As shown in the diagram, rectangle ABCD is a | vedclass

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(C) $1$. Area of rectangle $ABCD = 	ext{length} 	imes 	ext{breadth} = 20 	imes 14 = 280 \, cm^2$.$2$. Area of the semicircle with diameter $BC = 1

Vedclass · Accepted Answer

$49$

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(C) $1$. Area of rectangle $ABCD = 	ext{length} 	imes 	ext{breadth} = 20 	imes 14 = 280 \, cm^2$.$2$. Area of the semicircle with diameter $BC = 14 \, cm$ (radius $r = 7 \, cm$): $\frac{1}{2} \pi r^2 = \frac{1}{2} 	imes \frac{22}{7} 	imes 7 	imes 7 = 77 \, cm^2$.$3$. Area of the sector with centre $A$ and radius $AD = 14 \, cm$ (angle $	heta = 90^{\circ}$): $\frac{90}{360} 	imes \pi r^2 = \frac{1}{4} 	imes \frac{22}{7} 	imes 14 	imes 14 = 154 \, cm^2$.$4$. Area of the remaining sheet = Area of rectangle - (Area of semicircle + Area of sector) = $280 - (77 + 154) = 280 - 231 = 49 \, cm^2$.

As shown in the diagram, rectangle $ABCD$ is a metal sheet in which $CD = 20 \, cm$ and $BC = 14 \, cm$. From it, a semicircle with diameter $\overline{BC}$ and a sector with centre $A$ and radius $AD$ is cut off. Find the area of the remaining sheet in $cm^2$.

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