The Magic Of Construction. Try It Now.

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Let us draw the following.

Complete the larger chord $CD$. From the center $O$, draw perpendicular $OP$ to the smaller chord $AB$.

Join the two radii $OB$ and $OD$ as shown in the figure below.

Since $AB \parallel CD$, the line $OP \perp AB$, it is also perpendicular to $CD$.

$ABFE$ is a rectangle, therefore $FE = 24$

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$OP$ bisects both $AB$ and $EF$.

$\therefore QE = QF = 24 \div 2 = 12$

Also,

$FD = EC = EF - AB = 28 - 24 = 4$

We also have:

$QD = QF + FD = 12 + 4 = 16$

and

$OP = OQ + 4$

Using right angled $\triangle OQD$, we have

$OD^2 = OQ^2 + QD^2$

$\Rightarrow r^2 = OQ^2 + 16^2$ $...(i)$

Similarly, using right angled $\triangle OPB$

$OB^2 = OP^2 + PB^2$

$\Rightarrow r^2 = (OQ+4)^2 + 12^2$ $...(ii)$

Combining $(ii)$ and $(i)$

$OQ^2 + 16^2 = (OQ+4)^2 + 12^2$

$\Rightarrow OQ^2 + 16^2 = OQ^2 + 8 OQ + 4^2 + 12^2$

$\Rightarrow 8OQ = 16^2 - 12^2 - 4^2$

$\Rightarrow OQ = 12$

Substituting the value of $OQ$ in equation $(i)$ we get:

$r^2 = 12^2 + 16^2$

$r^2 = 4^2 (3^2 + 4^2)$

$r = \sqrt{4^2 \times 25} = 4 \times 5 = 20$