Question

two parallel chords of length 30 cm and 16 cm are drawn on the opposite sides of the centre of a cricle of radius 17 cm . find the distance between the chords with diagram

Answer 1

The chord is bisected by a perpendicular drawn from the circle's centre to the chord.

$=>AB=AE+AE ...(AE=EB)$

$=>30=2AE$

$=>AE=15cm$

$=>EB=AE=15cm$

Now,

In Δ PEB,

∠PEB $=90^o$

By using Pythagoras theorem,

$=>PB^2=PE^2+EB^2$

$=>17^2=PE^2+15^2$

$=>PE^2=17^2-15^2$

$=>PE^2=289-225$

$=>PE=\sqrt{64}=8$

Now,

$CF=FD$

$=>CD=CF+FD$

$=>CD=CF+CF$

$=>CD=2CF$

$=>CF=\frac{16}{2}$

$=>CF=FD=8$

So, In Δ$PFD$,

∠$PFD =$ $90^o$

$=>PD^2=PF^2+FD^2$

$=>17^2=PF^2+8^2$

$=>PF^2=289-64$

$=>PF^2=225$

$=>PF=\sqrt{225}=15cm$

As the chords are parallel to each other the distance between them,

Therefore,

$=>EF=EP+PF$

$=>EF=8+15$

$=>EF=23cm$

Hence, the distance between the chords is $23cm.$