Question

PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ =
16 cm and RS = 12 cm, find the distance between PQ and RS, if they lie
(i) on the same side of the centre O,

(ii) on the opposite sides of the centre O.
​

Answer 1

Answer:

2&14

Step-by-step explanation:

We join OQ & OS, drop perpendicular from O to PQ & RS.

The perpendiculars meet PQ & RS at M & N respectively.

Since OM & ON are perpendiculars to PQ & RS who are

parallel lines, M, N & O will be on the same straight line

and disance between PQ & RS is MN.........(i) and ∠ONQ=90

o

=∠OMQ......(ii)

Again M & N are mid points of PQ & RS respectively since OM⊥PQ & ON⊥RS

respectively and the perpendicular, dropped from the center of a circle to any of its chord,

bisects the latter.

So QM=

2

1

PQ=

2

1

×16 cm =8 cm and SN=

2

1

RS=

2

1

×12 cm=6 cm.

∴Δ ONQ & Δ OMQ are right triangles with OS & OQ as hypotenuses.(from ii)

So, by Pythagoras theorem, we get ON=

OS

2

−SN

2

=

10

2

−6

2

cm =8 cm and OM=

OQ

2

−QM

2

=

10

2

−8

2

cm =6 cm.

Now two cases arise- (i) PQ & RS are to the opposite side of the centre O.(fig I)

Here MN=OM+ON=(6+8)cm=14$$ cm (from i) or

(ii) PQ & RS are to the same side of the centre O. (fig II)

Here MN=ON−OM=(8−6) cm=2 cm.

So the distance between PQ & RS =14 cm when PQ & RS are to the opposite side of the centre O

and the distance between PQ & RS =2 cm when PQ & RS are to the same side of the centre O.