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

Given:

A triangle has sides $35\ cm, 54\ cm$ and $61\ cm$ long.

To do:

We have to find its area and the smallest of its altitudes.

Solution:

The sides of the triangle are $a=35\ cm, b=54\ cm$ and $c=61\ cm$.

Therefore,

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

$=\frac{35+54+61}{2}$

$=\frac{150}{2}$

$=75$

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

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

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

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

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

$=7 \times 60 \sqrt{5}$

$=420 \times 2.236 \mathrm{~cm}^{2}$

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

Length of the smallest altitude $=\frac{\text { Area } \times 2}{\text { Base }}$

$=\frac{939.14 \times 2}{61}$

$=30.79 \mathrm{~cm}$

The area of the triangle is $939.14\ cm^2$ and the length of the smallest altitude is $30.79\ cm$.

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Updated on: 10-Oct-2022

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