## Practice problem for the x and y coordinates of a trapezoid.

In this post, we will introduce a practice problem for the x and y coordinates of a Trapezoid.

As we can see in the next slide image a given trapezoidal area is drawn with the left side coinciding with the y-axis with length a and the lower side coinciding with the x-axis with length h. The right side of the Trapezoidal area is with length a.

### How to find the X-bar for the trapezoidal area?

The area can be considered as composed of two triangles and there is another position which is to consider it as a rectangle and with a triangle. For the x Coordinate, I will use two triangles. The first triangle is with base a and the height is h, the base coincides with the y-axis.

The second triangle is with base h and the height is b, the base coincides with the X-axis.

As we know the x-bar for the first triangle is equal to h/3 while the x-bar for the second triangle is 2/3 h. The areas of the two triangles are estimated based on the base and height of each one.

The next step is to estimate the product of area by the x bar of each triangle to get the final value of the first moment of area for the Trapezoidal area.

The final value of the first moment of the area can be found equal to (h^2/6)*(a+2b).

We divide the estimated first moment of the area of the trapezoid by its area we can find out that the required x-bar of the trapezoidal area is equal to (h/3)*(a+2b)/(a+b). Thus we have completed part 1 for the Practice problem for the x and y coordinates of a trapezoid.

### How to find the y-bar for the trapezoidal area? first option.

To estimate the y bar of the trapezoidal area, we will do it with two options. The first option is to consider the trapezoidal area as composed of the two triangles as we have done while estimating the x-bar.

From the first option, the area of the first triangle is (1/2)*b*h while for its y cg, we further divide it into two parts, the first part is with base b and with y bar equals 2/3b, while the second part is with a base of (a-b) and its y cg is (b+1/3(a-b), which can be adjusted to (2b+a)/3.

We will proceed to finalize the y-bar of the first triangle please refer to the next slide image for more details.

We will multiply each part by its y cg value for the first triangle to finalize the first moment of area for the first triangle, after adjusting the terms we can find that the value is equal to h/6*(a^2+ab). Please refer to the next slide image for more details.

For the second triangle area A2 the area is equal to (1/2*b*h) and its y cg is (b/ cg is (b/3). The next step is to sum the first moment of the area of the first triangle with the first moment of the area of the second triangle then divide by the area of the trapezoid. The final value of the Y-bar is equal to (a^2+ab+b^2)/(3)*(a+b). Thus we have completed part 2 of the Practice problem for the x and y coordinates of a trapezoid.

### How to find the y-bar for the trapezoidal area? second option.

The second option is to consider the trapezoid as composed of a rectangle and a triangle we estimate both the area and the y cg of each area about the x-axis. Then we sum the first moment of the area of the triangle and the rectangle. A step-by-step guide can be seen from the next slide image.

We divide the first moment of the area by the total area of the Trapezoid. Y bar can be found as matching to the value as estimated by Option -1. Thanks a lot.

For the estimation of the area and the Cg of trapezium, please refer to post 8.

Efunda gives a useful link for areas and Cg can be viewed.