In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $ and $ \mathrm{BM} $ is an altitude. If $ \mathrm{AM}=\mathrm{BM}=12 $, find $ \mathrm{AC} $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \) and \( \mathrm{BM} \) is an altitude.

\( \mathrm{AM}=\mathrm{BM}=12 \)

To do:

We have to find \( \mathrm{AC} \).

Solution:

In $\triangle ABC$ and $\triangle ABM$,

$AB=AB$

$\angle A=\angle A$

$\angle B=\angle M= 90^o$

Therefore,

$\frac{AB}{AM}=\frac{BC}{BM}$

$\frac{AB}{12}=\frac{BC}{12}$

$AB=BC$......(i)

By Pythagoras theorem,

$AB^2 =AM^2+BM^2$

$AB^2=AM^2+AM^2$ 

$AB^2=2AM^2$

$AB^2=BC^2=2AM^2$

In $\triangle ABC$,

$AC^2=AB^2+BC^2$ 

$AC^2=2AM^2+2AM^2$

$AC^2=4AM^2$

$AC^2=(2AM)^2$

$AC=2AM$

$=2\times12$

$=24$.

Hence, $\mathrm{AC}=24$.

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Updated on: 10-Oct-2022

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