UPSC 2015 CDS I Math Paper

In ΔPQR, PQ = QR = PR = r
Therefore, ΔPQR is an equilateral triangle.
Similarly, ΔPSQ is an equilateral triangle
∴ ∠RPS = 60° + 60° =120°
∠POR = 90° (PQRS is a rhombus)
In ΔPOR
sin‌60∘=‌‌

OR

r

⇒OR=‌‌

√3

2

r‌cos‌60∘=‌‌

OP

r

⇒OP=‌‌

1

2

r
Area common to the circles = 2(Area of section PSQR – Area of ΔPSR)
=‌‌

2

3

πr2−‌‌

√3

2

r2
Therefore, One-third of the sum ofthe areas of the two circles - Area common to the circles
=‌‌

1

3

×2πr2−(‌‌

2

3

πr2−‌‌

√3

2

r2)
=

√3

2

r2