The sides of a triangle are $35cm,$ $54cm$ and $61cm,$ respectively. The length of its longest altitude is

The correct option is C: $24\sqrt{5}cm$

Three sides of the given triangle are $35cm,$ $54cm$ and $61cm.$

Here $a=35,$ $b=54,$ $c=61$
semi-perimeter, $s=\frac{a+b+c}{2}=\frac{35+54+61}{2}=75$
Area of triangle by Herons formula,

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

$=\sqrt{75(75-35)(75-54)(75-61)}$

$=420\sqrt{5}c{m}^2$
Now, Area of the triangle is also given as

$A=\frac{1}{2}\times base\left(b\right)\times height\left(h\right)$
Where, $h$ is the longest altitude.

Therefore, $\frac{1}{2}\times b\times h=420\sqrt{5}$

Hence, $h=24\sqrt{5}cm$