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Help, I'm having a hard time with this geometry problem.

 

In the diagram, right-angled triangles $AED$ and $BFC$ are constructed inside rectangle $ABCD$ so that $F$ lies on $DE$. If $AE = 7$, $ED = 24$, and $BF = 15$, what is the length of $AB$?

 

 Sep 10, 2023
 #1
avatar+128732 
+1

 

 

 

AD = sqrt ( AE^2 + DE^2)  =sqrt (7^2 + 24^2)  = sqrt (625)  = 25 = BC

 

FC =  sqrt ( BC^2 - FB^2)  =sqrt (25^2  - 15^2)  =sqrt ( 625 -225)    = sqrt (400)  = 20

 

In triangle  BFC  draw altitude FG

Let GC = x    Let BG = 25 -x

 

FG^2 = FB^2 - BG^2 =  FB^2 - (25-x)^2  = 15^2 - (25 -x)^2

FG^2 =  FC^2  - CG^2 = FC^2 - x^2  =  20^2 - x ^2

 

So

FG^2 = FG^2

15^2 - (25-x)^2  =  20^2 - x^2

225 - 625 + 50x - x^2  = 400 - x^2

50x = 800

x = 16 = CG

So  FG =  sqrt (FC^2 - CG^2) = sqrt ( 20^2 - 16^2)  = sqrt (400 - 256) = sqrt (144)  = 12

 

Extend FG  through F  and  let it  intersect   AD  at  I

Triangle IDF is a right triangle with angle DIF = 90

Note that DI = CG

 

tan angle ADE  = AE /DE = 7/24

tan angle ADE = tan angle IDF  = FI /DI  = FI /16

 

So

FI /16  = 7/24

FI = (7/24) (16)  = (2/3) (7) =  14/3

 

AB = FG + FI  =  12 + 14/3 =   50 / 3

 

cool cool cool

 Sep 11, 2023
edited by CPhill  Sep 11, 2023

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