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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=2516=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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