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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. Then  the distance between PQ and RS, if they lie.
(i) on the same side of the centre O and
(ii) on the opposite of the centre O  are respectively


A
8 cm & 14 cm
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B
4 cm & 14 cm
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C
2 cm & 14 cm
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D
2 cm & 28 cm
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Solution

The correct option is C $$2$$ cm & $$14$$ cm
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 $$\angle ONQ={ 90 }^{ o }=\angle OMQ$$......(ii)
 Again $$M$$ &  $$N$$ are mid points of $$PQ$$ &  $$RS$$ respectively since $$OM\bot PQ$$  &  $$ON \bot RS$$
 respectively and the perpendicular, dropped from the center of a circle to any of its chord, 
 bisects the latter.
So $$QM=\dfrac { 1 }{ 2 } $$PQ$$=\frac { 1 }{ 2 } \times 16$$ cm $$=8$$ cm and $$SN=\dfrac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 12$$ cm$$=6$$ cm.
$$  \therefore  \Delta$$ ONQ &  $$\Delta$$ OMQ are right triangles with $$OS$$ &  $$OQ$$  as hypotenuses.(from ii)
So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ 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.

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Maths

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