PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If $P Q = 16 c m$ and $R S = 12 c m$ . Then find the distance between PQ and RS, if they lie.
(i) on the same side of the centre O. (ii) on the opposite of the centre O.

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8 cm and 14 cm

4 cm and 14 cm.

2 cm and 14 cm

2 cm and 28 cm.

answer is C.

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Detailed Solution

First, we place the given details

and

, O is the center of the circle in the image of the circle. The radius of the circle is 10 cm. We draw a perpendicular line from the center O to both PQ and RS. They cut the chords at points M and N. So,

We know that the perpendicular from the center of the circle to the chord of the circle bisects the chord.
In the first figure, the chords PQ and RS are on the same side of the center.

In the second image the chords PQ and RS are on the opposite sides of the centre.

Now OS and OQ are the radius. So,

O S = O Q = 10

. OM and ON bisects the chords PQ and RS. So,

P M = M Q = 12 P Q = 162 = 8

and

R N = N S = 12 R S = 122 = 6

So, we got two right-angle triangles in the form of

Δ O N S, Δ O M Q

.
Using Pythagoras’ theorem, we get

b a s e 2 + h e i g h t 2 = h y p o t e n u s e 2

So, for those two triangles

Δ O N S

and

Δ O M Q

we got

O N 2 + N S 2 = O S 2

and

O M 2 + M Q 2 = O Q 2

respectively.
We put the values to get

(O N) 2 + (N S) 2 = (O S) 2 ⇒ (O N) 2 + (6) 2 = (10) 2 ⇒ (O N) 2 = 100 − 36 = 64 = 82 ⇒ O N = 8

And,

(O M) 2 + (M Q) 2 = (O Q) 2 ⇒ (O M) 2 + (8) 2 = (10) 2 ⇒ (O M) 2 = 100 − 64 = 36 = 62 ⇒ O M = 6

Now we will have two Anss for two different images.
The distance between PQ and RS is MN which is equal to

M N = O N ± O M

.
Putting values, we get

M N = O N ± O M = 8 ± 6 = 14, 2

The value of 14 is for the first image where PQ and RS are on the same side of the centre and the value of 2 is for the second image where PQ and RS are on the opposite sides of the centre.
The correct option is 3.

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