Question

O and O’ are the centres of the circles of radius r as shown in figures (i) and (ii) respectively.

(i)	(ii)

Find the ratio of area of shaded region in figure (i) to that of area of shaded region in figure (ii).

Answer 1

(i)

Area of segment = Area of sector – Area of triangle BOD

= `πr^2 xx (120^circ)/(360^circ) - 2 xx 1/2 xx BD xx OM`

= `(πr^2)/3 - 2 xx 1/2 xx 2 xx BM xx OM`

= `(πr^2)/3 - 2 xx r  sin 60^circ xx  r  cos 60^circ`

=  `(πr^2)/3 - 2r^2 xx sqrt(3)/2 xx 1/2`

= `(πr^2)/3 - (sqrt(3)r^2)/2`

∴ Shaded area of figure (i) = `2 xx r^2 (π/3 - sqrt(3)/2)`

(ii)

Area of segment = Area of sector – Area of triangle

= `πr^2 xx (60^circ)/(360^circ) - sqrt(3)/4 r^2`   ...[∵ Area of equilateral triangle is `sqrt(3)/4` × (side)²]

= `(πr^2)/6 - sqrt(3)/4 r^2`

= `r^2 (π/6 - sqrt(3)/4)`

∴ Shaded area of figure (ii) = `2 xx r^2 (π/6  - sqrt(3)/4)`

Required ratio

= `(2 xx r^2 (π/3 - sqrt(3)/4))/(2 xx r^2 (π/6 - sqrt(3)/4)`

= `(4π - 3sqrt(3))/(2π - 3sqrt(3))`