+1541 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 $PQRS$.
+128475 P Q
4
X
8
S R
PXQ and RXS are similar triangles
The scale factor from RXS and PXQ = sqrt (8/4) = sqrt (2) / 1
Thus the height of RXS = sqrt (2) h / ( 1 + sqrt 2) where h is the height of the trapezoid
And the height of PXQ = h /( (1 + sqrt 2)
And the base of PXQ = base of RXS/sqrt 2
Using PXQ
Area = (1/2) (base of RXS) /sqrt (2) * h / ( 1 + sqrt 2)
4 = (1/2) (base RXS) * h / ( 2 +sqrt 2)
8(2 + sqrt 2) / base RXS = h
Area of trapezoid = (1/2) h (sum of bases)
Area of trapezoid = (1/2)[ 8 ( 2 + sqrt 2) ] / base of RXS ] [ base of RXS + base PXQ ] =
4 (2 + sqrt 2) / [ base of RXS ] * [ base of RXS * ( 1 + 1/ sqrt 2) ] =
4 [ 2 + sqrt 2 ] [ 2 + sqrt 2 ] / / 2 =
2 [ 2 + sqrt 2] ^2 =
2 [ 4 + 4sqrt 2 + 2] =
2 [ 6 + 4sqrt 2] =
12 + 8sqrt 2
