Question

ABCD is a diameter of a circle of radius 6cm .The lengths AB,BC,CD are equal . semicircles are drawn on AB and BD as diameter as shown in the given figure . Find the area of shaded region ​

Answer 1

Answer:

Since, Length of AB, BC and CD are equal.

Radius of circle = 6 cm

Now, AD = 2 × 6 = 12 cm

⇒ AB + BC + CD = 12

⇒3AB = 12

$⇒ \: AB \: = \: \frac{12}{3}$

⇒AB=4cm

⇒AB=BC=CD=4cm

Radius of semicircle AB=2cm

Radius of semicircle BC=4cm

Radius of semicircle AD=6cm

Area of the shaded region = Area of semicircle (AB+AD) − Area of semicircle (BD)

$⇒  Area \: of \: shaded \: region = \: 0.5\pi( {2}^{2} + {6}^{2}) \: - 0.5\pi {(4)}^{2}$

$⇒ Area \: of \: shaded \: region = \: 0.6\pi(4 + 36) - 0.5\pi \times 16$

$⇒ Area \: of \: shaded \: region = \: 20\pi - 8\pi$

$⇒ Area \: of \: shaded \: region = 12\pi {cm}^{2}$

Answer 2

Explanation of diagram [ please refer attached picture for diagram ] :-

=> ABCD is diameter of circle. Radius of circle is given as 6 cm. So diameter will be = 2×6 = 12 cm

=> It is mentioned in the question that AB,BC,CD are equal. This means that diameter of circle of length 12 cm is divided into 3 parts.

➡️12/3 = 4 cm

=> Diameter is divided into three parts each of 4 cm. AB= BC=CD=4 cm

=> Now we have to find the area of shaded region ( region 1 and 2 ).

=> Now look at semi-circle AGB. It's diameter is 4cm . So it's radius will be :-

➡️4/2 = 2 cm

=> Now look at semi-circle BED. It's radius is 4cm .

Now, area of shaded region = area of semi-circle AGB + ( area of semi-circle AFD - area of semi-circle BED)

area of semi-circle = 1/2 × π × r²

Where :- r = radius of semi-circle

=> area of semi-circle AGB = 1/2 × π × (2)²

=> area of semi-circle AGB = (4π)/2 cm²

=> area of semi-circle AFD = 1/2 × π × (6)²

=> area of semi-circle AFD = (36π)/2 cm²

=> area of semi-circle BED = 1/2 × π × (4)²

=> area of semi-circle BED = (16π)/2 cm²

=> area of shaded region = area of semi-circle AGB + ( area of semi-circle AFD - area of semi-circle BED)

=> area of shaded region = (4π)/2 + ( (36π)/2 - (16π)/2)

=> area of shaded region = 2π+ ( 18π - 8π)

=> area of shaded region = 2π+ 10π

=> area of shaded region = 12π cm²

Answer :

Area of shaded region = 12 π cm²