• May 2nd 2009, 04:29 PM
chrozer
Express the area of the shaded triangle in terms of "y".

I have attached a picture of the problem. As you can see, I have already tried to solve the problem, but to no avail. I cannot figure it out at all. Any ideas?
• May 2nd 2009, 05:21 PM
Plato
The answer is to be in terms of y.
Every point ont your line is $(4-2y,y)$
The width is $4-2y$ and the length is $y$.
What is the area?
• May 2nd 2009, 05:37 PM
chrozer
Quote:

Originally Posted by Plato
The answer is to be in terms of y.
Every point ont your line is $(4-2y,y)$
The width is $4-2y$ and the length is $y$.
What is the area?

So the width is $4y - 2y^2$? ok..I see it now.
• May 2nd 2009, 05:52 PM
Plato
Quote:

Originally Posted by chrozer
So the width is $4y - 2y^2$? ok..I see it now.

No the area is $4y - 2y^2$.
• May 2nd 2009, 06:43 PM
chrozer
Quote:

Originally Posted by Plato
No the area is $4y - 2y^2$.

Ok yeah...sorry that's what I meant. Thanks alot.