Question

$PQ$ and $RS$ are two parallel chords of a circle whose centre is $O$ and radius is $10cm$. If $PQ=16cm$ and $RS=12cm$. Find the distance between $PQ$ and $RS$ if they lie on the opposite side of the centre.

• A. 14 cm
• B. 12 cm
• C. 10 cm
• D. 11 cm

Answer 1

Answer: (A)

($14cm$)

Given $PQ=16cm$

$\Rightarrow PL=LQ=\frac{16}{2}=8cm$ as the perpendicular from the centre of the circle bisects the chord.

$△OQL$ is a right angled triangle.
From Pythagorean theorem, we have $O{Q}^2=O{L}^2+L{Q}^2$
$\Rightarrow O{L}^2=O{Q}^2-Q{L}^2$
$={10}^2-8^2$
$=100-64=36$
$\therefore OL=\sqrt{36}=6cm$

And also, $RS=12cm$

$\Rightarrow RM=MS=6cm$ [the perpendicular from the centre of the circle bisects the chord.]

Similarly, $△OSM$ is a right angled triangle.
Using Pythagoras theorem, $O{S}^2=O{M}^2+S{M}^2$
$\Rightarrow O{M}^2=O{S}^2-S{M}^2$
$={10}^2-6^2$
$=100-36$
$=64$
$\Rightarrow OM=\sqrt{64}$
$\therefore OM=8cm$

$\therefore$ $LM=LO+OM=8+6=14cm$

Hence, the distance between the two chords is $14cm$.