Question

The radius of a circle
of a circle with centre O is 13 cm. The distance of a chord from the
centre is 5 cm. Find the length of the chord,
​

Answer 1

Answer:

24cm

Step-by-step explanation:

Let AB be a chord anywhere inside the circle such that it connects A and B.

Join A and B to O.

Now, we know that AB is 5 cm away from the center so

we know that the length between the chord and Center would be 5cm.

Now Join O to point D on Chord AB

- We have two Right angle triangles now.

AD = DB

In Triangle OAD

OA^2 = DA^2 + DO^2

169= 25 + DA^2

$\sqrt{144} = da$

12=da

2×da =AB

2×12=AB

24cm = AB

Answer 2

Step-by-step explanation:

Let AB be the chord of a circle of radius 13 cm at a distance of 5cm from centre O.

Then, OA=13xm, OM=5cm

Using Pythagoras theorem,

OA²=OM²+AM²

i.e.,.

${13}^{2} = {5}^{2} + {AM}^{2}$

or

${AM}^{2} = {13}^{2} - {5}^{2}$

$= 169 - 25 = 144$

$AM = 12$

$AB = 2 \times 12 = 24cm$

perpendicular perpendicular from centre bisector of chord

length of the chord = 24 cm