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 $△ A O B$

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Solution

In $△ A O B, ∠ A O B = 90^{\circ}$
C is a point on AB such that AC = BC join OC

Since C is the mid point of hypotenuse of right $△ A O B$
$∴ A C = C B = O C = 6.5 c m [Since mid-point of the hypotenuse of a right triangle is equidistant from the vertices.] ∴ A B = 6.5 + 6.5 = 13 c m N o w i n r i g h t △ A O B A B^{2} = A O^{2} + O B^{2} \Rightarrow (13)^{2} = (12)^{2} + O B^{2} \Rightarrow 169 = 144 + O B^{2} \Rightarrow O B^{2} = 169 - 144 = 25 = (5)^{2} ∴ O B = 5 c m N o w a r e a o f △ A O B = \frac{1}{2} B a s e \times A l t i t u d e = \frac{1}{2} \times O B \times A O = \frac{1}{2} \times 5 \times 12 = 30 c m^{2}$

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