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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 $PQRS$.

 Mar 24, 2024
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         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) ] 

 

 

cool cool cool

 Mar 25, 2024

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