Question

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

Answer 1

Draw $OM\perp ABamON\perp CD$. join OB and OD.


$BM=\frac{AB}{2}=\frac{5}{2}$
(Perpendicular from the centre bisects the chord)
$ND=\frac{CD}{2}=\frac{11}{2}$
Let ON be x. Therefore, OM will be 6-x.
In $\Delta MOB$,
$O{M}^2+M{B}^2=O{B}^2(6-x{)}^2+{\left(\frac{5}{2}\right)}^2=O{B}^{2}36+x^2-12x+\frac{25}{4}=O{B}^2$ .....(1)
In $\Delta NOD$,
$O{N}^2+N{D}^2=O{D}^2{x}^2+{\left(\frac{11}{2}\right)}^2=O{D}^2{x}^2+\frac{121}{4}=O{D}^2$ ...(2)
We have OB=OD (Radii of the same circle)
Therefore, from equation (1) and (2).
$36+x^2-12x+\frac{25}{4}=x^2+\frac{121}{4}12x=36+\frac{25}{4}-\frac{121}{4}=\frac{144+25-121}{4}=\frac{48}{4}=12x=1$
From equation (2)
$(I{)}^2+\left(\frac{121}{4}\right)=O{D}^{2}O{D}^2=I+\frac{121}{4}=\frac{125}{4}OD=\frac{5}{2}\sqrt{5}$