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

2 thoughts on “find<br />the length of a chord<br />which is at a distance of 12cm from the center of a circle of radius 13cm.​”

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

   Reply

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,

   [tex]OA^{2}= OE^{2} + AE ^{2}[/tex]

   ⇒ [tex]13^{2} = 12^{2} + AE ^{2}[/tex]

   ⇒ 169= 144 + [tex]AE ^{2}[/tex]

   ⇒ 169- 144 = [tex]AE ^{2}[/tex]

   ⇒[tex]AE ^{2}[/tex]= 25

   ⇒ AE= [tex]\sqrt{25}[/tex]

   ⇒ 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

   Reply