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 The diagram below represents a photograph in a frame. The photo has a perimeter of 23 inches, and the area of the whole figure is 58.5 square inches. If the frame is 1 inch thick, what is the area of the photo in square inches?

 

 May 27, 2021
 #1
avatar+493 
+3

ok, lets say the length and width of the photo is l and w.

then, \(2\left(l+w\right)=23\)

and if the frame is 1 inch, \(\left(l+1+1\right)\left(w+1+1\right)=58.5\)

 

then we just solve the system of equations and get l = 7 and w = 4.5 so the area is 31.5

 May 27, 2021
edited by mworkhard222  May 27, 2021
 #2
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+1

Thanks for your help but how did you solve the equations? I can't seem to get a whole number?

Guest May 27, 2021
 #3
avatar+114069 
+1

Do you understand what mworkhard has presented so far?  I will assume so ....

 

2(L+W) = 23 (1)

 

(L+2)(W+2)=58.5  (2)

LW+2(W+L)+4=58.5

sub in (1)

LW+23+4=58.5

LW+27=58.5

LW=31.5

L=31.5/W

 

sub into (1) to find W

2(L+W) = 23 (1)

\(2(\frac{31.5}{W}+W) = 23 \\ \text{mult both sides by W}\\ 2W(\frac{31.5}{W}+W) = 23W \\ 63+2W^2=23W\\ 2w^2-23w+63=0\\ w=\frac{23\pm\sqrt{529-504}}{4}\\ w=\frac{23\pm\sqrt{25}}{4}\\ w=\frac{23\pm5}{4}\\ w=4.5\;or\; 7\)

 

Length = 7  width =4.5 inches

 May 28, 2021
 #4
avatar+493 
+1

omg how did i not see this! Here is how u can solve the system:

  1. Isolate \(l\) in \(2(l+w)=23 \) and get \(l=\frac{23}{2}-w\)
  2. Plug the solutions \(l=\frac{23}{2}-w\) into \(\left(l+1+1\right)\left(w+1+1\right)=58.5\)
  3. For \(\left(l+1+1\right)\left(w+1+1\right)=58.5 \) subsitute \(l\) with \(\frac{23}{2}-w: \:\:w=\frac{9}{2},l=7\)

 

Where I got my steps: https://www.symbolab.com/solver/step-bystep/2%5Cleft(l%2Bw%5Cright)%3D23%2C%20%5Cleft(l%2B1%2B1%5Cright)%5Cleft(w%2B1%2B1%5Cright)%3D58.5

 Jun 1, 2021

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