Question

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A chord of length 16 cm is drawn in a circle of radius 10 centimetre. Find the distance of the chord from the centre of the circle.

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Answer 1

$\huge\underline\mathrm{Answer-}$

Distance of chord from the centre is 6 cm.

$\huge\underline\mathrm{Explanation-}$

Given :

• Length of chord = 16 cm

• Radius of circle = 10 cm

To find :

• Distance of chord from the centre.

Solution :

We know that, if a perpendicular is drawn on the chord, then it bisect the chord.

So, length of chord = 8 cm.

Now, by using Pythagoras theorem,

OA² = OC² + AC²

=> 10² - 8² = OC²

=> 100 - 64 = OC²

=> OC² = 36

=> OC = ±√36

=> OC = ±6

Rejecting the negative value, we get,

OC = 6 cm.

•°• Distance of chord from the centre is 6 cm.

Answer 2

$\sf \fcolorbox{red}{pink}{ \huge{Solution :)}}$

Given ,

• The chord of length (AB) = 16 cm
• Radius of circle (AO) = 10 cm

We know that , the perpendicular drawn from the centre of the circle to the chord , bisects the chord

Thus , C is the midpoint of AB

AC = AB/2

AC = 16/2

AC = 8 units

Now , By Pythagoras theorem

$\large \sf \fbox{ {(AO) }^{2} = {(AC)}^{2} + {(OC)}^{2} }$

Substitute the known values , we get

(10)² = (8)² + (OC)²

100 = 64 + (OC)²

(OC)² = 36

(OC) = 6 cm {ignore negative value because distance can't be negative}

Hence , the distance of the chord from the centre of the circle is 6 cm

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