In the given figure, ABCD is a trapezium of area $24.5 {c m}^{2}$ . If $A D ∥ B C, ∠ D A B = 90^{0}, A D = 10 c m, B C = 4 c m$ and $A B E$ is quadrant of circle then find the area of the shaded region.

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Solution

Area of trapezium $= 24.5 c m^{2}$
$= \frac{1}{2} \times s u m o f p a r a l l e l s i d e s \times h e i g h t = 24.5$
$= h e i g h t = \frac{24.5 \times 2}{s u m o f p a r a l l e l s i d e s}$
$= h e i g h t = \frac{49}{10 + 4}$
$= h e i g h t = 3.5 c m$

Now BA=3.5 cm
Therefore area of quadrant $= \frac{π r^{2}}{4}$
$= 9.625 c m^{2}$

Then area of the shaded region $= 24.5 - 9.625 = 14.875 c m^{2}$

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In the given figure, ABCD is a trapezium of area 24.5cm2. If AD∥BC,∠DAB=900,AD=10cm,BC=4cm and ABE is quadrant of circle then find the area of the shaded region.

Area of trapezium=24.5cm2=12×sum of parallel sides×height=24.5=height=24.5×2sum of parallel sides=height=4910+4=height=3.5cmNow BA=3.5 cmTherefore area of quadrant=πr24=9.625cm2Then area of the shaded region =24.5−9.625=14.875cm2

In the given figure, ABCD is a trapezium of area $24.5 {c m}^{2}$ . If $A D ∥ B C, ∠ D A B = 90^{0}, A D = 10 c m, B C = 4 c m$ and $A B E$ is quadrant of circle then find the area of the shaded region.