Q4 of 77 Page 15

In Fig. 15.77, ∠AOB=90°, AC=BC, OA=12 cm and OC=6,5 cm. Find the area of Δ AOB.

Since,

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


Therefore,


CA = CB = OC


CA = CB = 6.5 cm


AB = 13 cm


In right (OAB)


We have,


AB2 = OB2 - OA2


132 = OB2 + 122


OB = 5 cm


Therefore,


Area (Δ AOB) = (OA * OB)


= (12 * 5)


= 30 cm2


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