SOLUTION: Can anyone seem to help me with this problem? I have come to a complete blank on it. Question 64 Address book The perimeter of the cover of an address book is 14 inches, a

Question 151683: Can anyone seem to help me with this problem? I have come to a complete blank on it.

Question 64
Address book
The perimeter of the cover of an address book is 14 inches, and the diagonal measures 5 inches. What are the length and width of the cover?
x^2 + x^2 = 5^2
x^4 = 25
x^4 - 25 = 14
I do not know what to do next. I followed the Pythagorean steps, but cannot figure out what has gone wrong.
Thanks.
Amanda

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The perimeter of the cover of an address book is 14 inches, and the diagonal measures 5 inches. What are the length and width of the cover?
:
Since we will be using Pythagoras, let the length and width = a & b
;
Perimeter is given so we have:
2a + 2b = 14
Simplify, divide by 2
a + b = 7
or
a = (7-b)
:
Enter Pythagoras:
a^2 + b^2 = 5^2
;
Substitute (7-b) for a, find b
(7-b)^2 + b^2 = 25
:
FOIL (7-b)
49 - 14b + b^2 + b^2 = 25
:
Arrange as a quadratic equation
2b^2 - 14b + 49 - 25 = 0
:
2b^2 - 14b + 24 = 0
Factor:
(2b - 8)(b - 3) = 0
Solutions
2b = 8
b = 4
and
b = 3
:
We can use either solution if b=4, then a=3, or vice-versa
:
Was this understandable to you?
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