Question

Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.

Answer 1

OA = 6 cm and OL $\perp$ AB, OL = 4 cm

$\because$ Perpendicular OL bisects the chord AB at L

$\therefore AL=LB=\frac{1}{2}AB$

Now in right $\Delta OAL$,

$O{A}^2=O{L}^2+A{L}^2$ (Pythagoras Theorem)

$(6{)}^2=(4{)}^2+A{L}^2$

$\Rightarrow 36=16+A{L}^2$

$\Rightarrow A{L}^2=36-16=20$

$\therefore AL=\sqrt{20}=\sqrt{4\times 5}=2\times 2.236$

$=4.472cm$

$\therefore ChordAB=4.472\times 2$

$=8.944=8.94cm$