In a circular table cover of radius 32cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in fig. Find the area of the design (shaded region).

In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

The area of an equilateral triangle ABC is 17320.5 cm^(2) . With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use pi=3.14 and sqrt(3)=1.73205 )

From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take pu = 22/ 7 )

The area of an equilateral triangle ABC is 17320.5 cm^2 . With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use pi=3.14 and sqrt3=1.73205 )

Area of an equilateral triangle is 17300 cm^2 . By taking each vertex of the triangle as centre, a circle is drawn at each vertex. The radius of thecircleis half of the length of the side of an equilateral triangle. Find the area of that portion of the triangle which is not included in the circle. (pi=3.14 and sqrt3=1.73 )

ABCD is a square of side 4 cm. With every comer as the centre, quadrant of a circle is drawn and at the centre a circle of 1 cm radius is drawn as shown in figure. Find the area of shaded portion.

In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20cm, find the area of the shaded region.(Use pi=3.14 )

In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.