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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

552

b

52

c

42

d

45

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 Image

Question Image

Let the radius of the circle is r and the length of OP be x.
Also, PQ = 6.

⇒ P O + O Q = 6 ⇒ x + O Q = 6 ⇒ O Q = 6 − x

Now perpendicular bisects a chord we get AP=PB.
We know that, AB = 5cm.

⇒ A P + P B = 5 ⇒ P B + P B = 5 ⇒ 2 P B = 5 ⇒ P B = 2.5 c m

Now for QD, because perpendicular is bisecting a chord such that:

C Q = Q D C D = 11 c m ⇒ C Q + Q D = 11 ⇒ Q D + Q D = 11 ⇒ 2 Q D = 11 ⇒ Q D = 5.5 c m

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= x−6..
Base =QD= 5.5 cm..
OD2 = OQ2 + QD2.
Now, we will substitute the values to get:

⇒ r 2 = (x − 6) 2 + 5 .5 2 ⇒ r 2 = x 2 − 12 x + 36 + 30 .25

. [2]
On comparing the equation [1] and [2] we get:

⇒ x 2 − 12 x + 36 + 30 .25= x 2 + 6.25 ⇒ x 2 − 12 x + 36 + 30 .25 − x 2 − 6.25 = 0 ⇒ − 12 x + 60 = 0 ⇒ − 12 x = − 60 ⇒ x = 5

Now we will substitute the value of x in (1) we get:

⇒ r 2 = x 2 + 6 .25 ⇒ r 2 = 52 + 6 .25 ⇒ r 2 = 25 + 6 .25 ⇒ r 2 = 31.25 ⇒ r 2 = 3125100 ⇒ r 2 = ± 1254 ⇒ r = 552 c m

Because radius can’t be negative so we have the radius of the circle is

552

cm.
Therefore, the correct option is 1.

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