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# Trapezoid

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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

Jan 16, 2024