Question 168149
let:
W1 = width of the large square.
L1 = length of the large square.
W2 = width of the large square.
L2 = length of the large square.
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you are given that w = the width of the large square:
W1 = w
since this is a square, then L1 = w also.
you have:
L1 = W1 = w
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you are given that .75*w = the width of the small square:
W2 = .75*w
since this is a square, then L2 = .75*w also.
you have:
L2 = W2 = .75*w
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area of the large square is L1*W1 = A1
area of the small square is L2*W2 = A2
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large square is all red (assumed since you are only given that the border is red).
small square is all blue (given).
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if you take the small square and put it in the middle of and on top of the large square, what you will have is a blue square in the middle with a red border.
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that, i believe, is what causes your problem to be solved as follows:
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you are given that the red border is 28 square meters.
if you take the area of the large square and subtract the area of the small square you will be left with the area of the border.
so, your equation is:
A1 - A2 = 28
since A1 = L1*W1, and A2 = L2*W2, this becomes:
L1*W1 - L2*W2 = 28
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since L1 = w, and W1 = w, this becomes:
w*w - L2*W2 = 28
since L2 = .75*w, and W2 = .75*w, this becomes:
w*w - (.75*w)*(.75*w) = 28
simplifying:
w^2 - .5625*w^2 = 28
(1-.5625)*w^2 = 28
.4375*w^2 = 28
w^2 = 28/.4375
w^2 = 64
w = 8
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if w = 8, then .75*w = 6
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since w = 8, then
L1 = 8
W1 = 8
since .75*w = 6, then
L2 = 6
W2 = 6
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your original equation is:
L1*W1 - L2*W2 = 28
this becomes:
8*8 - 6*6 = 28
this becomes:
64 - 36 = 28
which becomes:
28 = 28
since this statement is true, the value for w = 8 is correct.
the lengths of each side of the mat are:
8 inches for the large square.
6 inches for the small square.
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