Question

The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre.​

Answer 1

Given :

• AB = 24cm

• Radius of the circle = 13cm : OP

To Find :

Distance between chord and the centre.

Solution :

Step 1.

Draw a perpendicular from centre to AB at C.

( Perpendicular from centre bisects the chord)

Step 2.

Join O and A.

• OA = Radius = 13cm

• AC = BC = 12cn

Step 3.

With the help of pythogoras theorem which is :

$\pink{\bf{(H)² } = (P)² + (B)²}$

Here ;

• H = 13cm = OA

• H = 13cm = OAB = 12cm = AC

• H = 13cm = OAB = 12cm = ACP = To Find = OC

Step 4.

$\sf {OA}^{2} = {AC}^{2} + {OC}^{2} \\ \\ \sf {(13cm)}^{2} = {(12cm)}^{2} + {OC}^{2} \\ \\ \sf169 {cm}^{2} - 144 {cm}^{2} = {OC}^{2} \\ \\ \sf{25cm}^{2} = {OC}^{2} \\ \\ \sf \sqrt{ {(25cm)}^{2} } = OC \\ \\ \boxed{\pink{\sf5cm = OC}}$

Thus the distance between the centre and chord is 5cm.

I HOPE IT HELPS.