# Math

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.

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1. 50

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2. 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 $[ABP] + [BCP] + [CDP] + [DAP] = 8 + 12 + 18 + 12 = \boxed{50}$.

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3. Thanks for copying and pasting LaTeX and not bothering to put it so its legible.

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4. Hello,

Our challenge problems are designed for you to solve them to the best of your ability. If you need any guidance, ask on our message boards.

Sincerely,
AoPS

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5. Actually, the person above was not the real AoPS. Our message board sometimes have problems and we encourage you to post your answers here

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6. @above dont lie you know you are not supposed post AoPS answers here

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7. @fake aops it is "Our message board sometimes has problems" not "Our message board sometimes have problems"

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8. u guys should quit with the fake aops...
not that I care but you should... whatevs.

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9. Well said

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10. There was an honor code that everyone had to agree with, so posting AoPS's answers on here is "double-violating" the code

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11. yeah dont do that

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