Geometry-14 (CAT-2008)

Two circles, both of radii 1 cm, intersect such that the circumference of each one passes   through the centre of the circle of the other. What is the area (in sq cm) of the intersecting region?

(1) π/3 - √3/4            (2) 2π/3 + √3/2 (3) 4π/3 - √3/2 (4) 4π/3 + √3/2 (5) 2π/3 - √3/2

Solution follows here:

Solution:

Let A and B be the centres of the circles respectively. Let C and D be the points of intersection. Let line AB and line CD meet at E.

Funda here is: “The line joining centres of the circles perpendicularly bisects the line joining points of intersection” => CE = ½CD and CD ⊥ AB

By symmetry, E is midpoint of AB => AE = ½(AB) = ½(radius of circle with centre A) = ½(1) = ½

AC = radius = 1

Applying Pythagoras to the right triangle ACE, CE² = AC²-AE² = 1² – (½)² = ¾

=> CE = √3/2 => CD = 2*CE = √3

cos(∠CAE) = AE/AC = ½ => ∠CAE = 60⁰ => ∠CAD = 2*60⁰ = 120⁰

Area covered under arc CBD = 120⁰/360⁰ (π*1²) = π/3

Similarly, area covered under arc CAD = π/3

Area of intersection of the two circles

= (Area covered under arc CBD) + ( Area covered under arc CAD) - (Area of rhombus ACBD)

= (π/3)+( π/3) - (½ * product of diagonals) = 2π/3 – ½ * (AB)(CD) = 2π/3 – ½ * (1)(√3)

= 2π/3 – √3/2

Answer (5)