Answer :

Let's make a diagram for the given problem in which we have a circle having AB as diameter and O as center and having radius 5cm, and a tangent XAY at point A on the circle .

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

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

Now, ∠OAX =∠BAX= 90°

[Tangent and any point of a circle is perpendicular to the radius through the point of contact]

So we have

90 + ∠AEC = 180

∠AEC = 90°

Therefore, ∠OEC = 90°

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

Now in △OEB

By Pythagoras theorem [ i.e. (base)^{2} + (perpendicular)^{2} = (hypotenuse)^{2} ]

(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

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