In the given figure, ABCD is a square inside a circle with centre O. The centre of the circle coincides with O and diagonal AC is horizontal . . .

Question: In the given figure, ABCD is a square inside a circle with centre O. The centre of the circle coincides with O and diagonal AC is horizontal. If AP, DQ are vertical and AP=45 cm, DQ=25 cm then Find (a) the radius of the circle b) the side of the square c) the area of the shaded region. [Take √2 = 1.41 and π = 3.14]

Doubt by Muskan

Solution : 

AP = 45 cm (Given)

DQ = 25 cm (Given)

We know, diagonal of square are equal & bisect each other at 90°
 So AO=BO=CO=DO = x (Let)

Also, Let radius of the circle = r

So OP=OQ=r

OP=OD+DQ=r [OQ=OD+DQ]

r=OP=x+25
Let side of the square = a

a) Join OP
∠PAO=90° (Given)
Now, In Rt. ∆PAO
By Pythagoras Theorem
OP² = AP² + AO²
(x+25)² = 45² + x²
x²+625+50x = 2025+x²
50x=2025-625
50x =1400
x=28 cm
r = x+25
r = 28+25
r = 53 cm

b) ∠AOD=90°
(Diagonals of Square bisect each other at 90°)

Now, In Rt. ∆AOD
By Pythagoras Theorem
AD² = AO² + DO²
a² = x² + x²
a² = 2x²
a = x√2
a = 28√2 [x=28]
a = 28√2 cm
a = 28×1.41
a=39.48 cm

c) Area of shaded region

= Area of circle - area of square
= πr² - (a)²
= 3.14(53)² - (28√2)²
= 3.14×2809-1568
=8820.26-1568
=7252.26 cm²