In the accompanying figure, AB is one of the diameters of the circle and OC is perpendicular to it through the center O. If AC is $7 \sqrt{2}$ cm, then what is the area of the circle in sq. cm.?
Solution
AO = OC = radius
$\Delta AOC$ is a right angle triangle.
$OC^2 + OA^2 = AC^2$
$r^2 + r^2 = (7\sqrt{2})^2$
$r = 7$cm
Area of circle $= \dfrac{22}{7} \times 7 \times 7 = 154$cm$^2$