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Solution:

We have a circle with center O, radius 20 and a chord AB of length 32

Attachment:
circle2.png
circle2.png [ 10.28 KiB | Viewed 995 times ]

Let us draw a line OC perpendicular to chord AB from center O

We know that a perpendicular from the center to a chord will bisect the chord. This means \(AC=CB=\frac{32}{2}=16\)

Thus, we need the length \(OC\)

Attachment:
circle3.png
circle3.png [ 17.71 KiB | Viewed 977 times ]

Applying Pythagoras theorem on triangle \(OCB\), we can write \(OB^2=OC^2+CB^2\)

\(⇒ 20^2=OC^2+16^2 \)
\(⇒ 400=OC^2+256 \)
\(⇒ OC^2=400-256\)
\(⇒ OC^2=144\)
\(⇒ OC=12\)

Hence the right answer is Option E
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