Question

Question 2
Two chords AB and CD of lengths 5 cm 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 6 cm, find the radius of the circle.

Answer 1

Given,
AB = 5 cm
CD = 11cm
Draw $OM\perp AB$ and $ON\perp CD$, 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 $\Delta MOB$,
$O{M}^2+M{B}^2=O{B}^2$ [using pythagoras theorem]
$(6-x{)}^2+{\left(\frac{5}{2}\right)}^2=O{B}^2$
$36+x^2-12x\frac{25}{4}=O{B}^2......\left(1\right)$

In $\Delta NOD$,
$O{N}^2+N{D}^2=O{D}^2$ [using pythagoras theorem]
$x^2+{\left(\frac{11}{2}\right)}^2=O{D}^2$
$x^2+\frac{121}{4}=O{D}^2.......\left(2\right)$

OB = OD (radius of circle)
From (1)and (2)
$36+x^2-12x+\frac{25}{4}=x^2+\frac{121}{4}$
$12x=36+\frac{25}{4}-\frac{121}{4}$
$12x=\frac{144+25-121}{4}=\frac{48}{4}=12$
x=1
From equation (2),
$(1{)}^2+\left(\frac{121}{4}\right)=O{D}^2$
$O{D}^2=1+\frac{121}{4}=\frac{125}{4}$
$OD=\frac{5}{2}\sqrt{5}$
Therefore, the radius of the circle is $\frac{5}{2}\sqrt{5}cm$.