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The figure shows a circle of diameter $AB$ and radius $6.5$ cm.

If chord $CA$ is $5$ cm long, find the area of $\triangle ABC$. __________

Given that,

•  $\text{AB}=2\ast6.5=13$ cm.
•  $\text{CA}=5$ cm

We know that, Angle in a semicircle is a right angle. So $\angle ACB=90^\circ$.

Now, we can apply the Pythagorean theorem, in $\triangle ACB$.

$(AB)^{2}=(CA)^{2}+(CB)^{2}$

$\Rightarrow (13)^{2}=(5)^{2}+(CB)^{2}$

$\Rightarrow (CB)^{2}=169-25$

$\Rightarrow (CB)^{2}=144$

$\Rightarrow \boxed{CB=12\text{cm}}$

$\therefore$ The area of $\triangle ABC=\frac{1}{2}\times Base\times Height$

$=\frac{1}{2}\times 5\times 12$

$=30 \text{cm}^{2}$

Correct Answer :$30$

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