Question

Two chords $AB$ and $CD$ of lengths $5cm$ and $11cm$ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between $AB$ and $CD$ is $6cm$, find the radius of the circle.

• A. $\frac{5\sqrt{5}}{2}$
• B. $\frac{11\sqrt{5}}{2}$
• C. $\frac{5\sqrt{5}}{3}$
• D. $\frac{5\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{5\sqrt{5}}{2}$

We know $AM=\frac{AB}{2}=\frac{5}{2}cm$

and $CN=\frac{CD}{2}=\frac{11}{2}cm$

Let $MOZ\times cm\Rightarrow ON=6-xcm$

using pythoqeros in $\Delta AOM$, we get

$A{D}^2=A{M}^2+M{O}^2\Rightarrow R^2=x^2+{\left(\frac{5}{2}\right)}^2$ .... (1)

Using pythaqeros in $\Delta CO{N}_0$ we get

$C{O}^2=O{N}^2+C{N}^2+R^2=(6-x^2)+{\left(\frac{11}{2}\right)}^2$ ... (2)

On subtracting (2) from (1) , we gert

$x^2-(6-x{)}^2+{\left(\frac{5}{2}\right)}^2-{\left(\frac{11}{2}\right)}^2=0$

$(x+6-x)(x-6+x)+\frac{25}{4}-\frac{121}{4}=0$

$6(2x-6)=\frac{96}{4}=4$

$2x=10$

$x=5cm$

on putting value of x in equation (1) , we get

$R^2=(5{)}^2+{\left(\frac{5}{2}\right)}^2=25+\frac{25}{4}=\frac{125}{4}$

$\Rightarrow R=\frac{5\sqrt{5}}{2}cm$