Question 119836
A rectangular parcel of land is 40 ft wide. The length of a diagonal between opposite corners is 18 ft more than the length of the parcel. What is the length of the parcel? (Round the answer to one decimal place.)
:
Here's what we know:
W = 40
L is unknown
Diagonal = L + 18
:
Using Pythagorus: a^2 + b^2 = c^2
Let a = 40
Let b = L
Let c = (L+18)
:
40^2 + L^2 = (L+18)^2
:
FOIL (L+18)(L+18)
1600 + L^2 = L^2 + 36L + 324
:
Arrange equation; L's on the left, numerical values on the right
L^2 - L^2 - 36L = 324 - 1600
-36L = -1276
:
L = -1276/-36
:
L = +35.4 ft is the length.
:
Check the solution by finding the hypotenuse:
h = Sqrt(40^2 + 35.4^2) 
h = Sqrt(1600 + 1253.2)
h = 53.4 ft which is 18 ft greater than the length