Question

find
the length of a chord
which is at a distance of 12cm from the center of a circle of radius 13cm.​

Answer 1

Answer:

OA2 = OM2 + AM2

⇒ 132 = 122 + AM2

⇒ AM2 = 169 - 144 = 25 ⇒ AM = 5cm.

As the perpendicular from the centre of a chord bisects the chord.Therefore,

AB = 2AM = 2 x 5 = 10cm

Answer 2

Answer:

Given:

A circle C(O,r) in which:

1. AB is a chord.
2. Distance of AB from the centre= 12cm
3. Radius = r= 13 cm

To find: The length of AB

Construction:

1. Join OA= radius=r= 13 cm
2. Draw OE ⊥ AB such that OE= 12 cm (given)

Proof:

In circle C(O,r),    

ΔOEA is a right triangle, right angled at E.            [By construction, OE ⊥ AB]

∴ By Pythagoras theorem in ΔOEA,

$OA^{2}= OE^{2} + AE ^{2}$

⇒ $13^{2} = 12^{2} + AE ^{2}$

⇒ 169= 144 + $AE ^{2}$

⇒ 169- 144 = $AE ^{2}$

⇒$AE ^{2}$= 25

⇒ AE= $\sqrt{25}$

⇒ AE= 5 cm

Now, AB= 2AE                   [Perpendicular drawn from the centre to a chord,

                                            bisects the chord]

∴ AB= 2(5) cm

⇒ AB= 10 cm

Hence, the length of chord AB= 10 cm

P.F.A the figure drawn below:

Hope you got that.

Thank You