SOLUTION: Two rectangular rooms each have an area of 240m^2. If the length of one of the rooms is x metres and the other room is 4 metres longer, write down the width of each room in x
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-> SOLUTION: Two rectangular rooms each have an area of 240m^2. If the length of one of the rooms is x metres and the other room is 4 metres longer, write down the width of each room in x
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Question 1201227: Two rectangular rooms each have an area of 240m^2. If the length of one of the rooms is x metres and the other room is 4 metres longer, write down the width of each room in x
I have done this: 240/x and 240/x+4
It is the second part of the question that I do not understand.
If the widths of the rooms differ by 3 meters, form an equation in x and show that this reduces to x^2+4x-320=0
Solve this equation and hence find the difference between the perimeters of the rooms. Found 2 solutions by greenestamps, josgarithmetic:Answer by greenestamps(13203) (Show Source):
Be careful; that's not really what you have done. What you have done is "I have done this: 240/x and 240/(x+4)"
Proper use of parentheses is important!
Now for your question....
You have the expressions for the widths of the two rooms. Now write the equation that says the difference in the widths is 3 meters -- remembering that the greater width is with the smaller length:
Then solve; the first step is probably multiplying everything by the least common denominator, x(x+4).
Solve by factoring -- find two numbers whose product is 320 and whose difference is 4.
If you're not particularly good at mental math, here is how you might go about finding those two numbers. One "easy" factorization of 320 is 32*10; then note that if we divide the 32 by 2 and multiply the 10 by 2 (so that the product remains the same) we get 16 and 20 -- which is what we need. So
or
Clearly the negative solution makes no sense in the problem, so x=16.
So in one room the width is x = 16 and the length is 240/16 = 15; in the other the width is x+4 = 20 and the length is 240/20 = 12.
Then the perimeter of the first room is 2(16+15) = 62 and the perimeter of the second room is 2(20+12) = 64; so the difference between the perimeters of the two rooms is 2 meters.
There is a useful side note for this problem. It is an example of a general principle that, for a given area, a rectangle will have a smaller perimeter if there is a small difference between the length and width and a larger perimeter if there is a large difference.
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