Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

## Area of the Trapezium – PRMO 2017, Problem 30

Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

- $9$
- $40$
- $13$
- $20$

**Key Concepts**

Geometry

Triangle

Trapezium

## Check the Answer

Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

## Try with Hints

Let $x, y, z, w$ be areas of the four triangles as shown in figure.

then area of \(\triangle ADB\)= Area of \(\triangle ACB\)

$\Rightarrow x+y=x+w \Rightarrow y=w$

Also \(\frac{AE}{EC}\)=\(\frac{area of \triangle ADE}{area of \triangle DEF}\)=\(\frac{area of \triangle AEB}{area of \triangle BEC}\)

$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$

$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$

and $y w=z^{2} r^{2}=1296$

As $( z \leq y \leq x)$

Therefore $ \frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Therefore area of trapezium $\mathrm{ABCD}$

$=x+y+z+w=z r^{2}+2 z r+z$

$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$

Therefore Answer is $13 .$

## Other useful links

- https://www.cheenta.com/ordered-pairs-prmo-2019-problem-18/
- https://www.youtube.com/watch?v=7lDy7gLdCtU