SOLUTION: Could you please help me solve this word problem?
Crafts. A solar collector is 2.5 m long by 2.0 m wide. It is held in place by a frame of uniform width around its outside edge.
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Crafts. A solar collector is 2.5 m long by 2.0 m wide. It is held in place by a frame of uniform width around its outside edge.
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Question 98127: Could you please help me solve this word problem?
Crafts. A solar collector is 2.5 m long by 2.0 m wide. It is held in place by a frame of uniform width around its outside edge. If the exposed collector is 4m^2, what is the width of the frame to the nearest tenth of a centimeter?
Thank you Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A solar collector is 2.5 m long by 2.0 m wide. It is held in place by a frame of uniform width around its outside edge. If the exposed collector is 4m^2, what is the width of the frame to the nearest tenth of a centimeter?
:
Let x = width of the outside frame (in meters)
:
Dimensions of the exposed collector, (2.5 - 2x) by (2 - 2x) = 4
Area:
(2.5 - 2x) * (2 - 2x) = 4
FOIL
5 - 5x - 4x + 4x^2 = 4
:
4x^2 - 9x + 5 - 4 = 0
:
4x^2 - 9x + 1 = 0
:
Use the quadratic formula for this: a = 4, b = -9, c = 1
:
:
:
:
Solution 1:
:
x =
x = 2.13278 meters * 100 = 213.3 cm, not a possible solution
:
Solution 2:
:
x =
x = .11721 meters * 100 = 11.7 cm, a reasonable answer
:
:
Check solution by finding the area of exposed collector
2x = 2(.117) = .234 m
(2.5 - .234) * (2 - .234) =
2.266 * 1.766 = 4.00 sq m confirms our solution