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A semi-circle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semi-circle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semi-circle (in sq. cm) will be:
In triangle CDO, which is a right anged triangle, we can use pythagoras theorem.
$$6^2$$ + $$(r-2)^2$$ = $$r^2$$
=> 36 + $$r^2 - 4r + 4$$ = $$r^2$$
=> 4r = 40
=> r = 10
=> Area = $$\pi *10*10/2$$ = $$50\pi $$
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