Question

In a circle with center P, a chord of length 16 cm is at a distance of 8 cm from the centre. Find the radius of this circle.

Answer 1

Step-by-step explanation:

chord AB=16cm

let p bisect the chord AB at O

OP=8cm

PA=radius

OA=8cm(OA=OB)

therefore OPA is r8 angle triangle

therefore OP= ✅OA^2 + OP^2

=✅8×8 + 8×8

=✅128

=11.31 cm

Answer 2

$\underline{\large{\sf Answer :}}$

Here we have given,

A chord of length 16 cm And the distance of chord from the center is 8 cm.

we know if the line is drawn from the center of the circle to a chord is always perpendicular to the chord

let us consider, the chord is

AB = 16cm, perpendicular drawn from the center is MN = 8cm, Now join the points MB (radius of a circle)

Therefore in triangle BMN ,

/_MNB=90°

So, by Pythagoras theorem

(MB)² = (MN)² + (BN)²

(MB)² = (8)² + (1/2(AB))²

= (8)² + (1/2(16))²

= 64 + 64 = 128

MB = radius = √(128) = 8√2

Hence the radius of given circle is 8√2 cm