# Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.  As, ΔPQR is in the semicircle and we know angle in the semicircle is a right angle, therefore PQR is a right-angled triangle.
[ If an angle is inscribed in a semi-circle, that angle measures 90 degrees.]
Here, QR is the hypotenuse of ΔPQR as lines from two ends of diameter always make a right angle when they meet at the circumference of that circle.

Hence by Pythagoras theorem we get,
Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

QR2 = PQ2 + PR2

QR2 = 242 + 72

QR2 = 576 + 49

QR = 25 cm

Diameter = 25 cm

Area of right angled triangle  triangle =  = 12 × 7
= 84 cm2 = 245.3125 sq cm

Hence,

Area of shaded region = Area of semicircle - area of ΔPQR
=  245.3125 – 84

Area of shaded region = 161.3125 sq cm

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