Question

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

Answer 1

Let r be the radius of the circle with centre O. Two parallel chords AB = 5 cm, CD = 11 cm

Let $OL\perp ABandOM\perp CD$

$\therefore LM=6cm$ Let $OM=x,then$

$OL=6-x$

Now in right $\Delta OAL$,

$O{A}^2=O{L}^2+A{L}^2$

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

$\left\{\begin{array}{cc}\because OL\perp AB& \\ \therefore Lismidpoint& \end{array}\right\}$

$=36-12x+x^2+\frac{25}{4}................\left(i\right)$

Similarly in right $\Delta OCM$,

$r^2=x^2+{\left(\frac{11}{2}\right)}^2=x^2+\frac{121}{4}...............\left(ii\right)$

From (i) and (ii),

$x^2+\frac{121}{4}=36-12x+x^2+\frac{25}{4}$ (Since both are radii of the circle)

$\Rightarrow \frac{121}{4}-\frac{25}{4}-36=-12x$

$\Rightarrow \frac{96}{4}-\frac{36}{1}=-12x$

$\Rightarrow 12x=36-24=12$

$x=\frac{12}{12}=1$

$\therefore r^2=C{M}^2+O{M}^2$

$={\left(\frac{11}{2}\right)}^2+(1{)}^2$

$=\frac{121}{4}+1=\frac{125}{4}cm$

$\therefore r=\sqrt{\frac{125}{4}}=\frac{\sqrt{125}}{2}=\frac{\sqrt{25\times 5}}{2}cm$ $=\frac{5}{2}\sqrt{5}cm$