SOLUTION: Please someone help me quick! My son has 3 more test to take this summer and we are really having a hard time. It's been way too long since I have had math. The problem is: The

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Question 5367: Please someone help me quick! My son has 3 more test to take this summer and we are really having a hard time. It's been way too long since I have had math.
The problem is:
The perimeter of a rectangular field is 60 meters. Its area is 200 square meters. Find its dimensions.
Thank you in advance for your help. I LOVE THIS SITE!!!!!!!!!!
Mom in trouble deep!

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = width
y = length

Two equations:
2(x) + 2(y) = 60 Perimeter
xy = 200 Area

In the first equation, life is a lot easier if you divide both sides by 2 and write:
x + y = 30

From this you can solve for y by subtracting x from each side:
x -x + y = 30 - x
y = 30 - x

Now substitute this y = 30 -x into the other equation:
xy = 200
x(30 - x) = 200

Distributive property:
30x+-+x%5E2+=+200

This is quadratic because of the x%5E2+ term, so you must set it equal to zero. To avoid the negative x%5E2 term, take it to the right side by adding %2Bx%5E2 and +-+30x+ to each side.

+30x+-+x%5E2+%2B+x%5E2+-+30x+=+x%5E2+-+30x+%2B+200
0+=+x%5E2+-+30x+%2B+200

You can factor this or use the quadratic formula. Can you think of two numbers whose product is 200 and whose sum is 30? (Try 20 and 10!)
x%5E2+-+30x+%2B+200+=+0
%28x-20%29%28x-+10%29+=+0+
x=+20 or x=+10

Remember that y = 30 - x.
If x = 20, then y = 10.
If x = 10, then y = 20.

It is appropriate to think of the width as being 10 meters and the length 20 meters.

R^2 at SCC