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In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the area of the shaded region. Take π = 3.14.

In the figure, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the area of the shaded region. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Mensuration

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Answer

We know that,

Side of square ABCD = 4 cm

Radius of each quadrant circle (r) = 1 cm

Given,

Diameter of circle in the center = 2 cm

∴ Radius of circle in center (r1) = 22\dfrac{2}{2} = 1 cm.

From figure,

Area of shaded region = Area of square – Area of 4 quadrants – Area of circle at center

= side2−4×πr24−πr12=42−πr2−πr12=16−3.14(1)2−3.14(1)2=16−3.14−3.14=16−6.28=9.72 cm2.= \text{ side}^2 - 4 \times \dfrac{πr^2}{4} - πr1^2 \\[1em] = 4^2 - πr^2 - πr1^2 \\[1em] = 16 - 3.14(1)^2 - 3.14(1)^2 \\[1em] = 16 - 3.14 - 3.14 \\[1em] = 16 - 6.28 \\[1em] = 9.72 \text{ cm}^2.

Hence, area of shaded region = 9.72 cm2.

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