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In the given diagram, PQ and RS are comm...

In the given diagram, PQ and RS are common tangents to the two circles with centres C and D. Find the length of PQ and hence area of trapexium RSDC.

Text Solution

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Draw CM||RS so that `squareCMSR`becomes a rectangle, Now, we have CR=2 cm, DS=7 cm and CD=13 cm
`:." "DM=DS-MS`
`=DS-CR`
`=7-2=5 cm`
In right `DeltaDMC,` by Pythagoras theorem,
`CM^(2)=CD^(2)-DM^(2)`
`=(13)^(2)-(5)^(2)=(12)^(2)`
`:." "CM=12 cm`
`implies" "RS=12 cm" "`(`because`opposite sides of rectangle are equal)
`:." "PQ=12 cm" "`(`because`length of common tangents to two circles are always same)
`=(1)/(2)xxh`(sum of parallel sides)
`=(1)/(2)xxCM(CR+DS)`
`=(1)/(2)xx12(2+7)=54 cm^(2)`
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