Fig shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm- ABCD is a square and - CEF is an isosceles right angled triangle whose equal sides are 6 cm long- Find the area of the shaded region-

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Mathematics

Area of a Sector of a Circle

Fig shows a k...

Question

Fig shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and $Δ C E F$ is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

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Solution

Area of the shaded region = area of the quadrant BCD + area of triangle EFC. $= \frac{22}{7} \times \frac{42 \times 42}{4} + \frac{6 \times 6}{2} = 1404 c m^{2}$

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Fig shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and ΔCEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

Area of the shaded region = area of the quadrant BCD + area of triangle EFC.=227×42×424+6×62=1404 cm2

Fig shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and $Δ C E F$ is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

Area of the shaded region = area of the quadrant BCD + area of triangle EFC. $= \frac{22}{7} \times \frac{42 \times 42}{4} + \frac{6 \times 6}{2} = 1404 c m^{2}$