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In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded portion.

In the figure, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded portion. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Mensuration

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Answer

Since, ABCD is a rectangle.

AD = BC = 7 cm.

CD = AB = 14 cm.

From figure,

AD and BC are diameters of smaller circles.

Radius = 72\dfrac{7}{2} = 3.5 cm.

Area of each small semi-circle = πr22\dfrac{πr^2}{2}

= 227×(3.5)2×12\dfrac{22}{7} \times (3.5)^2 \times \dfrac{1}{2}

= 19.25 cm2.

From figure,

CD is the diameter of the larger semi-circle.

Radius = 142\dfrac{14}{2} = 7 cm.

Area of larger semi-circle = πr22\dfrac{πr^2}{2}

= 227×(7)2×12\dfrac{22}{7} \times (7)^2 \times \dfrac{1}{2}

= 77 cm2.

Area of rectangle ABCD = AB × BC

= 14 × 7

= 98 cm2.

From figure,

Area of shaded region = Area of rectangle ABCD + 2 × Area of each smaller semi-circle - Area of larger semi-circle

= 98 + (2 × 19.25) - 77

= 98 + 38.5 - 77

= 59.5 cm2.

Hence, area of shaded region = 59.5 cm2.

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