Question 57
If the area of the circle in the figure is $$36\pi$$ $$cm^{2}$$ and ABCD is a square, then the area of $$\triangle$$ ACD, in $$cm^{2}$$, is:
Solution
Area of circle = $$36\pi\ $$
==> $$\pi r^2=36\pi\ \ $$
This gives radius of circle = 6 cm.
Diagonal of the square = Diameter of the circle.
$$S\sqrt{\ 2}=12$$
S = $$6\sqrt{\ 2}$$cm.
Area of triangle ACD = half of the area of square = $$\frac{1}{2}\left(6\sqrt{\ 2}\right)^2=36$$ sq. cm.
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