+1439 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 $PQRS$.
+129895 P Q
4
X
8
S R
Triangle PXQ similar to triangle RXS
Ratio of sides of RXS to PXQ = sqrt [ 8/ 4 ] = sqrt (2)
Note that sin PXS = sin PXQ = sin QXR
Area of triangle PXS = (1/2) PX * SX sin PXS = (1/2) PX * sqrt (2) QX *sin PXS =
sqrt 2 [ (1/2) PX * QX * sin PXS ] = sqrt (2) [ (1/2) PX * QX * sin PXQ[ = sqrt (2) [ area PXQ]
Similarly
Area of triangle QXR = (1/2) QX * XR * sin QXR = (1/2) QX * sqrt (2) PX * sin QXR =
sqrt (2) [ (1/2) QX * PX * sin QXR ] = sqrt (2) [ (1/2) QX * PX * sin (PXQ) ] = sqrt (2) [ area PXQ]
So [ PQRS ] = [ PQX] + [ QXR ] + 2 sqrt (2) [ PXQ] = 4 + 8 + 2sqrt (2) [ 4 ] =
12 + 8 sqrt (2) = 4 [ 3 + 2sqrt (2) ] = 4 [ 3 + sqrt (8) ]
