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Q) PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn such that PQ = QR = RS. Semicircles are drawn on PQ and QS as diameters as shown in figure. Find the perimeter and area of the shaded region.

PQRS is a diameter of a

Ans: 

Given that the radius = 6 cm

∴  Diameter PQRS = 2 x 6 = 12 cm

Since PQRS = PQ + QR + RS and its given that PQ = QR = RS

∴  12 = 3 PQ

∴ PQ = PQRS is a diameter of a = 4 cm

∴  PQ = QR = RS = 4 cm

Let’s start with calculating Perimeter of Shaded region:

∵ We know that the perimeter of a circle is given by =  2 π r or  π d

∴ the perimeter of a half circle is given by:  π r or π PQRS is a diameter of a

Now the Perimeter of shaded region =  Perimeter of Semi-circle of dia. PS + Perimeter of Semi-circle of dia. PQ + Perimeter of Semi-circle of dia. QS

=  π PQRS is a diameter of a + π PQRS is a diameter of a + π PQRS is a diameter of a

Here PS = 12 cm, PQ = 4 cm, QS = QR + RS = 4 + 4 = 8 cm

∴ Perimeter of shaded region =  π PQRS is a diameter of a + π PQRS is a diameter of a + π PQRS is a diameter of a

=  6 π + 2 π + 4 π = 12 π

= 12 x PQRS is a diameter of a = PQRS is a diameter of a = 37.714

∴ Perimeter of shaded region = 37.714 cm

Next, we will calculate Area of the Shaded region:

∵ We know that the area of a circle is given by =  π r2 or  π PQRS is a diameter of a

∴ the area of a half circle is given by:  π PQRS is a diameter of a or π PQRS is a diameter of a

Area of shaded region =  Area of Semi-circle of dia. PS + Area of Semi-circle of dia. PQ – Area of Semi-circle of dia. QS

∴ Area of shaded region =  π PQRS is a diameter of a +  π PQRS is a diameter of a – π PQRS is a diameter of a

Here PS = 12 cm, PQ = 4 cm, QS = 8 cm

∴ Area of shaded region =  π PQRS is a diameter of aPQRS is a diameter of a  – π PQRS is a diameter of a

=  π PQRS is a diameter of a

=  π PQRS is a diameter of a

=  12 π = 12 x PQRS is a diameter of a = PQRS is a diameter of a = 37.714

∴ Area of shaded region = 37.714 cm2

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