Answer :

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.**

QR^{2} = PQ^{2} + PR^{2}

QR^{2} = 24^{2} + 7^{2}

QR^{2} = 576 + 49

QR = 25 cm

Diameter = 25 cmOr, radius = 12.5cm

Area of right angled triangle triangle =

= 12 × 7

= 84 cm^{2}

= 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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