+1525 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 $PQRS$.
+1894 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! :)
+1894 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! :)
