Question

In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords.

• A. $12cm$
• B. $18cm$
• C. $16cm$
• D. $14cm$

Answer 1

The correct option is D $14cm$


Draw $OE$ and $OF$ which is depicts the perpendicular distance of the chords from the centre.

Since, the perpendicular from the centre to the chord bisects the chord,

$AF=FB=\frac{AB}{2}=\frac{16}{2}$

$AF=FB=8cm$

$CE=ED=\frac{CD}{2}=\frac{12}{2}$

$CF=ED=6cm$

In $△OFB$,
$O{F}^2+F{B}^2=O{B}^2$
$O{F}^2+8^2={10}^2$
$O{F}^2=100-64$
$O{F}^2=36$
$OF=\sqrt{36}$
$OF=6cm$

In $△OED$,
$O{E}^2+E{D}^2=O{D}^2$
$O{E}^2+6^2={10}^2$
$O{E}^2=100-36$
$O{E}^2=64$
$OE=\sqrt{64}$
$OE=8cm$

$\Rightarrow FE=OF+OE$
$\Rightarrow FE=6+8$
$\Rightarrow FE=14cm$

$\therefore$ The distance between the two chords is $14cm$.