Question

In a circle with radius 13cm, two equal chords are at a distance of 5cm from the centre. Find the length of the chords.

Answer 1

Answer:

Perpendicular drawn from the centre of the circle to the chord bisects the chord. Thus, AB = CD = 24 cm as chords of a circle equidistant from the centre are congruent.

Step-by-step explanation:

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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