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# pls help

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In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of APAB is 16 square units, and the area of ▲PCD is 25 square units. What is the area of trapezoid ABCD?

Aug 30, 2022

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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 = 50

Aug 30, 2022