In the given figure- ABCD is a trapezium of area 24-5 cm2- If AD - BC- DAB - 90- AD - 10 cm- BC - 4 cm and ABE is quadrant of a circle then find the area of the shaded region-

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Standard X

Mathematics

Areas of Combination of Plane Figures

In the given ...

Question

In the given figure, ABCD is a trapezium of area 24.5 $c m^{2}$ . If AD || BC, $△ D A B = 90^{\circ}$ , AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle then find the area of the shaded region.

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Solution

Sol:
Area of the trapezium = h/2 (a + b) where a and b are the parallel sides and h is the perpendicualr distance between the parallel sides.
Area of the trapezium ABCD = AB/2 (AD + BC)
24.5 = (AB/2) (10 + 4)
49 = AB (14)
AB = 3.5 cm

Radius of the quadrant of the circle = AB = 3.5 cm
Area of the quadrant of the circle = (1/4) (πr²) = (1/4) (22/7 x 3.5 x 3.5) = 9.625 cm²

Area of the shaded region = Area of the trapezium - Area of the quadrant of the circle
= 24.5 - 9.625
= 14.875 cm²

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In the given figure, ABCD is a trapezium of area 24.5 cm2. If AD || BC, △DAB=90∘, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle then find the area of the shaded region.

In the given figure, ABCD is a trapezium of area 24.5 $c m^{2}$ . If AD || BC, $△ D A B = 90^{\circ}$ , AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle then find the area of the shaded region.