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Find the length of a chord which is at a distance of $4\ cm$ from the centre of the circle of radius $6\ cm$.
Given:
A chord is at a distance of $4\ cm$ from the centre of a circle of radius $6\ cm$.
To do:
We have to find the length of the chord.
Solution:
Let $AB$ be a chord of a circle with radius $6\ cm$.
$OL \perp AB$
This implies,
$OA = 6\ cm$
$OL = 4\ cm$
$OL$ divides $AB$ into two equal parts.
$AL = LB$
In right angled triangle $OAL$,
$OA^2 = OL^2 + AL^2$ (Pythagoras Theorem)
$(6)^2 = (4)^2 + AL^2$
$36 = 16 + AL^2$
$AL^2 = 36 - 16 = 20$
$\Rightarrow AL = \sqrt{20}$
$= \sqrt{4\times5}$
$= 2 \times 2.236$
$AB = 2 \times AL = 2 \times 2 \times 2.236$
$= 4 \times 2.236$
$= 8.94\ cm$
The length of the chord is $8.94\ cm$.
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