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A gateway is decorated as shown in the figure. There are four semicircles. BC the diameter of the largest semicircle of length 84 cm. The centers of the three semicircles lie on BC. ABC is an isosceles triangle with AB = AC. If BO = OC, find the area of the shaded area.

A gateway is decorated as shown in the figure. There are four semicircles. BC the diameter of the largest semicircle of length 84 cm. The centers of the three semicircles lie on BC. ABC is an isosceles triangle with AB = AC. If BO = OC, find the area of the shaded area. Correct Answer 1932

Given:

BC Is the diameter of the largest semicircle of length 84 cm. The centers of the three semicircles lie on BC. ABC is an isosceles triangle with AB = AC. If BO = OC

Concept Used:

Area of a right-angle triangle is ½ × Base × Height

Area of a semicircle with radius r is ½ × π × r2

Radius = ½ × Diameter

Calculation:

In ΔABC, AB = AC and ∠BAC is a semicircular angle

ΔABC is a right-angle isosceles triangle, with base BC = 84 cm, the diameter of the semicircle and height is AO, which is the radius of the semicircle

AO = ½ × 84

⇒ AO = 42

The area of the ΔABC is ½ × 84 × 42 cm2

⇒ 1764 cm2

Now OB = OC that means the three smaller semicircles on BC are of equal diameter, that means, of equal radius

The diameter of the three small semicircles are 84/3 = 28 cm

The radius of the semicircles are 14 cm

The radius of the bigger semicircle is 84/2 = 42 cm

Area of the shaded region = Area of the bigger semicircle + 3 × area of the small semicircle – the area of the triangle

⇒ ½ × π × 422 + 3 × ½ × π × 142 – 1764

⇒ ½ × π × (422 + 3 × 142) – 1764

⇒ ½ × π × 142 × (9 + 3) – 1764

⇒ ½ × (22/7) × 14 × 14 × 12 – 1764

⇒ 22 × 14 × 12 – 1764

⇒ 3696 – 1764

⇒ 1932

∴ The area of the shaded region is 1932 cm2.

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