Question -
Answer -
Solution:
Here, P is in the semi-circle and so,
P = 90°
So, it can be concluded that QR is hypotenuseof the circle and is equal to the diameter of the circle.
∴ QR = D
Using Pythagorean theorem,
QR2 = PR2+PQ2
Or, QR2 = 72+242
QR= 25 cm = Diameter
Hence, the radius of the circle = 25/2 cm
Now, the area of the semicircle = (πR2)/2
= (22/7)×(25/2)×(25/2)/2 cm2
= 13750/56 cm2 = 245.54 cm2
Also, area of the ΔPQR = ½×PR×PQ
=(½)×7×24 cm2
= 84 cm2
Hence, the area of the shaded region = 245.54cm2-84 cm2
= 161.54 cm2