GEBRA: How to Create Quadratic Equations for a Given Area of a Rectangle

[trulyfalse]

Hello PF! I'm having trouble approaching this problem. Any assistance would be greatly appreciated.

Homework Statement

A rectangle with area of 35 cm² is formed by cutting off strips of equal width from a rectangular piece of paper. The rectangular piece of paper is of 7cm width and 9cm length.

Homework Equations

ax²+bx+c

The Attempt at a Solution

I know that I have to create two separate equations and substitute to solve, however I'm really not certain how to create those equations (more-so what to base them on) . Perhaps a push in the right direction is all I require...

 

[eumyang]

I think you need to clarify what you mean by "cutting off strips of equal width." It sounds ambiguous to me.

 

[trulyfalse]

Sorry, I didn't include the actual question. It asks:
a) what is the width of each strip?
b) what are the dimensions of the new rectangle?

In my textbook, the diagram shows a rectangle with sides labeled 9cm and 7cm, and an inner rectangle of area 35 cm squared with unknown dimensions. The strips equal width are labeled as x and are shown to be the distance from one side of the inner rectangle to the closest side of the outer rectangle. I hope that clarifies my question; unfortunately I cannot upload the diagram right now.

 

[rollcast]

Is this what the diagram looks like

 

[trulyfalse]

Yes, that's correct.

 

[rollcast]

Ok think what the lengths of inner rectangle can be expressed as.

Hint you need to use the length of the corresponding outer side and x.

 

[trulyfalse]

Thank you! I think I figured it out:
A=lw
35=lw
35=(7-2x)(9-2x)
35=63-18x-14x+4x²
0=4x²-32x+28

I then factored out the polynomial:
4x²-4x-28x+28
4x(x-1)28(x-1)
(4x-28)(x-1)
x=1,7

And since a negative dimension is illogical, 7cm is extraneous and 1 cm is the width of the strip. After that I substituted the value back into the equation, and that yielded 7cm and 5cm as answers.

Thanks man! You've been a huge help!

 

[rollcast]

Thank you! I think I figured it out:
A=lw
35=lw
35=(7-2x)(9-2x)
35=63-18x-14x+4x²
0=4x²-32x+28

I then factored out the polynomial:
4x²-4x-28x+28
4x(x-1)28(x-1)
(4x-28)(x-1)
x=1,7

And since a negative dimension is illogical, 7cm is extraneous and 1 cm is the width of the strip. After that I substituted the value back into the equation, and that yielded 7cm and 5cm as answers.

Thanks man! You've been a huge help!

No problem. We did these in school last year and seeing the initial formula to get started is the hardest part. If you keep on trying questions you will eventually get a feel for what types of problem come up regularly and how to deal with them.

AL