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Three vertices of parallelogram $ABCD$ lie on rectangle $EFGH$. $BC=AF=10$, $CG=8$, and $AB=28$. Find $FG$.

 

 Mar 25, 2021
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BG  = sqrt  [ BC^2  -CG^2 ]  = sqrt [ 10^2 - 8^2 ]  =  sqrt  [100 - 64 ]  = sqrt [ 36] = 6

 

FB  =sqrt   [ AB^2  - AF^2 ] =sqrt [28^2  - 10^2 ] =  sqrt [ 784  -100 ]  =sqrt [684]  = 6sqrt (19)

 

FG =  FB + BG   =   6  + 6sqrt (19)  =  6 ( 1 + sqrt (19) ) ≈  32.15

 

 

cool cool cool

 Mar 25, 2021

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