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?

cos(∠CAE) = AE/AC = ½ => ∠CAE = 60

(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

**the line joining points of intersection” => CE = ½CD and CD ⊥ AB**__perpendicularly bisects__
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, CE

^{2}= AC^{2}-AE^{2}= 1^{2}– (½)^{2}= ¾
=> CE = √3/2 => CD = 2*CE = √3

cos(∠CAE) = AE/AC = ½ => ∠CAE = 60

^{0}=> ∠CAD = 2*60

^{0 }=

^{ }120

^{0}

Area covered under arc CBD = 120

^{0}/360^{0}(π*1^{2}) = 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)**
Can u xpln wat is dis area covered under arc CBD n CAD??

ReplyDeleteArea covered by an arc is called as Sector. Area of sector is ((theta)/360) * pi * r^2, where theta is the angle of the sector.

ReplyDelete