Question

Two chords $AB,$ $CD$ of lengths $5cm$ and $11cm$ respectively of a circle are parallel. If the distance between $AB$ and $CD$ is $3cm,$ find the radius of the circle.

Answer 1

Given: $AB=5cm,$ $CD=11cm$

Distance between $AB$ and $CD=3cm$

Join $OB$ and $OD.$

$OL$ and $OM$ are the perpendiculars on $CD$ and $AB$ respectively which bisect $CD$ and $AB.$

Let $OL=x,$ Then $OM=(x+3)$

Now in right $\Delta OLD$, we have

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

$=x^2+(5.5{)}^2$ [$\because LD=\frac{1}{2}CD=5.5cm$]

Similarly in right $\Delta OMB$, we have

$O{B}^2=O{M}^2+M{B}^2=(x+3{)}^2+(2.5{)}^2$ [$\because MB=\frac{1}{2}AB=2.5cm$]

But $OD=OB$ (Radii of the circle)

$\therefore (x+3{)}^2+(2.5{)}^2=x^2+(5.5{)}^2$

$\Rightarrow x^2+6x+9+6.25=x^2+30.25$ [$\because (a+b{)}^2=a^2+b^2+2ab$]

$\Rightarrow 6x=30.25-6.25-9=15$

$\Rightarrow x=\frac{15}{6}=\frac{5}{2}=2.5$

Now, $O{D}^2=x^2+(5.5{)}^2$ [from above]

$=(2.5{)}^2+30.25=6.25+30.25=36.50$

$OD=\sqrt{36.50}=\sqrt{\frac{3650}{100}}=\sqrt{\frac{146}{4}}$

$\therefore$ Radius $=OD=OB=\frac{\sqrt{146}}{2}cm$