Thread: Another area problem

1. Another area problem

I am also having problems with this ...

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

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 $\displaystyle a\cdot b.$

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

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:

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

4. Thank you for your help. Your insight into the understanding of mathematical concepts is appreciated.

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:

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