Centre of gravity of a right angled triangle?! is this right?

• Jun 3rd 2009, 04:55 AM
gva0324
Centre of gravity of a right angled triangle?! is this right?
If I have a triangle with the coodinates A(0,0) B (30,0) and C (30,40)

Is the centre of gravity 1/3 of the base and 1/3 of the height?

would this make it (10, 40/3) ??
• Jun 3rd 2009, 05:33 AM
aidan
Quote:

Originally Posted by gva0324
If I have a triangle with the coodinates A(0,0) B (30,0) and C (30,40)

Is the centre of gravity 1/3 of the base and 1/3 of the height?

would this make it (10, 40/3) ??

The center of gravity will balance the triangle in the x direction, as well as in the y direction.

The area of the triangle is

$\frac{30 \times 40}{2} = 600$

You now need a triangle such that $\frac{X \times Y}{2} = 300$

The ratio of the main triangle's height to the base is $\frac {40}{30}$
You need to have $\frac{Y}{X} = \frac{4}{3}$ or $Y \, = \, 1.33333 X$

$\frac {X \times 1.33333 X}{2}=300$

$X^2 = 450$

$X = \sqrt{450} = 21.2132$

The center of gravity will occur along the triangle with the x coordinate equal to 21.2132.

You will need to do the appropriate calculations for the y coordinate.

You want to cut the large triangle so that the upper or smaller triangle is half the area of the larger triangle.