Question

AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is

• A. 6 cm
• B. $5\sqrt{2}cm$
• C. 7 cm
• D. $3\sqrt{5}cm$

Answer 1

The correct option is D

$3\sqrt{5}cm$

AB and CD are two parallel chords of a circle with centre O.

Let r be the radius of the circle.

AB = 6 cm, CD = 12 cm and LM = 3 cm

Join OC and OA.

Let OM = x, then OL = x + 3

Now in right $\Delta OCM$, $O{C}^2=O{M}^2+C{M}^2=(x{)}^2+{\left(\frac{CD}{2}\right)}^2$

$\Rightarrow r^2=x^2+{\left(\frac{12}{2}\right)}^2=x^2+36$ .......(i)

In right $\angle OAL$, $O{A}^2=O{L}^2+A{L}^2$

$\Rightarrow r^2=(x+3{)}^2+{\left(\frac{6}{2}\right)}^2=(x+3{)}^2+(3{)}^2$

$=(x+3{)}^2+9$ ............(ii)

From (i) and (ii), $(x+3{)}^2+9=x^2+36$

$\Rightarrow x^2+6x+9+9=x^2+36$

$\Rightarrow 6x=36-18=18\Rightarrow x=\frac{18}{6}=3$

$\therefore (Radius{)}^2=r^2=x^2+36$

$=(3{)}^2+36=9+36=45$

$=9\times 5$

$\therefore r=\sqrt{9\times 5}=3\sqrt{5}cm$