Question

Two parallel chords are drawn on either side of the centre of a circle of diameter 30 cm. If the length of one chord is 24 cm and the distance between the two chords is 21 cm, then find the length of another chord.
$\left[3Marks\right]$

Answer 1

Radius of the circle $=\frac{30}{2}=15cm$

Chord AB || chord CD.

Length of one chord AB = 24 cm and distance MN between the two chords = 21cm.

$\because$ By Theorem- The perpendicular from the centre to the chord bisects it
$\left[1Mark\right]$

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

In right $\Delta OAM$,

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

$\Rightarrow (15{)}^2=O{M}^2+(12{)}^2\Rightarrow 225=O{M}^2+144$

$\therefore O{M}^2=225-144=81=(9{)}^2$

$\therefore$ OM = 9 cm. But MN = 21 cm
$\left[1Mark\right]$

$\therefore$ ON = MN - OM = 21 - 9 = 12 cm

Similarly in right $\Delta OCN$,

$O{C}^2=O{N}^2+C{N}^2\Rightarrow (15{)}^2=(12{)}^2+C{N}^2$
$\left[1Mark\right]$

$\Rightarrow 225=144+C{N}^2$

$\Rightarrow C{N}^2=225-144=81=(9{)}^2$

$\therefore CN=9cm$

$\because CN=\frac{1}{2}CD$

$\therefore CD=2\times CN=2\times 9=18cm$