+0

# help plz

0
1
1

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
+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   Sep 11, 2023
edited by CPhill  Sep 11, 2023