Area of the Trapezium | PRMO-2017 | Question 30

Contents

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

But try the problem first...

Answer:$13$

Source

Try with Hints

First hint

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 ..........

Second Hint

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........

Final Step

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 $13 .$