### Video Transcript

The diagram shows a uniform plane figure. Given that the grid is composed of unit squares, find the coordinates of the figureβs center of mass.

We begin by noticing that the figure on the grid is made up of a rectangle and a semicircle as shown. We are asked to find the figureβs center of mass. One way of doing this is to find the center of mass of the rectangle and semicircle first. As the grid is composed of unit squares, we begin by finding the coordinates of the four corners of our rectangle. The four vertices or corners have coordinates two, two; two, four; seven, two; and seven, four.

We know that the center of mass of a rectangle will be halfway along its width and halfway up its height. The π₯-coordinate will be the average of the four π₯-values and the π¦-coordinate the average of the four π¦-values. The π₯-coordinate will therefore be equal to two plus two plus seven plus seven all divided by four. Since the width of the rectangle is parallel to the π₯-axis, this is equivalent to two plus seven divided by two. We need to find the midpoint of two and seven, which is equal to 4.5.

We can repeat this process to find the π¦-coordinate. This time, we need to find the average of two and four. This is equal to three. The center of mass of the rectangle lies at the point with coordinates 4.5, three. When dealing with a semicircle, finding the center of mass is more complicated. The center of mass of the semicircle drawn is a distance π units along from the bottom-left corner, where π is the radius of the circle. The center of mass lies a perpendicular distance β from the base of the semicircle as shown, where β is equal to four π over three π.

In this question, we see that the radius of the semicircle is two units. Since the bottom vertex of our semicircle has coordinate seven, one, we can calculate the π¦-coordinate of the center of mass of the semicircle by adding two to one. This is equal to three. The value of β is equal to four multiplied by two divided by three π, as the radius is equal to two. This simplifies to eight over three π. The π₯-coordinate of the center of mass of the semicircle is therefore equal to seven plus eight over three π. We now have the centers of mass of both the rectangle and semicircle. Next, we need to recall how we find the center of mass of a composite shape or lamina.

If we have two shapes with areas π΄ sub one and π΄ sub two, then the π₯-coordinate of their center of mass is π΄ sub one π₯ sub one plus π΄ sub two π₯ sub two all divided by π΄ sub one plus π΄ sub two, where π₯ sub one and π₯ sub two are the π₯-coordinates of the centers of mass of the individual shapes. In the same way, the π¦-coordinate of the center of mass is equal to π΄ sub one π¦ sub one plus π΄ sub two π¦ sub two divided by π΄ sub one plus π΄ sub two, where π¦ sub one and π¦ sub two are the π¦-coordinates of the individual centers of mass.

We have already calculated π₯ sub one and π¦ sub one together with π₯ sub two and π¦ sub two. The area of the rectangle, π΄ sub one, is equal to five units multiplied by two units. This is equal to 10 square units.

We can calculate the area of any semicircle by using the formula ππ squared divided by two. In this question, we know that the radius of the semicircle is two units. The area of the semicircle is therefore equal to two π square units. This is our value of π΄ sub two. We can now substitute in our values to find the center of mass of the figure.

The π₯-coordinate is equal to 10 multiplied by 4.5 plus two π multiplied by seven plus eight over three π all divided by 10 plus two π. Distributing the parentheses on the numerator gives us 45 plus 14π plus 16 over three, which in turn is equal to 151 over three plus 14π. We can eliminate the fraction from the numerator by multiplying each of the four terms by three. This gives us 151 plus 42π over 30 plus six π. The π₯-coordinate of the center of mass is equal to 42π plus 151 divided by six π plus 30.

We can find the π¦-coordinate using the same method. This is equal to 10 multiplied by three plus two π multiplied by three all divided by 10 plus two π. Whilst we could multiply 10 by three and two π by three, we notice that both terms on the numerator have a common factor of three. Factoring this out, we have three multiplied by 10 plus two π all divided by 10 plus two π. By canceling 10 plus two π from the numerator and denominator, we are left with an answer of three. The π¦-coordinate of the center of mass of the figure is three.

It is worth noting at this point that our value of π¦ one was equal to the value of π¦ two. The π¦-coordinate of the center of mass of the rectangle and the semicircle were both equal to three. This means that the center of mass of the whole figure will also be three. We can therefore conclude that the coordinates of the figureβs center of mass are 42π plus 151 over six π plus 30, three.