Question

In a circle with a radius of 25cm, a chord lies at the distance at 7 cm from the centre find

.the length of that chord.​

Answer 1

Answer:

Refer the attachment

️‍️

In the figure, AB is the chord .

AM = BM ( the perpendicular from the centre bisects the chord)

OA = 25cm

OM = 7cm

Now,

By Pythagoras theorem,

${AM}^{2} + {MO}^{2} = {AO}^{2}$

$= > {AM}^{2} = {AO}^{2} - {MO}^{2}$

$= > {AM}^{2} = \sqrt{( {AO}^{2} - {MO}^{2} }$

$= > AM = \sqrt{( {25}^{2} - {7}^{2}) }$

$= > AM = \sqrt{(625 - 49)}$

$= > AM = \sqrt{576}$

$= > AM = 24cm$

Now ,

Chord AB = AM + BM

BM = AM

=> AB = AM + AM

=> AB = 24+24

=> AB = 48 cm