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(
120∘
360∘
×πr2−
1
2
×2×
√3
2
r×
r
2
)
=
2
3
πr2−
√3
2
r2 Therefore, One-third of the sum ofthe areas of the two circles - Area common to the circles =