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Find the sum of areas enclosed between arc AB and line AB, between arc BC and line BC and between arc CD and line CD where ABCD is isosceles trapezium inscribed in a circle of radius 20 cm so that the non-parallel sides are 20 cm each and AD is the diameter of the circle?
- a)200 π - 320 √3
- b)200 π - 300 √3
- c)200 π - 350 √3
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Find the sum of areas enclosed between arc AB and line AB, between ar...
Since AD is the diameter of the circle, we know that ∠ABD = 90°. In ΔABD, since AB = ½ AD, we can conclude that ∠D = 30° and ∠A = 60°.
Drop BE ^ AD and CF ^ AD. Since ΔBEA is a 30° - 60° - 90° triangle,
BE = 10√3 cm and AE = 10 cm.
Since ABCD is an isosceles trapezium, we get AE = FD = 10 cm and EF = BC = 20 cm.
Thus the area of trapezium ABCD = ½ × (20 + 40) × 10√3 = 300√3 cm2
Sum of areas enclosed between arc AB and line AB, between arc BC and line CD and arc CD
=Area of semi circle - Area of trapezium ABCD
=½ π (20)2 - 300√3
=200 π - 300 √3
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Find the sum of areas enclosed between arc AB and line AB, between arc BC and line BC and between arc CD and line CD where ABCD is isosceles trapezium inscribed in a circle of radius 20 cm so that the non-parallel sides are 20 cm each and AD is the diameter of the circle?a)200 π - 320 √3b)200 π - 300 √3c)200 π - 350 √3d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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