In Fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the

1.	In Fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.
Answer» 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 (ΔOAB) We have, AB² = OB² - OA² 13² = OB² + 12² OB = 5 cm Therefore, Area (Δ AOB) = \(\frac{1}{2}\)(OA x OB) =\(\frac{1}{2}\)(12 x 5) = 30 cm²

In Fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.

Answer»

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 (ΔOAB)

We have,

AB² = OB² - OA²

13² = OB² + 12²

OB = 5 cm

Therefore,

Area (Δ AOB) = \(\frac{1}{2}\)(OA x OB)

=\(\frac{1}{2}\)(12 x 5)

= 30 cm²