Question

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?

Answer 1

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`