Question

A chord of length 24 cm is at a distance of 5 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the centre.

• A. 13 cm
• B. 10 cm
• C. 5 cm
• D. 21 cm

Answer 1

The correct option is B

10 cm

A circle with centre O and radius OA

Chord AB = 24 cm

OM $\perp$ AB which bisects it at M.

$\therefore$ OM = 5 cm and

$AM=\frac{1}{2}AB=\frac{1}{2}\times 24=12cm$

Now in right $\Delta OAM$,

$O{A}^2=A{M}^2+O{M}^2$ (Pythagoras Theoram)

$=(12{)}^2+(5{)}^2=144+25=169$

$\therefore OA=\sqrt{169}=13cm$

Again OA = 13 cm and distance from the centre is 12 cm.

Again in right $\Delta ONA$,

$\therefore A{O}^2=O{N}^2+N{A}^2$

$\Rightarrow (13{)}^2=(12{)}^2+N{A}^{2}169=144+N{A}^2$

$\Rightarrow N{A}^2=169-144=25=(5{)}^2$

$\Rightarrow NA=5cm$

Hence $AC=2\times NA=2\times 5=10cm$