Question

The radius of the circle is 6cm.What is the perpendicular distance from the centre of the circle if the length of the chord is 8cm?​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

We have that:

• Radius of Circle C(O,r) = 6 cm
• Length of the Chord = 8 cm

⚘ As We know that when Perpendicular is drawn onto the chord from the centre of the circle then Chord will bisect Equal in two Parts.

Let the Chord AB, when line drawn from centre 'O' then AB bisect on point M.

• OA = OB = 6 cm [Radii of Circle]
• AM = BM [Given]
• OM = OM [Common]
• AM = BM = 4 cm

⚘ Now, We can use Pythagorean theorem to find the Distance From the centre of the circle as:-

$\colon\Rightarrow {\sf{ (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 }} \\ \\ \colon\Rightarrow {\sf{ (OA)^2 = (AM)^2 + (OM)^2 }} \\ \\ \colon\Rightarrow {\sf{ (6)^2 = (4)^2 + (OM)^2 }} \\ \\ \colon\Rightarrow {\sf{ 36 = 16 + OM^2 }} \\ \\ \colon\Rightarrow {\sf{ 36 - 16 = OM^2 }} \\ \\ \colon\Rightarrow {\sf{ OM = \sqrt{20} }} \\ \\ \colon\Rightarrow {\underline{\boxed{\sf{ OM = 2 \sqrt{5} cm_{(Distance \ from \ Centre)} }}}}$

Hence,

${\pmb{\underline{\sf{ The \ Distance \ b/w \ Chord }}}} \\ {\pmb{\underline{\sf{and \ Centre \ of \ the \ Circle \ is \ 2 \sqrt{5} cm . }}}}$

Answer 2

$\sf ANSWER - \red{2\sqrt{5} cm}$

Given :

• The radius of the circle is 6cm.

To find :

• What is the perpendicular distance from the centre of the circle if the length of the chord is 8cm .

Solution :

Given chord = Let AB = 8 cm

Radius OA = 6 cm

.

Now,

The perpendicular from the centre of the circle to the bisects .

AP = PB = 4 cm

In rt , Angle D Triangle OAP = OA² - AP²

= 36 - 16 = 20 cm

$\sf OP = \sqrt{20} = 2 \sqrt{5} \: cm$