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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`)

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#### 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

AB^{2} + AC^{2 }= BC^{2}

⇒ x^{2} + x^{2} = 84^{2}

⇒ 2x^{2} = 84 x 84

⇒ x^{2} = 84 x 42

∴ Area of Δ ABC = `1/2` x AB x AC

= `1/2` x 84 cm x 42 cm

= 1764 cm^{2}

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 42^{2} cm^{2} + 3 xx 1/2π xx 14^{2} cm^{2} - 1764 cm^{2}

= `1/2` π ( 42^{2} + 3 xx 14^{2} ) cm^{2} - 1764 cm^{2}

= `1/2 xx 22/7` x 142 ( 9 + 3) cm^{2} - 1764 cm^{2}

= 3696 cm^{2} - 1764 cm^{2 }= 1932 cm^{2}

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