In fig, $∠ A B C = 90^{o}, A C = B C, O A = 12 c m$ and $O C = 6.5 c m$ . Find the area of $△ A O B .$

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Solution

Since mid-point of the hypotenuse of a right triangle is equidistant from the vertices.

$∴ C A = C B = O C \Rightarrow C A = C B = 6.5 c m \Rightarrow A B = 13 c m$

In right triangle $O A B$ , we have

$A B^{2} = O B^{2} + O A^{2} \Rightarrow 13^{2} = O B^{2} + 12^{2} \Rightarrow O B = 5$

$∴ a r (△ A O B) = \frac{1}{2} (O A \times O B) = \frac{1}{2} (12 \times 15) = 30 c m^{2}$

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Similar questions

In the figure, $∠ A O B = 90^{\circ}, A C = B C, O A = 12$ cm and $O C = 6.5 c m$ . Find the area of $△ A O B$

△ A B C, ∠ A B C = 90^{o}, A D = D C, A B = 12 c m

B C = 6.5 c m

△ A D B

∠ A O B = 90^{o}, A C = B C, O A = 10 c m

O C = 13 c m

△ A O B

A B C

∠ A = 90^{o}

A N ⊥ B C

B C = 12 c m

A C = 5 c m

△ A N C

△ A B C

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