Centroid of non-right triangle

[Gauss M.D.]

Homework Statement

See attachment

Homework Equations

The Attempt at a Solution

If the y-coordinate of the center of mass is given by (1/A)*∫y dA, how come the solution uses bh/2 as the area? This triangle isn't right angled so that area formula should not hold. What am I missing?

 

[ap123]

how come the solution uses bh/2 as the area? This triangle isn't right angled so that area formula should not hold. What am I missing?

The area of any triangle is given by bh/2, not just a right-angled one.

 

[Dieinhell100]

The area of any triangle is given by bh/2, not just a right-angled one.

What he said.

A right triangle is basically like half of a square, cut down in a diagonal. The area of a square is b*h (or side^2). So a right triangle's area should be bh/2.

In the case of non-right triangles, this still holds true. Cut a parallelogram diagonally and you get two triangles (just like the square, except without 90 degree angles). The area of a parallelogram is b*h, just like the square's area. Therefore the triangle's area is still bh/2.

 

[i_mt]

What he said.

A right triangle is basically like half of a square, cut down in a diagonal. The area of a square is b*h (or side^2). So a right triangle's area should be bh/2.

In the case of non-right triangles, this still holds true. Cut a parallelogram diagonally and you get two triangles (just like the square, except without 90 degree angles). The area of a parallelogram is b*h, just like the square's area. Therefore the triangle's area is still bh/2.

A right triangle is not always; "half of a square, cut down in a diagonal". This is only true for equilateral right triangles. All right triangles are; "half of a rectangle, cut on a diagonal".

 

[haruspex]

This triangle isn't right angled so that area formula should not hold.

Drop a perpendicular from the top of the triangle to the base, cutting it into two right triangles. Say this meets the base a distance c from the lower left corner.
The two right triangles have areas ch and (b-c)h. Add them up.

 

[robphy]

Imagine cutting your triangle into horizontal strips, as in your diagram.
Then slide each strip in the stack to make one side [approximately] vertical.
The base didn't change. The height didn't change. The area hasn't changed.
Refine your strips. Repeat.

 

[kuruman]

The formula works with scalene triangles too. Suppose you are looking for the area of the green triangle ABC. First find the area of the big right triangle ADB and subtract from it the area of the smaller blue right triangle CDB.

$A_{\text{ADB}}=\frac{1}{2}(BD)\times(AC+CD)$
$A_{\text{CDB}}=\frac{1}{2}(BD)\times(CD)$
$A_{\text{ABC}}=\frac{1}{2}(BD)\times(AC+CD)-\frac{1}{2}(BD)\times(CD)=\frac{1}{2}(BD)\times(AC)=\frac{1}{2}(\text{height}\times\text{base}).$

 

[i_mt]

Drop a perpendicular from the top of the triangle to the base, cutting it into two right triangles. Say this meets the base a distance c from the lower left corner.
The two right triangles have areas ch and (b-c)h. Add them up.

For this geometric construction to work consistantly, obtuse triangles must be positioned such that the longest side, i.e. that side opposite the obtuse angle must form the base. With this qualification this is construction is correct.