Question

$\overline{AB}and\overline{CD}$ are two parallel chords of a circle, which are on opposite sides of the centre, such that $\overline{AB}=10cm$, $\overline{CD}=24cm$ and the distance between them is

Answer 1

$OA=OB=OC=OD$

= radius of circle

$\to EO+OF=EF$

$AB=10cm$

,$AE=EB=5cm$

$CD=24cm$

$CF=FD=12cm$

In $△AEO$ ( rt angle triangle)

$\therefore A{O}^2=A{E}^2+O{E}^2$

$A{O}^2=5^2+O{E}^2------\left(1\right)$

In $△COF$ ( rt angle triangle)

$O{C}^2=O{F}^2+C{F}^2$

$O{C}^2=O{F}^2+{12}^2------\left(2\right)$

Since $AO=OC$ and comparing $\left(1\right)$ & $\left(2\right)$, we get

In $\left(1\right)$ & $\left(2\right)$ let radius $=13cm$

$O{E}^2={13}^2-5^2=169-25$

$=\sqrt{144}$

$OE=12cm$

$O{F}^2={13}^2-{12}^2$

$=164-144=25$

$OF=5cm$

as $EF=OE+OF=12+5$

$EF=17cm$