Find the area of a triangle whose sides are respectively $150\ cm, 120\ cm$ and $200\ cm$.

Given:

The sides of a triangle are  $150\ cm, 120\ cm$ and $200\ cm$.

To do:

We have to find the area of the triangle. 

Solution:

Let $a=120\ cm, b=150\ cm$ and $c=200\ cm$

Therefore,

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

$=\frac{120+150+200}{2}$

$=\frac{470}{2}$

$=235$

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

$=\sqrt{235(235-120)(235-150)(235-200)}$

$=\sqrt{235 \times 115 \times 85 \times 35}$

$=\sqrt{5 \times 47 \times 5 \times 23 \times 5 \times 17 \times 5 \times 7}$

$=5 \times 5 \sqrt{47 \times 23 \times 17 \times 7} \mathrm{~cm}^{2}$

$=25 \sqrt{47 \times 23 \times 17 \times 7}$

$=25 \times \sqrt{128639} \mathrm{~cm}^{2}$

$=25 \times 358.66 \mathrm{~cm}^{2}$

$=8966.57 \mathrm{~cm}^{2}$

The area of the triangle is $8966.57\ cm^2$.

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Updated on: 10-Oct-2022

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