Q. 54.3( 12 Votes )

4 cm

B. 5 cm

C. 6 cm

D. 8 cm

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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