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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 $4,$ and the area of triangle $PSR$ is $8.$  Find the area of trapezoid $PQRS$.

 Jun 8, 2024

Best Answer 

 #1
avatar+759 
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First, let's note that PXQ and RXS are similar triangles. 

 

The scale factor of the two is \( \sqrt{8/4} = \sqrt{2}\)

 

This means that the height of RXS is \( \frac{h\sqrt{2}}{ 1 + \sqrt{2}}\) where h is the height of the trapezoid. 

We have \(\frac{ h}{1 + \sqrt{2} }\) for the height of PXQ

The base of PXQ is \(\text{base of RXS}/\sqrt{2}\)

 

Now, using this information from PXQ, we can write

\(\text{Area} = \frac{(1/2) (\text{base of RXS})}{\sqrt{2} * h ( 1 + \sqrt{2})} \\ 4 = (1/2) (\text{base RXS}) * h / ( 2 +\sqrt{ 2}) \\ 8(2 + \sqrt{2}) / \text{base RXS} = h\)

 

We know the area of the trapezoid is \((1/2) h (\text{sum of bases})\)

 

We have 

\( 4 [ 2 +\sqrt {2}] [ 2 + \sqrt {2} ] / / 2 \\ 2 [ 2 + \sqrt {2}] ^2 \\ 2 [ 4 + 4\sqrt {2}+ 2] = \\ 2 [ 6 + 4\sqrt {2}] = \\ 12 + 8\sqrt {2} \)

 

Thanks! :)

 Jun 8, 2024
 #1
avatar+759 
+1
Best Answer

First, let's note that PXQ and RXS are similar triangles. 

 

The scale factor of the two is \( \sqrt{8/4} = \sqrt{2}\)

 

This means that the height of RXS is \( \frac{h\sqrt{2}}{ 1 + \sqrt{2}}\) where h is the height of the trapezoid. 

We have \(\frac{ h}{1 + \sqrt{2} }\) for the height of PXQ

The base of PXQ is \(\text{base of RXS}/\sqrt{2}\)

 

Now, using this information from PXQ, we can write

\(\text{Area} = \frac{(1/2) (\text{base of RXS})}{\sqrt{2} * h ( 1 + \sqrt{2})} \\ 4 = (1/2) (\text{base RXS}) * h / ( 2 +\sqrt{ 2}) \\ 8(2 + \sqrt{2}) / \text{base RXS} = h\)

 

We know the area of the trapezoid is \((1/2) h (\text{sum of bases})\)

 

We have 

\( 4 [ 2 +\sqrt {2}] [ 2 + \sqrt {2} ] / / 2 \\ 2 [ 2 + \sqrt {2}] ^2 \\ 2 [ 4 + 4\sqrt {2}+ 2] = \\ 2 [ 6 + 4\sqrt {2}] = \\ 12 + 8\sqrt {2} \)

 

Thanks! :)

NotThatSmart Jun 8, 2024

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