A chord of a circle of radius 20, cm subtends | vedclass

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(D) Given: Radius $r = 20\, cm$,central angle $	heta = 90^{\circ}$.Area of the major sector $= \frac{360^{\circ} - 90^{\circ}}{360^{\circ}} 	imes \p

Vedclass · Accepted Answer

$1142.5$

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(D) Given: Radius $r = 20\, cm$,central angle $	heta = 90^{\circ}$.Area of the major sector $= \frac{360^{\circ} - 90^{\circ}}{360^{\circ}} 	imes \pi r^2 = \frac{270}{360} 	imes 3.14 	imes (20)^2 = \frac{3}{4} 	imes 3.14 	imes 400 = 3 	imes 3.14 	imes 100 = 942\, cm^2$.Area of $	riangle OAB = \frac{1}{2} 	imes r^2 	imes \sin(90^{\circ}) = \frac{1}{2} 	imes 20 	imes 20 	imes 1 = 200\, cm^2$.Area of the major segment = Area of the major sector + Area of $	riangle OAB = 942 + 200 = 1142\, cm^2$.

$A$ chord of a circle of radius $20\, cm$ subtends an angle of $90^{\circ}$ at the centre. Find the area of the corresponding major segment of the circle. (Use $\pi=3.14$) (in $cm^2$)

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