Question

The sides of a triangular field are in the ratio $5:3:4$ and its perimeter is $180$ m, then find the altitude of the triangle corresponding to its largest side (in $m.$)

Answer 1

Let the sides of the triangular field be $5a,3a,4a$.

Perimeter of the triangle $=$ Sum of all sides $=5a+3a+4a=12a$

Given, perimeter of the triangular fied $=180m$

$\Rightarrow 12a=180m$

$\Rightarrow a=15m$

So, the sides of the triangular field are $5a=75m$

$3a=45m$

$4a=60m$.

$\Rightarrow$ Area of triangle with sides $a,b,c$ is $\sqrt{s(s-a)(s-b)(s-c)}$.

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

$\Rightarrow$ For the triangular field with sides $75m$, $45m$ and $60m$,

$s=\frac{75+45+60}{2}$

$=90m$

$\Rightarrow$ Area of the triangular field with sides $75$ m, $45$ m and $60$ m

$=\sqrt{90(90-75)(90-45)(90-60)}$

$=\sqrt{90\times 15\times 45\times 30}$

$=\sqrt{15\times 6\times 15\times 15\times 3\times 15\times 2}$

$=15\times 15\times 6=1350m^2$

Also, Area of triangle $=\frac{1}{2}\times$ base $\times$ height

Considering the height on the largest side of the triangle $75m$ as $h$,

$\Rightarrow \frac{1}{2}\times 75\times h=1350sq.m$

$\Rightarrow h=36m$