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

The correct option is (c) or 0.5.

Let the radius of the circle be $R=11$ cm. Since CD is a diameter, its length is:
$CD=2R=22$ cm.

We are given that AB and CD intersect at a point E inside the circle, with $CE=7$ cm. Since CD is a diameter, $ED=CD-CE=22-7=15$ cm.

By the Intersecting Chords Theorem for chords AB and CD intersecting at E:
$AE\cdot BE=CE\cdot ED$
$AE\cdot BE=7\cdot 15=105$.

We are also given the length of chord AB as:
$AB=AE+BE=20.5$ cm.

We want to find the absolute difference between the lengths of BE and AE, i.e., $|BE-AE|$. Using the algebraic identity:
${(BE-AE)}^2={(BE+AE)}^2-4(AE\cdot BE)$
${(BE-AE)}^2={\left(20.5\right)}^2-4\left(105\right)$
${(BE-AE)}^2=420.25-420=0.25$.

Taking the square root of both sides, we get:
$|BE-AE|=\sqrt{0.25}=0.5$ cm.