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

Solution:

Given,

The sides of the triangle are

a = 35 cm

b = 54 cm

c = 61 cm

Perimeter 2s = a + b + c

2s = 35 + 54 + 61 cm

Semi perimeter s = 75 cm

By using Heron's Formula,

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

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

$=939.14 \mathrm{~cm}^{2}$

The altitude will be smallest provided the side corresponding to this altitude is longest.

The longest side = 61 cm

Area of the triangle = 1/2 × h × 61

$1 / 2 \times \mathrm{h} \times 61=939.14 \mathrm{~cm}^{2}$

$\mathrm{h}=30.79 \mathrm{~cm}$

Hence the length of the smallest altitude is $30.79 \mathrm{~cm}$