The question is below?

Let ABCD be a rhombus with AC and BD as diagonals of length 16 and 30 respectively. Let N be a point on AB, and let P and Q be feet of perpendiculars from N to AC and BD, respectively. Then what is the minimum value of PQ?

1 Answer
Jul 11, 2018

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Let ABCD be rhombus having diagonals AC=16cmandBD=30cm.The diagonals bisect each other at perpendicularly at O.N is any point on side AB.NPandNQ are perpendiculars drawn from N to ACandBD respectively.So NQOP is a rectangle.Hence its diagonals PQandON are equal.We are to find out the minimum length of PQ that is minimum length of ON. The length of ONorPQ will be minimum when ON is perpendicular to AB.

Now AB=sqrt(OA^2+OB^2)=sqrt(8^2+15^2)=17cm

When ON is perpendicular to AB for minimum of PQ, the area of DeltaAOB=1/2xxABxxON_"min"

=>DeltaAOB=1/2xx17xxPQ_"min"

Otherwise the area of

DeltaAOB=1/2xxOAxxOB=1/2*8*15=60cm^2

Equating these two areas we get

PQ_"min"=120/17cm