Question

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 ${\displaystyle \pi =\frac{22}{7}}$)

Answer 1

Here, the radius of larger semicircle = ${\displaystyle \frac{84}{2}=42cm}$

And, the radius of smaller semi-circle = ${\displaystyle \frac{84}{3\times 2}=14cm}$

Area of the shaded region  =${\displaystyle \frac{\pi {\left(42\right)}^2}{2}+3\times \frac{\pi {\left(14\right)}^2}{2}-\frac{1}{2}\times 84\times 42}$

${\displaystyle =\frac{22}{7}[21\times 42+3\times 14\times 7]-42\times 42}$

${\displaystyle =22[3\times 42+42]-42\times 42}$

${\displaystyle =42\times [88-42]}$

${\displaystyle =1932c{m}^2}$

Answer 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 = ${\displaystyle \frac{1}{2}}$ x AB x AC
= ${\displaystyle \frac{1}{2}}$ x 84 cm x 42 cm
= 1764 cm²

Radius of semicircle with BC as diameter = ${\displaystyle \frac{1}{2}}$ x 84 = 42 cm

Diameter of each of three equal semicircles = ${\displaystyle \frac{1}{3}}$ 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

= ${\displaystyle \frac{1}{2}}$ π ( 42² + 3 xx 14² ) cm² - 1764 cm²

= ${\displaystyle \frac{1}{2}\times \frac{22}{7}}$ x 142 ( 9 + 3) cm² - 1764 cm²

= 3696 cm² - 1764 cm² = 1932 cm²