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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 a - Mathematics

Sum

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
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 cm= 1932 cm2

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ICSE Class 10 Mathematics
Chapter 17 Mensuration
Exercise | Q 10
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