Question

in the figure AB and CD are two chords of circle with Centre O at a distance of 6 cm and 8 cm from O if the radius of the circle is 10 cm find the length of the chord

Answer 1

the answer of ur question

Answer 2

Answer:

AB = 16 cm.

CD = 12 cm.

Step-by-step explanation:

We have,

AB and CD as two chords of the circle.

O is the centre of the circle.

Let, us say the perpendicular dropped from O to AB, touches AB at L .

and,

The perpendicular dropped from O to CD, touches CD at M.

So,

As we know that the perpendicular dropped from the centre of the circle to the chord always bisects it.

So,

In triangle OLA, using the Pythagoras theorem, we get,

$OA^{2}=OL^{2}+LA^{2}\\So,\\10^{2}=6^{2}+LA^{2}\\LA=8\ cm$

So, the length of chord AB is given by,

AB = 2LA

AB = 16 cm.

Similarly,

In triangle OMC, using the Pythagoras theorem, we get,

$OC^{2}=OM^{2}+MC^{2}\\So,\\10^{2}=8^{2}+MC^{2}\\MC=6\ cm$

So, the length of chord CD is given by,

CD = 2MC

CD = 12 cm.

Therefore, the length of the chords AB and CD are 16 cm and 12 cm respectively.