Question

Two chords $AB$ and $CD$ of a circle are each at distances $4cm$ from the centre Then find $AB-CD$

Answer 1

Given:$AB$ and $CD$ are chords of a circle and they are equidistant $4$cm from the center $O$.

$OP$ and $OQ$ are perpendiculars$=4$cm from $O$ to $AB$ and $CD$ respectively.

$\therefore$ they bisect $AB$ and $CD$

$\Rightarrow AP=\frac{1}{2}AB$ and $DQ=\frac{1}{2}DC$

$\Rightarrow \angle OPA=\angle OQD={90}^{\circ }$

Now, between the $△AOP$ and $△DOQ$ we have

$OP=OQ=4$cm (given)

$\angle OPA=\angle OQD={90}^{\circ }$

$\Rightarrow OB=OD$ (radii of the same circle)

$\therefore △BOP$ is identical to $△DOQ$ by RHS test

$\therefore AP=DQ\Rightarrow AB=DC$

$\Rightarrow AB-DC=0$