Question 102728: Please help this is really hard!!
The answer is Length=16 meters;width=7 meters. How will I get it??
Problem: There is a rectangle whose perimeter is 18 centimeters. If its length is decreased by 5 centimeters and its width is increased by 12 centimeters, its area is doubled. Find its length and width.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! You've given the wrong answers for this problem. The first clue is that the answers you
provided for Length and Width are in meters while the dimensions in your problem are in
centimeters. Another check is that the perimeter of the rectangle is supposed to be 18 cm.
However, your answers of Length = 16 and Width = 7 would give a perimeter of 16 + 7 + 16 + 7
which totals 46.
.
So let's ignore your answer and work the problem using the information given.
.
Begin by calling L the original length of the rectangle and calling W the original width.
The perimeter of the rectangle is the distance around it, and this is L + W + L + W which
combines to 2L + 2W. You are told that this perimeter equals 18 cm. So we can write the
equation:
.
2L + 2W = 18
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and we can reduce this a little by dividing both sides (all terms) by 2 to get:
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L + W = 9
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Let's now solve this for one of the quantities in terms of the other. Let's say we solve
for L in terms of W. Do this by subtracting W from both sides of this equation to get:
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L = 9 - W
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One final step in this first part. Let's find the area of the original rectangle. We
know that the area (A) of a rectangle is found by multiplying the length L by the width W.
Therefore we can write the equation:
.
A = L*W
.
But we found that for this rectangle L is equal to 9 - W. Therefore, we can substitute
9 - W for L in the area equation and we get that the area of the original rectangle
is:
.
A = L*W = (9 - W)*W = 9W - W^2
.
Now to the second part of the problem. If we decrease the original length by 5 cm, the
new length is L - 5. Then if we increase the original width by 12 cm the new width is
W + 12. Therefore, the area of the new rectangle is found by multiplying the new length by
the new width which can be written as:
.
New Area = (L - 5)(W + 12)
.
But recall that L is equal to 9 - W, so replace L by 9 - W in the New Area equation to
get:
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New Area = (9 - W - 5)(W + 12)
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Combine the numbers 9 and -5 in the first set of parentheses and the equation reduces to:
.
New Area = (4 - W)(W + 12)
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Multiply out the right side and the equation becomes:
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New Area = 4W + 48 - W^2 - 12W
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and combining like terms 4W and - 12W the equation reduces to:
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New Area = -W^2 - 8W + 48
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The problem tells you that this new area equals twice the old area. Recall that earlier we
found that the old area was 9W - W^2 so twice the old area would be 18W - 2W^2. Now we
can write the equation that the new area (-W^2 - 8W + 48) equals twice the old area (18W - 2W^2) ...
.
-W^2 - 8W + 48 = 18W - 2W^2
.
Let's get rid of the two terms on the right side. First get rid of the - 2W^2 by adding 2W^2
to both sides to get:
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W^2 - 8W + 48 = 18W
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Next get rid of the 18W on the right side by subtracting 18W from both sides:
.
W^2 - 26W + 48 = 0
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Next factor the left side:
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(W - 24)(W - 2) = 0
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Note that this equation will be true if either of the factors on the left side is zero because
a multiplication by zero on the left side will make the left side zero and therefore
equal to the right side.
.
By setting the first factor (W - 24) equal to zero and solving for W we get that W = +24 is
a possible answer. But is it really??? If the original width is 24, then the original
perimeter has to be much bigger than 18. So toss out the answer W = 24.
.
By setting the second factor equal to zero we find that W = +2. That looks a lot better. Going
way back to the beginning of our work we had found that L = 9 - W and if W is 2, then
L = 9 - 2 = 7.
.
So it looks as if the dimensions of the original rectangle is 7 cm long and 2 cm wide.
.
Let's check. The perimeter of the original rectangle would be 7 + 2 + 7 + 2 = 18 cm. That's OK.
The Area of the original rectangle is 7 times 2 or 14 square cm.
.
Now decrease the original length by 5 cm and you get 2 cm. Then increase the original
width by 12 cm and you get 2 + 12 = 14 cm. So the new figure has dimensions of 2 cm by 14 cm.
The area of this new figure is the product of 2 cm by 14 cm which equals 28 square cm.
And this area (28) is twice the square cm of the original rectangle (14). So everything
works out and our answers of 2 cm for the original width and 7 cm for the original
length are correct.
.
A lot of work ... hope this helps you to see your way through this problem.
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