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1. Could sombody please tell me how to solve this word problem

A square photograph has to b trimmed so that it can fit into a rectangular picture frame. The frame is 2cm narrower than the fotograph and 5cm shorter once trimmed the fotograph has an area of 54cm squared.

What was the area of the fotograph before it was trimmed?

Thanks for any help.
2. first find the lengths (L cm) of the square:

54 = (L-2)(L-5) = L² - 7L + 10

=> L² - 7L - 44 = 0

solve this to find L (remembering L>0)

then original area = L²
3. (Original post by Jbowat)
Could sombody please tell me how to solve this word problem

A square photograph has to b trimmed so that it can fit into a rectangular picture frame. The frame is 2cm narrower than the fotograph and 5cm shorter once trimmed the fotograph has an area of 54cm squared.

What was the area of the fotograph before it was trimmed?

Thanks for any help.
I couldnt really work it out doing espaws mathod but this is what i did:

xy = 54
(y+5)(x+2)
=xy +2y + 5x +10
=54 +2y + 5x + 10
=64 +2y + 5x

Then by sing trial and error i guessed 2 values of x and y to make 54. So using x = 6 and y = 9.

64 + 18 + 30 = 112

So then by multiplying (y+5)(x+2) = 14 x 8 = 112

So the x and y values satisfy the original area which is 112cm^2
4. (Original post by ryan750)
I couldnt really work it out doing espaws mathod but this is what i did:
This might help:-
Let's call the photographs length x.
So the frame is 2cm narrower - ie, x-2 WIDTH
And it is 5cm shorter. ie, x-5 LENGTH
We know that the area of the frame is going to be the width x the length which is
(x-2)(x-5) = 54 (from the question)
So this is what we need to solve. Expand out and simply:-
x^2 - 7x -44 =0 Factorise:-
(x-11)(x+4)=0 So x, the original length of the photograph is=11 or -4. (obviously the positive value!)
So its original area (remember it was a square) would be 11 x 11 = 121
5. (Original post by ryan750)
I couldnt really work it out doing espaws mathod but this is what i did:

xy = 54
(y+5)(x+2)
=xy +2y + 5x +10
=54 +2y + 5x + 10
=64 +2y + 5x

Then by sing trial and error i guessed 2 values of x and y to make 54. So using x = 6 and y = 9.

64 + 18 + 30 = 112

So then by multiplying (y+5)(x+2) = 14 x 8 = 112

So the x and y values satisfy the original area which is 112cm^2
Sorry i was wrong - didn't read that it was a square to begin with - so it is 11 x 11 = 121

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