Question

AB and CD are two parallel chords of a circle of radius $3$ cm. If $AB=4$ cms and $CD=5$ cm. Then the distance between them (in cm) is

• A. $\frac{\sqrt{5}}{2}+\sqrt{11}$
• B. $\sqrt{5}+\sqrt{11}$
• C. $\sqrt{5}+\frac{\sqrt{11}}{2}$
• D. $\sqrt{2}+\frac{\sqrt{11}}{\sqrt{5}}$

Answer 1

The correct option is C $\sqrt{5}+\frac{\sqrt{11}}{2}$

Let $AB$ and $CD$ be two parallel chords.
Radius $=3$cm

Now from $△OFB$,
$O{F}^2=9-4=5,\therefore OF=\sqrt{5}$

From $△OEC$,
$O{E}^2=9-\frac{25}{4}=\frac{11}{4}\Rightarrow OE=\frac{\sqrt{11}}{2}$

So distance between $AB$ and $CD$
$=OF+OE=\sqrt{5}+\frac{\sqrt{11}}{2}$