Question

In the following figure, if chords AB and CD are equal to 6 cm, then the distances of the chords from the centre of the circle of radius 5 cm are ___ and ___ respectively.

Answer 1

Given that chords AB and CD are equal to 6 cm.
Draw perpendiculars from the centre of the circle to these chords.
We know the property that the perpendicular drawn from the centre of a circle to the chord bisects the chord.
$\therefore AE=BE=DF=CF=\frac{6}{2}=3cm.$

Considering $△OEB$ and $△OFD,$ we have
OB=OD, (Radii of the same circle)

$\angle OEB=\angle OFD={90}^{\circ }$

BE=DF.

$\text{Thus, by RHS criterion,}△OEB\cong △OFD.$

$⟹OE=OF$ (By C.P.C.T.)
But, using Pythagoras' theorem in $△OEB,$ we have
$O{E}^2+B{E}^2=O{B}^2.$
$⟹O{E}^2+3^2=5^2⟹OE=4cm$
Thus, $OE=OF=4cm.$