# Find The Hypotenuse.

joe shmo
Posers and Puzzles 29 Apr '16 22:43
1. joe shmo
Strange Egg
29 Apr '16 22:43
1) Triangle "ABC" is a right triangle with hypotenuse "AC" of 60.
2) Point "E" lies on line "BC" such that right triangle "ABE" has a hypotenuse "AE" of 52.
3) Point "D" lies on line "AB" such that right triangle "DBC" has a hypotenuse "DC" of 39.

Find Length "DE"?
2. wolfgang59
01 May '16 09:19
Originally posted by joe shmo
1) Triangle "ABC" is a right triangle with hypotenuse "AC" of 60.
2) Point "E" lies on line "BC" such that right triangle "ABE" has a hypotenuse "AE" of 52.
3) Point "D" lies on line "AB" such that right triangle "DBC" has a hypotenuse "DC" of 39.

Find Length "DE"?
mmmm
Deceptively tricky.
I have 5 unknowns and only 4 equations.
Must be a little geometric trick in there that I have forgotten.
ðŸ™‚
3. wolfgang59
01 May '16 09:271 edit
But that's ok.
("DE" )^2 = (52*52)-(21*99)
4. joe shmo
Strange Egg
01 May '16 15:092 edits
Originally posted by wolfgang59
But that's ok.
("DE" )^2 = (52*52)-(21*99)
Yep, you got it! It is certainly more tricky than it first appears.

Let:
AB = A
BC = B
EC = b

A^2+B^2 = 60^2 (1)

A^2 + (B-b)^2 = 52^2 (2)

(A-a)^2 + B^2 = 39^2 (3)

Let:

(A-a) = F
(B-b) = G

Substitute F and G into (2)&(3)

System:

A^2 + B^2 = 60^2 (1)

A^2 + G^2 = 52^2 ( 2' )

F^2 + B^2 = 39^2 ( 3' )

( 2' )+( 3' ) :

A^2 + G^2 + F^2 + B^2 = 52^2 + 39^2 (4)

(4) - (1):

A^2 + G^2 + F^2 + B^2 -A^2 -B^2 = 52^2 + 39^2-60^2

G^2+F^2 = 52^2 + 39^2-60^2 = 625

DE = Sqrt(G^2+F^2) = 25