Two chords AB and CD of lengths 5cm and 11cm are respectively parallel. If the distance between AB and CD is 3cm, find the radius if the circle.
Here, it isn't specified that the chords are on the opposite sides of centre. How to solve this??

dear student,

Two chords AB and CD of lengths 5cm and 11cm respectively. If the distance  between AB and CD is 3cm , find the radius of the circle. – DronStudy  Questions

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I THINK THIS IS THE SPECIFIC DIAGRAM NEEDED
Let r be the radius of given circle and its centre be O. Draw OM 
perpendicular AB and ON perpendicular CD
Since, OM perpendicular AB, ON perpendicular CD
and AB||CD
Therefore, points M, O and N are collinear.So, MN = 6cm
Let, OM = x cm.Then,ON = (6 - x)cm.
Join OA and OC. Then OA = OC = r.
As the perpendicular from the centre to a chord of the circle 
bisects the chord.
∴   AM = BM = 1/2AB = 1/2 x 5 = 2.5cm.
   CN = DN = 1/2CD = 1/2 x 11 = 5.5cm.
In right triangles OAM and OCN, we have
 OA2 = OM+ AMand OC= ON2 + CN2
  r= x2 + (5/2)2  ...(i)
 r2 = (6-x)+ (11/2)2 ....(ii)
From (i) and (ii),we have
  x+ (5/2)= (6-x)2 + (11/2)2
  x2 + 25/4 = 36 + x2 - 12x + 121/4
⇒   4x2 + 25 = 144 + 4x- 48x + 121
⇒    48x = 240
⇒  x  = 240/48 ⇒ x = 5
Putting the value of x in euation (i), we get
     r2 = 52 + (5/2)2  ⇒ r2 = 25 + 25/4
 ⇒  r= 125/4  ⇒    r = 5√5/2 cm
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Case- I: When we take chords on the same side.


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Case- II: When we take chords on the opposite side.

Let r be the radius of given circle and its centre be O. Draw OM 
perpendicular AB and ON perpendicular CD
Since, OM perpendicular AB, ON perpendicular CD
and AB||CD
Therefore, points M, O and N are collinear.So, MN = 6cm
Let, OM = x cm.Then,ON = (6 - x)cm.
Join OA and OC. Then OA = OC = r.
As the perpendicular from the centre to a chord of the circle 
bisects the chord.
∴   AM = BM = 1/2AB = 1/2 x 5 = 2.5cm.
   CN = DN = 1/2CD = 1/2 x 11 = 5.5cm.
In right triangles OAM and OCN, we have
 OA2 = OM+ AMand OC= ON2 + CN2
  r= x2 + (5/2)2  ...(i)
 r2 = (6-x)+ (11/2)2 ....(ii)
From (i) and (ii),we have
  x+ (5/2)= (6-x)2 + (11/2)2
  x2 + 25/4 = 36 + x2 - 12x + 121/4
⇒   4x2 + 25 = 144 + 4x- 48x + 121
⇒    48x = 240
⇒  x  = 240/48 ⇒ x = 5
Putting the value of x in euation (i), we get
     r2 = 52 + (5/2)2  ⇒ r2 = 25 + 25/4
 ⇒  r= 125/4  ⇒    r = 5√5/2 cm
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I think like this.
In the question, they have not specified whether the chords are on the 'same side of the circle' or on the 'opposite sides and are parallel to each other'.
So I think both these two cases are possible for this question.
But the 'radius' will be different in both these cases.
Hence both these are possible and both are correct.
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