In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.

Area of square = (side) 2 = 14 2 = 196 cm 2 . From figure, Diameter of both semicircle = side of square = 14 cm. ∴ Radius (r) = = 7 cm. Area of both the semi-circles = Area of shaded region = Area of square - Area of semi-circles = 196 - 154 = 42 cm 2 . From figure, Perimeter = arc DPA + DC + arc CPB + AB = πr + 14 + πr + 14 = 2πr + 28 = + 28 = 44 + 28 = 72 cm. Hence, area of shaded region = 42 cm 2 and perimeter = 72 cm.

In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.

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In the given figure, find the area of the unshaded portion within the rectangle.
(Take π = 3.14)

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Mathematics

In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.

Mensuration

Related Questions

In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.

In the given figure, find the area of the unshaded portion within the rectangle.
(Take π = 3.14)

Mathematics

In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.

Mensuration

Related Questions

In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.

In the given figure, find the area of the unshaded portion within the rectangle.(Take π = 3.14)

In the given figure, find the area of the unshaded portion within the rectangle.
(Take π = 3.14)