A doorway is decorated as shown in the figure. There are four semi-circles. BC, the diameter of the larger semi-circle is of length 84 cm. Centres of the three equal semicircles lie on BC. ABC is an isosceles triangle with AB = AC. If BO = OC, find the area of the shaded region. (Take `pi = 22/7`)
Solution 1
Here, the radius of larger semicircle = `84/2 = 42 cm`
And, the radius of smaller semi-circle = `84/(3xx2) = 14 cm`
Area of the shaded region =` (pi(42)^2)/2 + 3xx(pi(14)^2)/2 - 1/2 xx 84 xx 42`
`= 22/7 [21 xx 42 + 3 xx 14 xx 7] - 42 xx 42`
`= 22[3 xx 42 + 42] - 42 xx 42`
`= 42 xx [88 - 42]`
`= 1932 cm^2`
Solution 2
As angle in a semicircle is 90°,
∠ A = 90°
From Δ ABC,
by Pythagoras theorem, we get
AB2 + AC2 = BC2
⇒ x2 + x2 = 842
⇒ 2x2 = 84 x 84
⇒ x2 = 84 x 42
∴ Area of Δ ABC = `1/2` x AB x AC
= `1/2` x 84 cm x 42 cm
= 1764 cm2
Radius of semicircle with BC as diameter = `1/2` x 84 = 42 cm
Diameter of each of three equal semicircles = `1/3` x 84 = 28 cm
⇒ Radius of each of 3 equal semicircles = 14 cm
Area of the shaded region = Area of semicircle with 42 cm as Radius + Area of three equal semicircles of radius 14 cm - area of Δ ABC
= `1/2π xx 422 cm2 + 3 xx 1/2π xx 142 cm2 - 1764 cm2
= `1/2` π ( 422 + 3 xx 142 ) cm2 - 1764 cm2
= `1/2 xx 22/7` x 142 ( 9 + 3) cm2 - 1764 cm2
= 3696 cm2 - 1764 cm2 = 1932 cm2
