I have a problem for you.

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lceman23

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Sep 28, 2011
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I am in Intermediate Algebra in college and I have this word problem. I cannot figure out how to set it up. If someone can help me set it up I can probably solve it. The problem is this.

A rectangle and a square have the same area. The width of the rectangle is 6cm less than the side of the square and the length of the rectangle is 5 cm more than twice the side of the square. WHAT ARE THE DIMENSIONS OF THE RECTANGLE?
 
lceman23 said:
I am in Intermediate Algebra in college and I have this word problem. I cannot figure out how to set it up. If someone can help me set it up I can probably solve it. The problem is this.

A rectangle and a square have the same area. The width of the rectangle is 6cm less than the side of the square and the length of the rectangle is 5 cm more than twice the side of the square. WHAT ARE THE DIMENSIONS OF THE RECTANGLE?
Click to expand...

Rectangle Area = w * L


w,L are the side lengths which are parallel

Square Area = a*a

w=a-6

L=5+2a

Areas are Equal
 
Apparently I am still working it wrong. I know that the answer to the problem is width= 4 height= 25 cm.
I cannot figure out how to get that answer. Can anyone show me the work involved to get the answer so I know how to do it. Thanks.
 
burakaltr said:
Rectangle Area = w * L


w,L are the side lengths which are parallel

Square Area = a*a

w=a-6

L=5+2a

Areas are Equal
Click to expand...

a*a=(a-6)(5+2a)

a*a=5a+2 a**2 -30 -12a

a**2=2a**2 -7a - 30

0=a**2-7a-30

what 2 numbers make -30 when multiplied and -7 when added up ?
 
When I set it up I forgot to set a*a which didn't give me a**2 in my equation. Thanks for the help
 
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