In a $\triangle ABC, AB = 15\ cm, BC = 13\ cm$ and $AC = 14\ cm$. Find the area of $\triangle ABC$ and hence its altitude on $AC$.

Given:

In a $\triangle ABC, AB = 15\ cm, BC = 13\ cm$ and $AC = 14\ cm$.

To do:

We have to find the area of $\triangle ABC$ and hence its altitude on $AC$.

Solution:

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

$=\frac{15+13+14}{2}$

$=\frac{42}{2}$

$=21$

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

$=\sqrt{21(21-15)(21-13)(21-14)}$

$=\sqrt{21 \times 6 \times 8 \times 7}$

$=\sqrt{7 \times 3 \times 3 \times 2 \times 2 \times 2 \times 2 \times 7} \mathrm{~cm}^{2}$

$=7 \times 3 \times 2 \times 2$

$=84 \mathrm{~cm}^{2}$

Let $\mathrm{BL} \perp \mathrm{AC}$

Length of $\mathrm{BL}=\frac{\text { Area } \times 2}{\text { Base }}$

$=\frac{84 \times 2}{14} \mathrm{~cm}$

$=12 \mathrm{~cm}$

The area of $\triangle ABC$ is $84\ cm^2$ and its altitude on $AC$ is $12\ cm$.

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Updated on: 10-Oct-2022

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