<PRE><B>Question 202506</B>
i think i have the solution as follows:
let P = perimeter of first board.
L = length of first board
W = width of first board
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length of first board = 2 * width of first board - 3
L = 2W - 3
perimeter of first board = 2L + 2W
since L = 2W-3, then
P = 2*(2W-3) + 2W which equals 4W-6 + 2W which equals 6W-6
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length and width of the second board are reciprocals of corresponding length and width of the first board.
length of second board = 1/(2W-3)
width of second board = 1/W
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perimeter of second board is 1/5th perimeter of the first board.
perimeter of second board = P/5
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perimeter of the second board = (2/(2W-3) + 2/W)
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since the perimeter of the first board is 5 times the perimeter of the second board, this means that:
6W-6 = 5*(2/(2W-3) + 2/W)
which is the same as:
6W-6 = (10/(2W-3) + 10/W)
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if we multiply both sides of this equation by (2W-3), we get:
(6W-6)*(2W-3) = 10 + 10*(2W-3)/W
if we multiply both sides of this equation by W, we get:
(6W-6)*(2W-3)*W = 10*W + 10*(2W-3)
we can simplify this to become:
(6W-6)*(2W-3)*W = 10W + 20W - 30
which becomes:
(6W-6)*(2W-3)*W = 30W-30
if we divide both sides of this equation by (6W-6), we get:
(2W-3)*W = (30W-30)/(6W-6)
which becomes:
(2W-3)*W = 5
this can be simplified to:
2W^2 - 3W = 5
subtract 5 from both sides of this equation to get:
2W^2 - 3W - 5 = 0
which can be factored into:
(2W-5)*(W+1) = 0
which makes:
W = 5/2
or
W = -1.
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W can&#39;t be negative so the only possible answer is W = 5/2.
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if W = 5/2, then
P = 2W-3 = 5-3 = 2
we have:
P = 2
W = 5/2
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P = 2L + 2W gets P = 4 + 5 = 9
The perimeter of the first board is 9 feet.
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since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.
to see if that&#39;s correct, we substitute known values for L and W into the second board.
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the length of the second board is 1/L.
This becomes 1/2.
the width of the second board is 1/W
This bgecomes 1/(5/2) = (2/5)
let L2 = length of second board.
let W2 = width of second board.
L2 = (1/2)
W2 = (2/5)
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let P2= perimeter of second board.
then P2 = 9/5
since P2 = 2*L2 + 2*W2, we should get P2 = 9/5 using the dimensions of L2 and W2.
2 * L2 = 2 * (1/2) = 1
2 * W2 = 2 * (2/5) = 4/5
1 is the same as 5/5
5/5 + 4/5 = 9/5
P2 checks out good.
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the dimensions as given are good.
answer to the problem is:
Perimeter of the first rectangular board is 9 feet.
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