Question 88

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

Solution

In figure AE*BE=CE*DE    (The   intersecting chords theorem)

=> 7*15= x(20.5-x)                  (Assuming AE=x)

=> 210=x(41-2x)

=> $$2x^2$$-41x+210=0

=> x=10 or x=10.5   => AE=10 or AE=10.5      Hence BE = 20.5-10=10.5  or BE = 20.5-10.5=10

Required difference= 10.5-10=0.5

Video Solution

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