The bases of trapezoid ABCD are AB and CD. Let P be the intersection of diagonals AC and BD. If the areas of triangles ABP and CDP are 8 and 18, respectively, then find the area of trapezoid ABCD.
@fake aops it is "Our message board sometimes has problems" not "Our message board sometimes have problems"
50
Actually, the person above was not the real AoPS. Our message board sometimes have problems and we encourage you to post your answers here
Triangles $PAB$ and $PCD$ are similar, so
\[\frac{[PAB]}{[PCD]} = \left( \frac{AB}{CD} \right)^2 = \frac{x^2}{y^2}.\]
But we are given that $[PAB] = 8$ and $[PCD] = 18$, so
\[\frac{x^2}{y^2} = \frac{8}{18} = \frac{4}{9},\]
which means $x/y = 2/3$.
Then $AP/PC = AB/CD = 2/3$. Triangles $ABP$ and $BCP$ have the same height with respect to base $\overline{AC}$, so
\[\frac{[BCP]}{[ABP]} = \frac{CP}{AP} = \frac{3}{2},\]
which means $[BCP] = 3/2 \cdot [ABP] = 3/2 \cdot 8 = 12$.
Also, $BP/PD = AB/CD = 2/3$. Triangles $ABP$ and $ADP$ have the same height with respect to base $\overline{BD}$, so
\[\frac{[ADP]}{[ABP]} = \frac{DP}{BP} = \frac{3}{2},\]
which means $[ADP] = 3/2 \cdot [ABP] = 3/2 \cdot 8 = 12$.
Therefore, the area of trapezoid $ABCD$ is $[ABP] + [BCP] + [CDP] + [DAP] = 8 + 12 + 18 + 12 = \boxed{50}$.
There was an honor code that everyone had to agree with, so posting AoPS's answers on here is "double-violating" the code
yeah dont do that
Do not cheat you are basically very wasteful
@above dont lie you know you are not supposed post AoPS answers here
lmao u guys should quit with the fake aops...
you guys are wasting your money and your life for looking for answers online.
not that I care but you should... whatevs.