Question

Question 3
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Answer 1

Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Given,
AB = 6 cm and CD = 8 cm
Distance of smaller chord AB from the centre of the circle is 4 cm
i.e; OM = 4 cm
$MB=\frac{AB}{2}=\frac{6}{2}=3cm$ (Perpendicular from the centre bisects the chord)

In $\Delta OMB$
$O{M}^2+M{B}^2=O{B}^2$ [using pythagoras theorem]
$(4{)}^2+(3{)}^2=O{B}^2$
$16+9=O{B}^2$
$OB=\sqrt{25}$
OB=5cm

In $\Delta OND,$
OD =OB=5cm (Radii of the same circle)
$ND=\frac{CD}{2}=\frac{8}{2}=4cm$
$O{N}^2+N{D}^2=O{D}^2$ [using pythagoras theorem]
$O{N}^2+(4{)}^2=(5{)}^2$
$O{N}^2=25-16=9$
ON=3cm
Therefore, the distance of the bigger chord from the centre is 3 cm.