Question

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​

Answer 1

The idea that we use here is remarkably simple.

We know, the product of rectangles formed by two intersecting chords are always equal

So, AE x EB = CE x ED

AE x EB = 7 x 15

Also, we know that AB = 20.5 cms = AE + EB

So, the Sum of AE and EB must be 20.5 and their product must be equal to 7 x 15

7 x 15 = 105

The numbers must be close to each other, for their sum to be 20.5

From trial and error, we find their values to be 10 and 10.5 cms respectively

So, AE x EB = 7 x 15

10 x 10.5 = 7 x 15

Difference = 10.5 - 10 cms = 0.5 cm

Answer

0.5 cm

Answer 2

Answer:

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