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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 $PQRS$.

 Jan 16, 2024
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        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

 

cool cool cool

 Jan 16, 2024

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