# Help with Difficult Centroid Problem

• Oct 18th 2009, 01:10 PM
messianic
Help with Difficult Centroid Problem
Here is the problem,

Show by direct computation that the centroid of the triangle with vertices (0,0), (r,0), and (0,h) is the point (r/3, h/3). Verify that this point lies on the line from the vertex (0,0) to the midpoint of the opposite side of the triangle and twothirds of the way from the vertex of the midpoint.

I really don't even know where to begin..
• Oct 18th 2009, 01:38 PM
skeeter
Quote:

Originally Posted by messianic
Here is the problem,

Show by direct computation that the centroid of the triangle with vertices (0,0), (r,0), and (0,h) is the point (r/3, h/3). Verify that this point lies on the line from the vertex (0,0) to the midpoint of the opposite side of the triangle and twothirds of the way from the vertex of the midpoint.

I really don't even know where to begin..

what do you know about finding the centroid of a planar region?
• Oct 18th 2009, 01:54 PM
messianic
I know you first have to find the mass but I don't know where to start for that
• Oct 18th 2009, 02:05 PM
skeeter
Quote:

Originally Posted by messianic
I know you first have to find the mass but I don't know where to start for that

$\displaystyle \bar{x} = \frac{\int_a^b x \cdot f(x) \, dx}{\int_a^b f(x) \, dx}$

$\displaystyle \bar{y} = \frac{\int_a^b \frac{1}{2}[f(x)]^2 \, dx}{\int_a^b f(x) \, dx}$
• Oct 18th 2009, 02:08 PM
HallsofIvy
No, you don't find the "mass", a geometric figure does not have a "mass"!

You find the area of the figure. This is a triangle with base of length r and height h. What is its area?

You still have not answered the question skeeter asked. He asked if you knew how to find the centroid and you answered that you first had to find the "mass". After you have found the area (not mass) what do you do?

Since this is a right triangle, it will be particularly easy.