chord pq||rs , 0 is a central and radio 10cm. if pq=16 cm and rs=12 find the distance between pq and rs when they lie.
Answers
Answer:
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.
Ans- Option C.
I hope this helps you dear