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.

Asked by Pragya Singh | 1 year ago |  117

##### Solution :-

Here, P is in the semi-circle and so,

P = 90°

So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.

QR = D

Using Pythagorean theorem,

QR= PR2+PQ2

Or, QR= 72+242

QR= 25 cm = Diameter

Hence, the radius of the circle = $$\dfrac{25}{2}$$ cm

Now, the area of the semicircle = $$\dfrac{πR^2}{2}$$

= ($$\dfrac{22}{7}$$)×($$\dfrac{25}{2}$$)×($$\dfrac{25}{2}\dfrac{1}{2}$$) cm2

$$\dfrac{13750}{56}$$ cm= 245.54 cm2

Also, area of the ΔPQR = $$\dfrac{1}{2}$$×PR×PQ

=($$\dfrac{1}{2}$$)×7×24 cm2

= 84 cm2

Hence, the area of the shaded region = 245.54 cm2-84 cm2

= 161.54 cm2

Answered by Abhisek | 1 year ago

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