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Photo Frame Perimeter Question

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Joined: 16 Jul 2018
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16 Jul 2018, 19:29
I am struggling to break through this question and have revisited over multiple days. Any help would be appreciated.

A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?
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Re: Photo Frame Perimeter Question  [#permalink]

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16 Jul 2018, 19:57
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Let the dimensions of the photograph be L*B (L is length and B is breadth)

The photograph has 1 inch wide border on each side so in the dimension of the photograph and the border (1 inch on each side) will become (L+1+1) * (B+1+1) (See attached figure)
So Area = (L+2) * (B+2) = M (Given) ...(1)

Similarly when the Border is 2 inch on each side then Area will become (L+2+2) * (B+2+2) (see attached figure)
=> (L+4) * (B+4) = M + 52 (Given) ...(2)
[(L+2) + 2] * [(B+2) + 2] = M + 52
(L+2)*(B+2) + 2*(L+2) + 2(B+2) + 2*2 = M + 52
(L+2) * (B+2) = M from 1
=> M + 2* (L+2 + B+2) + 4 = M+52
=> 2*(L+B+4) = M + 52 - M - 4
=> 2(L+B+4) = 48
=> L + B + 4 = 24
=> L + B = 20
So, Perimeter = 2*(L+B) = 2*20 = 40 inches

Hope it helps!

More discussion here

Chelsann15 wrote:
I am struggling to break through this question and have revisited over multiple days. Any help would be appreciated.

A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

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16 Jul 2018, 20:43
Chelsann15 wrote:
I am struggling to break through this question and have revisited over multiple days. Any help would be appreciated.

A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

You can get a quick solution geometrically...
The area of just outer boundary is M+52-M=52..
Draw a line (red) to divide this boundary into 4 parts..
1) 2 parts are rectangle with sides (X+4) and 1
So area = $$(X+4)*1=X+4$$... Area of both parts = $$2(X+4)=2x+8$$
2) 2 parts are rectangle with sides (y+2) and 1
So area = $$(y+2)*1=y+2$$......area of both parts =$$2(y+2)=2y+4$$

Total area =$$52=2x+8+2y+4=2x+2y=12.$$..
So $$2x+2y=52-12=40........2(X+y)=40$$
2(X+y) is nothing but the perimeter of photograph.
Hence Ans 40

Hope it helps
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PicsArt_07-17-09.04.48.jpg [ 49.14 KiB | Viewed 163 times ]

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Re: Photo Frame Perimeter Question  [#permalink]

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18 Jul 2018, 08:42
Chelsann15 wrote:
I am struggling to break through this question and have revisited over multiple days. Any help would be appreciated.

A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?

Let $$l$$ and $$w$$ be the length and the width of the photograph.
From the condition we have $$(l+2)(w+2) = lw + 2l + 2w + 4 = M$$.
We also have $$(l+4)(w+4) = lw + 4l + 4w + 16 = M + 52$$.
Then $$(lw + 4l + 4w + 16 ) - (lw + 2l + 2w + 4) = 2l + 2w + 12 = (M+52)- M$$.
$$2(l+w) = 40$$
Here $$2(l+w)$$ is the perimeter of the photography.
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Re: Photo Frame Perimeter Question &nbs [#permalink] 18 Jul 2018, 08:42
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