- Joined
- 10 Dec 06
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29 Apr 16
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"?
01 May 16
Originally posted by joe shmommmm
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"?
Deceptively tricky.
I have 5 unknowns and only 4 equations.
Must be a little geometric trick in there that I have forgotten.
🙂
- Joined
- 10 Dec 06
- Moves
- 8528
01 May 16
2 edits
Originally posted by wolfgang59Yep, you got it! It is certainly more tricky than it first appears.
But that's ok.
("DE" )^2 = (52*52)-(21*99)
Let:
AB = A
BC = B
AD = a
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