Question

Please answer this question

Answer 1

given OP=OQ=10cm
it is known that tangents drawn from an external point to a circle are equal in length
so,
OP=OQ=10cm
therefore, triangleABC is aa equilateral triangle 
<POQ=60
now,
area of PAQP =area of the sector - area of the equilateral triangle POQ
​= <POQ/360*pir2-root3/4*(10)2
=60/360*pi(10)2-root3/4*(10)2
=100[pi-root3]
          6     4

area of the semicircle on diameter PQ= area of PAQP + area of semicircle
=1*pi(5)2 = 25pi sq units 
  2                 2
therefore , area of the shaded region
= 25pi - 100[pi-root3]
     2              6    4
= 25pi - 100pi + 25root3
     2         6
=25root3-25pi
                 6
=25[root3 - pi]
                    6
hope u understand this :)
 

Answer 2

happy republic day i think it help you