rectangle problem

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Motherof8
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rectangle problem

Postby Motherof8 » Tue Aug 23, 2011 12:28 am

Here's the problem: The perimeter of a rectangle is 56 ft. The diagonal is 20 ft. Find the dimensions. I made this formula:

2(x + y) = 56 x squared + y squared = 400 (20 squared) but I don't know what to do from there.

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stapel_eliz
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Postby stapel_eliz » Tue Aug 23, 2011 2:17 am

Motherof8 wrote:Here's the problem: The perimeter of a rectangle is 56 ft. The diagonal is 20 ft. Find the dimensions. I made this formula:

2(x + y) = 56 x squared + y squared = 400 (20 squared) but I don't know what to do from there.

One way to proceed might be to divide the first equation by 2, and then solve for one of the variables. Plug this into the other equation, and solve the resulting quadratic. :wink:

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maggiemagnet
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Re: rectangle problem

Postby maggiemagnet » Tue Aug 23, 2011 4:07 pm

You can see an example of this on the second half of this page.

:clap:

akihironihongo
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Re: rectangle problem

Postby akihironihongo » Wed Aug 24, 2011 9:03 am

Motherof8 wrote:Here's the problem: The perimeter of a rectangle is 56 ft. The diagonal is 20 ft. Find the dimensions. I made this formula:

2(x + y) = 56 x squared + y squared = 400 (20 squared) but I don't know what to do from there.


well first of all divide 56 by two to get 28. now you have o find the two numers that square to 400 and add to 28. now im pretty sure you can do this with an equation but i forgot it and thus did guess and check and found the dimentions to be 12 and 16.
PROOF: 12+16=28 and 12 squared is 144, 16 squared is 256, 144+256=400


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