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# Another trapezoid question.

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In trapezoid PQRS, $$\overline{PQ} \parallel \overline{RS}$$. Let X be the intersection of diagonals $$\overline{PR}$$ and $$\overline{QS}$$. The area of triangle PQX is 20, and the area of triangle RSX is 45. Find the area of trapezoid PQRS.

Apr 29, 2020

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In trapezoid $$PQRS$$, $$\overline{PQ} \parallel \overline{RS}$$.
Let $$X$$ be the intersection of diagonals $$\overline{PR}$$ and $$\overline{QS}$$.
The area of triangle $$PQX$$ is $$20$$, and the area of triangle $$RSX$$ is $$45$$.
Find the area of trapezoid $$PQRS$$.

$$\begin{array}{|rcll|} \hline \text{Let Area }A_1 &=& [PQX] \\ A_1 &=& 20 \\\\ \text{Let Area }A_2 &=& [RSX] \\ A_2 &=& 45 \\\\ \text{Let Area }A_3 &=& [PXS] \\\\ \text{Let Area }A_4 &=& [QXR] \\\\ \text{Let Area }A_5 &=& [PSR] \\ &=& A_2+A_3 \\\\ \text{Let Area }A_6 &=& [QSR] \\ A_6 &=& A_2 +A_4 \\ \hline \mathbf{A_5} &=& \mathbf{A_6} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{A_5} &=& \mathbf{A_6} \\ A_2+A_3 &=& A_2 +A_4 \\ \mathbf{A_3 }&=& \mathbf{A_4} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline A_3 = \dfrac{SX*h_1}{2} && A_1 = \dfrac{QX*h_1}{2} \\ \mathbf{\dfrac{A_3}{A_1} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ A_2 = \dfrac{SX*h_2}{2} && A_4 = \dfrac{QX*h_2}{2} \\ \mathbf{\dfrac{A_2}{A_4} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ \dfrac{SX}{QX}=\mathbf{\dfrac{A_3}{A_1} }&=&\mathbf{\dfrac{A_2}{A_4} } \quad | \quad A_4 = A_3 \\\\ \dfrac{A_3}{A_1} &=& \dfrac{A_2}{A_3} \\\\ A_3^2 &=& A_1A_2 \\ A_3^2 &=& 20*45 \\ A_3^2 &=& 900 \\ \mathbf{A_3} &=& \mathbf{30} \\\\ A_4 &=& A_3 \\ \mathbf{A_4} &=& \mathbf{30} \\ \hline \end{array}$$

$$\text{The area of trapezoid PQRS = A_1+A_2+A_3+A_4 = 20+45+30+30=\mathbf{125} }$$

Apr 29, 2020