Question

AB and CD are two parallel chords of a circle of radius 3 cm IF the length of AB is 4 cm and of CD is 5 cm then the distance between them is ( in cm)

• A. $\sqrt{5}+\frac{1}{\sqrt{11}}$
• B. $\frac{\sqrt{5}}{2}+\sqrt{11}$
• C. $\sqrt{5}+\frac{\sqrt{11}}{2}$
• D. $2\sqrt{5}+\sqrt{11}$

Answer 1

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

Let O be the center of the circle.

From O draw OM$\perp$AB and ON$\perp$CD

$\therefore AM=\frac{1}{2}AB=\frac{1}{2}\times 4=2cm$

And $CN=\frac{1}{2}CD=\frac{1}{2}CD=\frac{1}{2}\times 5=\frac{5}{2}cm$

In$△CNO$

$\Rightarrow (CO{)}^2=(ON{)}^2+(CN{)}^2$

$\Rightarrow (ON{)}^2=(CO{)}^2-(CN{)}^2$

$\Rightarrow (ON{)}^2=(3{)}^2-(\frac{5}{2}{)}^2$

$\Rightarrow (ON{)}^2=9-\frac{25}{4}$

$\Rightarrow (ON{)}^2=\frac{36-25}{4}$

$\Rightarrow (ON{)}^2=\frac{11}{4}$

$\Rightarrow ON=\frac{\sqrt{11}}{2}$

In$△AOM$

$\Rightarrow (AO{)}^2=(OM{)}^2+(AM{)}^2$

$\Rightarrow (OM{)}^2=(AO{)}^2-(AM{)}^2$

$\Rightarrow (OM{)}^2=(3{)}^2-(2{)}^2$

$\Rightarrow (OM{)}^2=9-4$

$\Rightarrow (OM{)}^2=5$

$\Rightarrow OM=\sqrt{5}$

In figure MN=ON+OM

$\therefore MN=\sqrt{5}+\frac{\sqrt{11}}{2}$