It can be observed that RQ is the diameter of the circle. Therefore, ∠RPQ will be 90º.

By applying Pythagoras theorem in ΔPQR,

RP^{2} + PQ^{2} = RQ^{2}

(7)^{2} + (24)^{2} = RQ^{2}

Radius of circle,^{ }

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Since RQ is the diameter of the circle, it divides the circle in two equal parts.

^{}

Area of ΔPQR = 1/2 x PQ x PR

=1/2 x 24 x 7

= 84 cm^{2 }

Area of shaded region = Area of semi-circle RPQOR − Area of ΔPQR