Area Problem

**norifurippu** · November 22nd 2009, 11:52 PM

Hello,
The Area of a rectangle is 360 square cm, if 12 is added to the length of the figure, and 5 is substracted from the width, the resulting Area is the same. What are the dimensions of the first rectangle?

Tried to solve it using a system of 2 equations but I'm confused about the "square cm".

Thanks in advance.

**alexmahone** · November 23rd 2009, 12:33 AM

Originally Posted by norifurippu

Hello,
The Area of a rectangle is 360 square cm, if 12 is added to the length of the figure, and 5 is substracted from the width, the resulting Area is the same. What are the dimensions of the first rectangle?

Tried to solve it using a system of 2 equations but I'm confused about the "square cm".

Thanks in advance.

Let the length of the first rectangle be x.
Let the width of the first rectangle be y.

$xy=360$

$(x+12)(y-5)=360$

$xy-5x+12y-60=360$

$-5x+12y-60=0$

$5x-12y+60=0$

$5x-12*\frac{360}{x}+60=0$

$x-12*\frac{72}{x}+12=0$

$x^2+12x-864=0$

$(x+36)(x-24)=0$

$x=-36$ or $x=24$

Rejecting the negative root,

$x=24$ cm

$y=\frac{360}{24}=15$ cm

The dimensions of the first rectangle are 24 cm and 15 cm.

**norifurippu** · November 23rd 2009, 10:02 PM

Thanks a lot!

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