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Q.


Two chords named AB and CD having lengths of 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, then what is the radius of the circle?

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a

5 5 2  

b

5 2  

c

4 2  

d

4 5   

answer is A.

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Detailed Solution

Given data, the length of the AB chord is = 5cm.
The length of the CD chord is 11 cm.
The distance between them is 6 cm.
We draw the perpendicular lines from the centre O to the AB and CD as shown.
Question ImageLet the radius of the circle is r and the length of OP be x.
Also, PQ = 6.
PO+OQ=6 x+OQ=6 OQ=6x  
Now perpendicular bisects a chord we get AP=PB.
We know that, AB = 5cm.
AP+PB=5 PB+PB=5 2PB=5 PB=2.5cm  
Now for QD, because perpendicular is bisecting a chord such that:
CQ=QD CD=11cm CQ+QD=11 QD+QD=11 2QD=11 QD=5.5cm  
Now we will apply the Pythagoras theorem on OPB.
Hypotenuse= OB=r.
Perpendicular=OP=x.
Base =PB = 2.5 cm.
OB2 = OP2 + PB2.
Now we will substitute the values to get:
r 2 =  x 2 + 2 .5 2 r 2 =  x 2 + 6.25  . [1]
For OQD,
Hypotenuse= OD=r.
Perpendicular=OQ= x6..
Base =QD= 5.5 cm..
OD2 = OQ2 + QD2.
Now, we will substitute the values to get:
r 2 = (x6 ) 2 + 5 .5 2 r 2 =  x 2 12x+36+ 30.25  . [2]
On comparing the equation [1] and [2] we get:
x 2 12x+36+ 30.25=  x 2 +6.25 x 2 12x+36+ 30.25 x 2 6.25=0 12x+60=0 12x=60 x=5  
Now we will substitute the value of x in (1) we get:
r 2 = x 2 + 6.25 r 2 = 5 2 + 6.25 r 2 =25+ 6.25 r 2 =31.25 r 2 =  3125 100 r 2 =± 125 4 r= 5 5 2 cm  
Because radius can’t be negative so we have the radius of the circle is 5 5 2   cm.
Therefore, the correct option is 1.
 
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Two chords named AB and CD having lengths of 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, then what is the radius of the circle?