Question

Hoy.....


At one end A of a diameter AB of a circle of radius 5cm, tangent XAY is drawn to a circle.The length of the chord CD parallel to XY and at a distance 8cm from A is ____ cm.​

Answer 1

Answer:

CD = 8 cm

Step-by-step explanation:

A chord CD is drawn which is parallel to XY at a distance of 8 cm from A.

∠BAX + ∠AEC = 180° [Sum of co-interior angles is 180°]

Now, ∠OAX =∠BAX= 90°

So we have

=> 90 + ∠AEC = 180

∠AEC = 90°

Therefore, ∠OEC = 90°

So, OE ⏊ CD and △OCB is a right-angled triangle.

Now in △OEB

(OE)² + (EC)² = (OC)²

And we have

OC = 5 cm [radius]

OE = AE - AO = 8 - 5 = 3 cm

⇒ (3)² + (EC)² = (5)²

⇒ 9 + (EC)² = 25

⇒ (EC)² = 25 - 9 = 16

⇒ EC = 4 cm

Also, CE = ED

∴ Since, perpendicular from center to the chord bisects the chord.

So,

CD = CE + ED

    = 2CE

   = 2(4)

  = 8 cm

Therefore,

Length of chord CD = 8 cm

Hope it helps!

Answer 2

Answer:

HELLO DEAR,

CD=2×MC

given that:-

radius of circle = 5cm=AO=OC

AM=8CM

and

AM=OM+AO

THEN

OM =AM-AO

Put the vlues in this equation given above

We Get

=> OM= (8-5) =3CM

we know that;-

OM is perpendicular to the chord CD.

In∆OCM <OMC=90°

using Pythagoras theorem

we get,

OC²=OM²+MC²

.

Hence,

CD= 2 ×CM  = 8 cm

I HOPE ITS HELP YOU DEAR,

THANKS