# Another area problem

• Jul 20th 2009, 10:18 PM
KevinVM20
Another area problem
I am also having problems with this ...

http://i561.photobucket.com/albums/s...vm35/area2.jpg
• Jul 20th 2009, 10:26 PM
AlephZero
Quote:

Originally Posted by KevinVM20
I am also having problems with this ...

http://i561.photobucket.com/albums/s...vm35/area2.jpg

This is just a simple application of the Pythagorean Theorem, $\displaystyle a^2 + b^2 = c^2.$ In this case, $\displaystyle a=12,$ and $\displaystyle c=13.$ Solve for $\displaystyle b$. The area of the rectangle is then just $\displaystyle a\cdot b.$
• Jul 20th 2009, 10:28 PM
earboth
Quote:

Originally Posted by KevinVM20
I am also having problems with this ...

http://i561.photobucket.com/albums/s...vm35/area2.jpg

You are dealing with 2 congruent right triangles. Use Pythagorean theorem to get the length of the rectangle's width.

Spoiler:

Let w denote the length of the width. Then you have:

$\displaystyle 13^2=w^2+12^2~\implies~w^2=25$
• Jul 21st 2009, 12:29 AM
KevinVM20
Thank you for your help. Your insight into the understanding of mathematical concepts is appreciated.

Quote:

Originally Posted by earboth
You are dealing with 2 congruent right triangles. Use Pythagorean theorem to get the length of the rectangle's width.

Spoiler:

Let w denote the length of the width. Then you have:

$\displaystyle 13^2=w^2+12^2~\implies~w^2=25$