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?

tmatthews May 27, 2021

#1**+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

mworkhard222 May 27, 2021

#3**+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

Melody May 28, 2021

#4**+1 **

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

- Isolate \(l\) in \(2(l+w)=23 \) and get \(l=\frac{23}{2}-w\)
- Plug the solutions \(l=\frac{23}{2}-w\) into \(\left(l+1+1\right)\left(w+1+1\right)=58.5\)
- 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

mworkhard222 Jun 1, 2021