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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