# 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?

Jul 11, 2018 Let ABCD be rhombus having diagonals $A C = 16 c m \mathmr{and} B D = 30 c m$.The diagonals bisect each other at perpendicularly at $O$.N is any point on side $A B$.$N P \mathmr{and} N Q$ are perpendiculars drawn from N to $A C \mathmr{and} B D$ respectively.So NQOP is a rectangle.Hence its diagonals $P Q \mathmr{and} O N$ are equal.We are to find out the minimum length of $P Q$ that is minimum length of $O N$. The length of $O N \mathmr{and} P Q$ will be minimum when $O N$ is perpendicular to $A B$.

Now $A B = \sqrt{O {A}^{2} + O {B}^{2}} = \sqrt{{8}^{2} + {15}^{2}} = 17 c m$

When ON is perpendicular to AB for minimum of PQ, the area of $\Delta A O B = \frac{1}{2} \times A B \times O {N}_{\text{min}}$

$\implies \Delta A O B = \frac{1}{2} \times 17 \times P {Q}_{\text{min}}$

Otherwise the area of

$\Delta A O B = \frac{1}{2} \times O A \times O B = \frac{1}{2} \cdot 8 \cdot 15 = 60 c {m}^{2}$

Equating these two areas we get

$P {Q}_{\text{min}} = \frac{120}{17} c m$