CAT 2019 Shift 1 Paper

Question : 92 of 100

Marks: +1, -0

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

Validate

Solution:

Refer to the diagram below:

Applying chord chord power theorem

A E × B E=C E × D E

⇒ A E × B E=7 × 15=105 . . .(1)

Also, it is given that

{AE}+ {BE}=20.5 . . .(2)

(A E-B E)^{2}=(A E+B E)^{2}-4 × A E × B E

⇒(A E-B E)^{2}=(20.5)^{2}-4 × 105

⇒(A E-B E)^{2}=420.25-420=0.25

⇒(A E-B E)=0.5

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