Question

in the figure 0 is the centre of the. circle such that mlA0c = 140° there find mlABC​

Answer 1

Step-by-step explanation:

In ΔABC,

Angle in a semicircle, ∠C=90o

Therefore, by Pythagoras theorem

BC2+AC2=AB2

BC2+122=132

BC2=169−144=25

BC=5

Area of shaded region = Area of semicircle – Area of triangle

= 21πr2−21bh

= 21×3.14×(213)2−21×5×12

= 2×43.14×13×13−30

Area of shaded region =36.39 cm2.

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