# 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)

#### 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

Is there an error in this question or solution?

#### APPEARS IN

ICSE Class 10 Mathematics
Chapter 17 Mensuration
Exercise | Q 10