Question

In circle with radius 13 cm, two equal chord are at a distance of 5 cm from the centre find the length of the chord

Answer 1

Answer:

Step-by-step explanation:

hyp^2 = alt^2 + base^2

13^2 = 5^2 + base^2

Base^2 = 169 - 25

            = 144

Base = 12

length of chord = 2 x Base = 24cm

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