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Quantitative Aptitude Question on Triangles, Circles & Quadrilaterals

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

A

2.5

B

3.5

C

0.5

D

1.5

Answer

0.5

Explanation

Solution

Referring to the diagram below:
In a circle of radius 11 cm, CD is a diameter and AB is a chord
Using the chord-chord power theorem, the product of AE and BE is equal to the product of CE and DE, given by AE x BE = 7 x 15 = 105 …(1)
Additionally, it is provided that AE + BE = 20.5 ...(2)
Applying the formula (AE - BE)2 = (AE + BE)2 - 4 x AE x BE, we find (AE - BE)2 = (20.5)2 - 4 x 105, leading to (AE - BE)2 = 420.25 - 420 = 0.25
Consequently, (AE - BE) equals 0.5