k{)}^2+(4k{)}^2$

By Pythagoras theorem, given triangle is a right-angled triangle.

$5k$ is hypotenuse. [Since, longest side]

$\Rightarrow$ if the base is $3k$ then its altitude is $4k$.

$\Rightarrow$ Area of the triangle $=\frac{1}{2}\times$ base $\times$ altitude

It is given that, Area of the triangle $=54c{m}^2$

$\Rightarrow \frac{1}{2}\times 3k\times 4k=54c{m}^2$

$\Rightarrow 6k^2=54c{m}^2$

$\Rightarrow k^2=9c{m}^2$

$\Rightarrow k=3cm$ [Since, length of the side is positive]

Therefore, the sides of the triangle are,

$3k=3\times 3cm=9cm$

$4k=4\times 3cm=12cm$

$5k=5\times 3cm=15cm$