Question

I in the figure 'o' is the centre of the circle o B = 5 cm AB is chord at distance of 3 cm from the centre find the length Of AB
​

Answer 1

Radius of given circle = OB = OA = 5 cm

OP = 3 cm

If a perpendicular is dropped from center of the circle on a chord, it bisects the chord.

→ AB = 2AP

In right angles triangle OAP, using pythagoras theorem:

AP = √OA² - OP²

AP = √5232 = 4 cm

AB = 2 x 4 cm = 8 cm

${ \green{ \boxed{ \pink{ \mathfrak{hope \: it \: helps \: you}}}}}$

${ \green{ \boxed{ \pink{ \mathfrak{please \: mark \: me \: as \: brainliest}}}}}$

Answer 2

Answer:

8cm

Explanation:

If the length of the radius OB is 5 cm and distance between center and chord OP is 3cm. AB divides into AP and BP.

So by Pythagoras Theorem,

BP²= BO²-PO²

BP²= 5²-3²

BP²= 25-9

BP²= 16

BP= 4

AP= PB= 4 cm (Beacuse, center divides the chord into equal halves)

AB= AP+BP

= 4 + 4

= 8 cm.

Therefore, length of the chord AB is 8cm. (Ans)