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Question

In Figure, ABCD is a trapezium of area 24.5 sq. cm. In it, ADBC, DAB=90o,AD=10 cm and BC=4 cm. If ABE is a quadrant of a circle, then find the area of the shaded region (upto the nearest interger). [Take π=227]
261088_993688284b40414abf515cebff036e23.png

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Solution

Given: trapezium ABCD, DAB=90, AD=10, BC=4 cm ,Area=24.5 sq.cm

Area of trapezium = 12 (Sum of parallel sides) (Perpendicular)

24.5=12×(10+4)×AB

AB=3.5

Area of quadrant ABE= θ360π(AB)2

Area of quadrant ABE= 90360×227(3.5)2

Area of quadrant ABE= 9.625cm2

Area of shaded portion = Area of trapezium Area of quadrant ABE

Area of shaded portion = 24.59.625

Area of shaded portion = 14.375 cm2 =14 cm2(rounding off to the nearest integer)

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