Question

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.

Answer 1

Sides of a triangle are 35 cm, 54 cm, 61 cm
$\therefore s=\frac{a+b+c}{2}=\frac{35+54+61}{2}$
$=\frac{150}{2}=75cm\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{75(75-35)(75-54)(75-61)}=\sqrt{75\times 40\times 21\times 14}\sqrt{5\times 5\times 3\times 5\times 2\times 2\times 2\times 3\times 7\times 7\times 2}=7\times 5\times 3\times 2\times 2\sqrt{5}c{m}^2=7\times 60\sqrt{5}c{m}^2=420\times 2.236c{m}^2=939.14c{m}^2$
Length in smallest altitude corresponds to the longest side of 61 cm and is equal to
$\frac{Area\times 2}{Base}=\frac{939.14\times 2}{61}=30.79cm$