Question

A triangle has sides of $35cm,54cm$ and $61cm$ long. Find its area and its smallest altitude.

Answer 1

Let say sides be $a=35cm$

$b=54cm$

$c=61cm$

$s=\frac{a+b+c}{2}=\frac{35+54+61}{2}$

$=\frac{150}{2}$

$=75cm$

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

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

$=\sqrt{75\times 40\times 21\times 14}$

$=\sqrt{25\times 3\times 4\times 10\times 7\times 3\times 7\times 2}$

$=\sqrt{5\times 5\times 3\times 3\times 2\times 2\times 7\times 7\times 2\times 2\times 5}$

$=5\times 3\times 2\times 7\times 2\times \sqrt{5}$

$=939.15c{m}^2$

Smallest altitude will be the altitude on greatest side

So, Area of triangle $=939.15$

$\Rightarrow \frac{1}{2}\times 61\times h=939.15$

$h=30.8cm\left(approx\right)$.