# Thread: Centre of gravity of a right angled triangle?! is this right?

1. ## 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) ??

2. 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

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

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

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

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

$\displaystyle X^2 = 450$

$\displaystyle X = \sqrt{450} = 21.2132$

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.

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# centre of gravity of right angled triangle formula

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