Question

A radius of a circle is 8cm and length of its one chord is 12cm.Find the distance of chord from the centre.​

Answer 1

Answer:

Given:

=> Radius of circle (OB) = 8 cm

=> Length of one chord (AB) = 12 cm

To Find:

=> Distance of chord from the centre (OM).

Formula used:

=> Pythagoras theorem

As we know, The perpendicular from center to chord bisects the chord.

$\sf{\therefore AM=BM=\dfrac{12}{2}=6\;cm}$

Now, by using Pythagoras theorem we will find distance of chord from centre.

=> OB² = OM² + MB²

=> 8² = OM² + 6²

=> 64 = OM² + 36

=> OM² = 28

=> OM = √28

Hence, distance of chord from the centre = √28 cm.

Answer 2

Answer:

$\large\bold\red{2\sqrt{7}\:cm}$

Step-by-step explanation:

We have been given that,

• Radius, OQ = 8 cm
• Chord, PQ = 12 cm

To find,

• The distance of chord from centre

Construction:

• Draw a perpendicular OR on PQ from center O.

Note:- Refer to the attachment for figure.

Now,

From the properties of circles,

• This Perpendicular will bisect the given chord.

Therefore,

$=>PR = QR = \dfrac{PQ}{2}=\dfrac{12}{2}\\\\=>QR=6\:cm$

Now,

In ∆OQR,

By Pythagoras Theorem,

We have,

$= > {OQ}^{2} = {OR}^{2} + {QR}^{2} \\ = > {OR}^{2} = {OQ}^{2} - {QR}^{2}$

Therefore,

Putting the respective values,

We get,

$= > {(OR)}^{2} = {8}^{2} - {6}^{2} \\ = > {(OR)}^{2} = 64 - 36 \\ = > {(OR)}^{2} = 28 \\ = > OR = \sqrt{28} \\ = > OR = 2 \sqrt{7}$

Hence,

The distance of chord from centre is $\large\bold{2\sqrt{7}\:cm}$