Question

B) In the adjacent figure, QR is the diameter of the circle with
centre 'O'. If PQ = 15 cm, PR=8 cm, find the area of the
shaded region. (Use r=3.14)
8 cm
15cm
8​

Answer 1

Answer:

Diagram is not given how one can solve it without diagram so you should join diagram first then ask it

Step-by-step explanation:

ok

Answer 2

Answer:

●step 1: in /\PQR /_ P = 90° ( as an angle in a semicircle is 90° ) therefore, using pythagoras theorum: find the hypotenuse QR= 17cm( this becomes the diameter n from tht the radius which is = 17/2cm of circle)

●step 2: now u hv the base n height of triPQR , find its area using 1/2 base× height= 1/2×8×15 = 60sq.cm☆

●step 3: find the area of semicircle PQOR using 1/2 pi-r-square = 113.4325 sq.cm☆

●step 4: area of shaded region = area of semicircle - area of triPQR

113.4325- 60 = 53.4325 sq.cm☆

( we can even solve this problem by using sector area - area of tri PQR)