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
B
4 cm & 14 cm
C
2 cm & 14 cm
D
2 cm & 28 cm

Solution

## The correct option is C $$2$$ cm & $$14$$ cmWe 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.Maths

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