In the figure,ABCD is a trapezium with AB par | vedclass

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(A) Given: $AB = 18 \, cm$,$DC = 32 \, cm$,height $(h) = 14 \, cm$,and radius of each arc $(r) = 7 \, cm$.Since $AB \parallel DC$,the sum of consecuti

Vedclass · Accepted Answer

$196$

Vedclass · Answer

(A) Given: $AB = 18 \, cm$,$DC = 32 \, cm$,height $(h) = 14 \, cm$,and radius of each arc $(r) = 7 \, cm$.Since $AB \parallel DC$,the sum of consecutive interior angles is $180^{\circ}$. Thus,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.Area of a sector with angle $	heta$ is given by $\frac{	heta}{360^{\circ}} 	imes \pi r^2$.Area of sectors at $A$ and $D = \frac{\angle A + \angle D}{360^{\circ}} 	imes \pi r^2 = \frac{180^{\circ}}{360^{\circ}} 	imes \frac{22}{7} 	imes (7)^2 = \frac{1}{2} 	imes 22 	imes 7 = 77 \, cm^2$.Similarly,the area of sectors at $B$ and $C = \frac{\angle B + \angle C}{360^{\circ}} 	imes \pi r^2 = \frac{180^{\circ}}{360^{\circ}} 	imes \frac{22}{7} 	imes (7)^2 = 77 \, cm^2$.Area of trapezium $ABCD = \frac{1}{2} 	imes (AB + DC) 	imes h = \frac{1}{2} 	imes (18 + 32) 	imes 14 = \frac{1}{2} 	imes 50 	imes 14 = 350 \, cm^2$.Area of the shaded region = Area of trapezium $-$ (Sum of areas of the four sectors) = $350 - (77 + 77) = 350 - 154 = 196 \, cm^2$.

In the figure,$ABCD$ is a trapezium with $AB \parallel DC$,$AB = 18 \, cm$,$DC = 32 \, cm$,and the distance between $AB$ and $DC = 14 \, cm$. If arcs of equal radii $7 \, cm$ with centers $A, B, C$,and $D$ have been drawn,find the area of the shaded region of the figure (in $cm^2$).

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