Question

radius of a circle is 25 cm and the distance of its chord from the centre is 4 cm what is the length of the chord?​

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Going off ʕ•ﻌ•ʔ​

Answer 1

Answer:

As perpendicular from centre bisects the chord,

⇒CA=AD=20cm

In ΔOAD, OD2=OA2+AD2(Pythagoras thereom)

⇒OA2=OD2−AD2

=252−202

=625−400=225

⇒OA=15cm

OA is distance of chord from O.

Step-by-step explanation:

bye bye

Answer 2

Step-by-step explanation:

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Here's ur solution -

Given,

r = 25cm

chord AB = 40cm

We know that,

perpendicular from centre bisects the chord.

So the Length of chord AM = 20cm

Since, ∆0AM is a right angled triangle.

We apply -

OA² = AM² + OM² ( Pythagoras theorem)

= 25² = 20² + 0M²

OM = √(25² - 20²)

OM = √(625 - 400)

OM = √225

OM = 15

Hence, the distance of the chord from centre is 15cm.

Hope it helps!