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Question

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.


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

Mathematics
Secondary School Mathematics X
Standard X

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