## Product Of Inertia Ixy for an isosceles Triangle.

### Brief content of the video.

We are going to estimate the product of inertia Ixy for an isosceles triangle at the left corner. The isosceles triangle is a special case of the triangle ABC, for which the two sides AC and CB are equal. If we drop a perpendicular to the base, we will find that the dimension a=b/2.

So, actually, we are going to use the expressions which we have estimated for the product of inertia for the triangle Ixy and then readjust and substitute the value of a =b/2. Since we have Ixy=(b*h^2/24)*(2a+b) and our a, which is the distance from our perpendicular to the left corner A is =b/2, and our two axes.

We can draw, this as the X-axis, this is the y-axis, and our CG. the estimation of the Product Of Inertia Ixy for an isosceles at the CG is explained as well. This is a part of the video with a closed caption in English.

If you wish you can watch the video on U-tube and here is the link.

### Product of inertia Ixy for an isosceles triangle.

1-To get the expression for the Product of inertia, Ixy for an isosceles triangle. We will use the words we have estimated for the Ixy for the triangle and readjust Ixy for an isosceles triangle.

The next step is substituting the value of (a ) as a=b/2. In the Ixy formula of an isosceles triangle, the Ixy formula for the triangle. This is the link to the related post. The expression for Ixy for an isosceles triangle is shown in the following formula.

For the case of an isosceles, we have a relationship between a and b values as we can substitute for a=b/2 in the previous expression for Ixy for the triangle as follows: The expression for the product of inertia Ixy for an isosceles at point a which is at the left corner at which we have the intersection between the two axes and y, of the isosceles triangle is shown in the next slide.

The product of inertia =base^2*height^2/(12).

### Product of inertia Ixy for an isosceles at the CG.

For the product of inertia. Ixy formula of the isosceles triangle at the CG, we will follow the parallel axis theorem, for which we will deduct the product of area by the square distance from the CG axis from the value of Ixy of an isosceles at the external axis that is coincided by the base. We have already estimated the Product of inertia Ixy for an isosceles at point a, which is equal to base^2*height^2/(12).

The product of inertia at the Cg of an isosceles triangle Ixcg=Ixy at point a- A*(x-cg)*(y-cg).

The distance xcg=(1/2b), while y cg=(1/3)h, where b is the base length, while h is the height of the isosceles.

The area of the triangle is (1/2)*(b)*(h).

We apply the formula, of the product of inertia for the triangle we will find the product of inertia for an isosceles at the CG. We will discover that the Ixy at the Cg equals zero.

The calculations for the estimation of the Product Of Inertia Ixy for an isosceles at the CG can be viewed from the next slide image.

This is the pdf file used in the illustration of this post.

For an external resource, the definition of the moment of inertia with solved problems for the 2nd moment of inertia.

For the next post, Moment of inertias Ix for the Trapezium. step-by-step introduction to the method of estimating the inertia for a trapezium.