SOLUTION: hi, can you help me with this problem? 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.

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Question 119836: hi, can you help me with this problem?
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.)
Answer by ankor@dixie-net.com(22740) (Show Source):
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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