pluto82 wrote:
Let the width of rectangle NQ = x
Length of rectangle PQ =y
(y/2)^2+(x+6)^2 = 25
Considering Pythagorean triples (3,4,5) => (3*5,4*5,5*5)
y/2 = 15 => y = 30
x+6 = 20 => x = 14
Area or rectangle = 30*14 = 420
Area of shaded region = Area of circle - 2*area of rectangle
= 625*pi - 840
Answer (D)
To determine the area of shaded region : {(Area of circle)- (Area of ABDC +Area of MNQP)}
We are given the radius (25 units).
Let R be mid point of MN such that OR =6 units (the shortest distance from O to segment MN is 6)
Let T be a point on PQ such that RT = x units
OT =OR + RT = 6+x units
Let TQ = y units.
OTQ is a right angled triangled at T
Using Pythagorean Principle :
OT ^2 +TQ ^2 = OQ
(6+x)^2 + y ^2 = 25 ^2
The triplets could be (15, 20) or (20 ,15). So length & breadth of the rectangles can be (9, 20) or (14,15).
Case 1:(9, 20*2)
Area of the rectangle : 360 units
Area of 2 rectangle : 2 * 360 : 720 units
Area of shaded region = area of circle - area of 2 rectangles.
========= 625 pie - 720
Case 2:(14 , 15*2)
Area of rectangle : 420 units
Area of 2 rectangles : 840 units
Area of shaded region :area of circle - area of 2 rectangles.
= 625 pie - 840.
Case 2 matches with option D