In the figure, $\angle AOB = 90^o, AC = BC, OA = 12\ cm$ and $OC = 6.5\ cm$. Find the area of $\triangle AOB$.
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Given:

$\angle AOB = 90^o, AC = BC, OA = 12\ cm$ and $OC = 6.5\ cm$.

To do:

We have to find the area of $\triangle AOB$.

Solution:

$C$ is the mid-point of hypotenuse.

This implies,

$AC = CB = OC = 6.5\ cm$

$AB = 6.5 + 6.5$

$= 13\ cm$

In $\triangle AOB$,

$AB^2 = AO^2 + OB^2$                (Pythagoras Theorem)

$(13)^2 = (12)^2 + OB^2$

$169 = 144 + OB^2$

$OB^2 = 169 - 144$

$= 25$

$= 5^2$

$\Rightarrow OB = 5\ cm$

Area of $\Delta \mathrm{AOB}=\frac{1}{2}\times$ Base $\times$ Altitude

$=\frac{1}{2} \times \mathrm{OB} \times \mathrm{AO}$

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

$=30 \mathrm{~cm}^{2}$

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Updated on: 10-Oct-2022

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