Question

$PQ$ and $RS$ are two parallel chords of a circle. with centre $C$ such that $PQ=8$ cm and $RS=16$ cm. If the chords are on the same side of the centre and the distance between them is $4$ cm, then the radius of the circle is:

• A. $3\sqrt{2}$ cm
• B. $3\sqrt{5}$ cm
• C. $4\sqrt{5}$ cm
• D. $5\sqrt{5}$ cm

Answer 1

Answer: (C)

$4\sqrt{5}$ cm

Given, chords $PQ=8$ cm and $RS=16$ cm and $AB=4$ cm
Let perpendicular from the centre meet the chord RS at B and PQ at A.
Since, the perpendicular from the centre of the circle to the chord bisects the chord,
$RB=BS=8$ cm and $PA=AQ=4$ cm.
Let $OB=x$ cm, then $OA=OB+AB=(x+4)$ cm.
In right triangle $OBS,$
$O{S}^2=O{B}^2+B{S}^2=O{B}^2+64$ (i)
In right triangle $OAQ,$
$O{Q}^2=O{A}^2+A{Q}^2=O{A}^2+16$ (ii)
$\therefore$ $OS$ and $OQ$ are the radii of the circle
$\therefore$ $O{B}^2+64=O{A}^2+16$
$\Rightarrow$ $x^2+64={(x+4)}^2+16$
$\Rightarrow$ $x^2+64=x^2+8x+16+16$
$\Rightarrow$ $8x=32\Rightarrow x=4$
$\therefore$ From eqn (i)
Radius of the circle$(OS{)}^2=16+64=80$
$\Rightarrow$ $OS=4\sqrt{5}$ cm