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Question

Four points A, B, C, and D are given on circle. Line segment AB and CD are parallel. Find the area of the figure formed by ABCD.

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Solution

Construction: Draw perpendiculars OE and OF onto AE and CD respectively from the centre O. In ΔOEB, OE2 + BE2 = OB2 [Using Pythagoras theorem] OE2 = OB2 - BE2 OE2 = 25 - 9 (Perpendicular drawn from centre to a chord bisects the chord, so BE = 3 cm) Therefore, OE = 4cm Similarly In ΔOFD, OF2 + FD2 = OD2 OF2 = OD2 - FD2 OF2=25−16=9 (Perpendicular drawn from centre to a chord bisects the chord, so FD = 4 cm) ⇒ OF = 3 cm Here, the figure formed by the points A, B, C, and D is a quadrilateral with one pair of sides parallel to each other. Hence, ABCD is a trapezium. We have, Area of trapezium = 12(perpendicular distance between two parallel sides)(sum of the lengths of the parallel sides) =12 x (OE + OF) x (AB + CD) = 12(4 + 3)(8 + 6) = 49 cm2

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