### Video Transcript

A uniform square lamina π΄π΅πΆπ· has a side length π. Another uniform lamina π΅πΆπΈ of the same density, shaped as an isosceles triangle, is attached to the square such that πΈ lies outside the square and π΅πΈ is equal to πΆπΈ. Given that the squareβs side length is five-thirds times the length of the triangleβs height, find the center of mass of the system.

Taking a look at our diagram, we see our square π΄π΅πΆπ· and the isosceles triangle attached to one side of the square. In this exercise, we want to find the center of mass of the square and triangle combined. To help us do that, weβre told that the square has a side length π and that the length of the triangleβs height, this distance here, is related to the square side length by this relationship: five-thirds times β equals π. Having written this down, letβs clear some space on screen and begin working on our solution of what is the center of mass π₯- and π¦-coordinate for these combined shapes.

For our overall strategy, we can begin by solving for the center of mass of the shapes individually, the square and the triangle by themselves. Once weβve done that, we can use the fact that a two-lamina center of mass, that is, the center of mass of two individual laminas combined together, is given by these relationships for the π₯- and π¦-coordinates of that center of mass. Here, π΄ one and π΄ two are the respective areas of our two laminas involved. π₯ one and π₯ two are the center of mass π₯-coordinates of these respective shapes. And π¦ one and π¦ two are their corresponding π¦ center of mass coordinates. In order to apply these relationships then, weβll need to solve for the centers of mass of our square and triangle as well as their areas.

Letβs start by considering our square. And if we consider the fact that this squareβs geometric center is located at its center of mass, we know that that is at a point here with both π₯- and π¦-coordinates of π over two. So letβs write those down as πΆππ π₯ square and πΆππ π¦ square coordinates. And then letβs consider the area of our square. Since it has side length π, that area is simply π squared. Now letβs move on to our triangle, calculating the center of mass π₯- and π¦-coordinates for it. We can recall that for such a shape, the center of mass π₯- and π¦-coordinates are equal to the average π₯- and π¦-coordinates, respectively, of the triangleβs vertices. Considering the coordinates at vertex π΅, we can see that these are π, π. At vertex πΆ, they are π and zero.

But then what about at vertex πΈ? We were told in our problem statement that this is an isosceles triangle. This means that the π¦- coordinate of point πΈ is midway between that of points πΆ and π΅. This tells us that itβs π over two. To solve for the π₯-coordinate though, weβll need to add the height β to the length π that we already have as the side length of our square. Considering this relationship between β and π, if we multiply both sides of the equation by three divided by five, then we find that β, the height of this isosceles triangle, is three-fifths π, which means that the π₯-coordinate of vertex πΈ will be π plus three-fifths π. Combining these gives us eight π divided by five.

Now that we know the coordinates of each of the three vertices of our triangle, we can use the fact that the center of mass π₯-coordinate is equal to the average of these three π₯-values and the center of mass π¦-coordinate is equal to the average of these three π¦-values. And note that this is true for any triangle whether isosceles or not. Beginning with our π₯-coordinate, this is equal to π plus π plus eight π over five all divided by three. This is equal to 18π over five divided by three or 18 over 15π, simplifying to six-fifths π. So, weβll record this value as our center of mass π₯-coordinate for the triangle and then clear this away so we can start to work on the center of mass π¦-coordinate.

This is equal to the average π¦-coordinate of the three vertices. Thatβs π plus zero plus π over two divided by three, which is three-halves π over three or simply π over two. And weβll then enter this value into our coordinate pair. The last thing we need to do before we can apply these two center of mass equations is calculate the area of our triangle. In general, triangle area is one-half base times height. Here weβve said that our height is three-fifths times π. And with that orientation, we can see that our base here is π. The area then is one-half π times three-fifths π or three-tenths π squared. Weβre now completely prepared to apply these two relationships to solve for our overall center of mass.

The center of mass of our two laminas combined has an π₯-coordinate given by the area of the square multiplied by the center of mass π₯-coordinate of the square plus the area of the triangle times the center of mass π₯-coordinate of that shape all divided by the areas of these two shapes added together. When we plug in all these values, we can note that a factor of π squared appears in both numerator and denominator in all terms. This means we can factor it out and then cancel it out, leaving us with this expression for our center of mass π₯-coordinate.

Note that in the numerator we can now factor out one power of π. And then if we add together one-half and three-tenths times six-fifths, thatβs equal to 43 divided by 50. And then in our denominator, one plus three-tenths is equal to 13 divided by 10. If we then multiply top and bottom by 10 divided by 13, our denominator simplifies to one, and we have π times forty-three fiftieths times ten thirteenths. This simplifies to forty-three sixty-fifths π. Thatβs the center of mass π₯-coordinate of our combined two-lamina shape. So we enter this value into that coordinate pair, and now we clear space to calculate our overall center of mass π¦-coordinate.

Similar to our π₯-coordinate, this equals the area of the square times the center of mass π¦-coordinate of that shape plus the area of the triangle times the center of mass π¦-coordinate of the triangle, and all this is divided by the sum of the areas of those two shapes. With those values substituted in, once again, we see that π squared can be factored out of numerator and denominator. Just like before then, that factor cancels out, and we see that, once more, one power of π can be factored out of the numerator.

When we go to add one-half to three-tenths times one-half, we get 13 over 20. And just like before, one plus three-tenths simplifies to thirteen-tenths. Multiplying numerator and denominator once more by 10 over 13, we see that our denominator simplifies to one while our numerator becomes π times one-half. This then is the center of mass π¦-coordinate of our two laminas together, which means that our final answer is that the center of mass coordinates of these two laminas is forty-three sixty-fifths π and π over two.