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# Geometry

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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

#1
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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
+759
+1

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