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The figure shows a circle of diameter $\text{AB}$ and radius $6.5$ cm. 

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

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 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 $\angle \text{ACB} = 90^\circ$.

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

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

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

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

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

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

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

$\qquad \qquad = \frac{1}{2}\times 5\times 12 = 30\; \text{cm}^{2}$

Correct Answer $:30$

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