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In the given figure, PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn on PQ and QS as diameters. If PS = 12 cm, find the perimeter and the area of the shaded region.

In the given figure, PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn on PQ and QS as diameters. If PS = 12 cm, find the perimeter and the area of the shaded region. Circumference & Area of a Circle, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Mensuration

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Answer

Given,

PS = 12 cm

PQ = QR = RS = 123\dfrac{12}{3} = 4 cm, QS = 8 cm.

Perimeter of shaded region = Arc PTS + Arc PBQ + Arc QES

Radius of semi-circle PTS = 122\dfrac{12}{2} = 6 cm.

Arc PTS = 12\dfrac{1}{2} × 2π.radius

= πr

= 3.14 × 6 = 18.84 cm.

Radius of semi-circle PBQ = 42\dfrac{4}{2} = 2 cm.

Arc PBQ = π.radius

= 3.14 × 2 = 6.28 cm.

Radius of semi-circle QES = 82\dfrac{8}{2} = 4 cm.

Arc QES = π.radius

= 3.14 × 4 = 12.56 cm.

Perimeter of shaded region = Arc PTS + Arc PBQ + Arc QES

= 18.84 + 6.28 + 12.56

= 37.68 cm.

Area of shaded region = Area of big semicircle + Area of small semicircle on PQ - Area of semicircle on QS

Calculating area of big semicircle,

Area of big semicircle=12π.(radius)2=12×3.14×62=12×3.14×36=3.14×18=56.52 cm2.\text{Area of big semicircle} = \dfrac{1}{2}π.\text{(radius)}^2 \\[1em] = \dfrac{1}{2} \times 3.14 \times 6^2 \\[1em] = \dfrac{1}{2} \times 3.14 \times 36 \\[1em] = 3.14 × 18 \\[1em] = 56.52 \text{ cm}^2.

Area of small semicircle on PQ = 12\dfrac{1}{2} π.(radius)2

= 12\dfrac{1}{2} × 3.14 × 2^2

= 3.14 × 2 = 6.28 cm2.

Area of semicircle on QS = 12\dfrac{1}{2} π.(radius)2

= 12\dfrac{1}{2} × 3.14 × 42

= 3.14 × 8 = 25.12 cm2

Area of shaded region = 56.52 + 6.28 - 25.12

= 37.68 cm2.

Hence, area of shaded region = 37.68 cm2 and perimeter = 37.68 cm.

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