# Question Video: Finding the Coordinates of the Center of Mass of a Uniform Triangular Lamina with a Discrete Mass at One of Its Sides Mathematics

A uniform lamina in the form of an equilateral triangle of side length 24 cm has a mass of 298 g. A mass of 149 g is attached to the lamina at one of the trisection points of line segment π΄π΅ as shown in the figure. Determine the coordinates of the systemβs center of gravity.

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### Video Transcript

A uniform lamina in the form of an equilateral triangle of side length 24 centimeters has a mass of 298 grams. A mass of 149 grams is attached to the lamina at one of the trisection points of line segment π΄π΅ as shown in the figure. Determine the coordinates of the systemβs center of gravity.

In this question, we are asked to find the coordinates of the systemβs center of gravity. The system is made up of an equilateral triangle of mass 298 grams and side length 24 centimeters, together with a mass of 149 grams attached at one of the trisection points of line segment π΄π΅. We will begin by finding the center of gravity of both components. The center of gravity of the 140-gram mass will be the point at which this is attached. The line segment π΄π΅ is 24 centimeters long. To find the trisection point, we need to calculate one-third of 24 centimeters. This is equal to eight centimeters. Therefore, the π₯-coordinate of the center of gravity is eight. As the mass lies on the π₯-axis, the π¦-coordinate will be zero. This means that the center of gravity of the 149-gram mass is eight, zero.

Our next step is to calculate the center of gravity of the 298-gram triangle. This will occur at the intersection of the three diagonals shown, where the diagonals are the perpendicular bisectors of π΄π΅, π΅πΆ, and π΄πΆ. One way of finding this point is to begin by considering the coordinates of vertices π΄, π΅, and πΆ. Point π΄ is at the origin so has coordinates zero, zero. As the triangle has side length 24 centimeters, point π΅ has coordinates 24, zero. Point πΆ will have an π₯-coordinate of 12, as this is the midpoint of zero and 24.

Working out the π¦-coordinate is more complicated, and we need to use our knowledge of right triangles. We can use either the Pythagorean theorem or the trigonometric ratios. The Pythagorean theorem states that π squared plus π squared is equal to π squared, where π is the length of the longest side or hypotenuse of the triangle and π and π are the lengths of the shorter sides. If we let the coordinate we are trying to calculate be π¦, we have 12 squared plus π¦ squared is equal to 24 squared. 12 squared is 144 and 24 squared is 576. Subtracting 144 from both sides, we have π¦ squared is 432. Since π¦ must be positive, taking the square root of both sides of this equation gives us π¦ is equal to 12 root three. Vertex πΆ of our triangle has coordinates 12, 12 root three.

We can now find the center of gravity of the triangle by finding the average of the π₯- and π¦-coordinates. The average of the π₯-coordinates is equal to 12 plus zero plus 24 all divided by three. This is equal to 36 divided by three, which in turn equals 12. The π₯-coordinate of the center of gravity of the triangle is 12. Repeating this for the π¦-coordinates, we have zero plus zero plus 12 root three divided by three. This is equal to four root three. The center of gravity of the triangle lies at the point 12, four root three.

An alternative way of calculating the π¦-coordinate would be to consider the green right triangle shown. We could use the trigonometric ratio tan of π is equal to the opposite over the adjacent, where the angle π is equal to 30 degrees as the hypotenuse bisects a 60-degree angle and the adjacent side is 12 centimeters. Substituting in these values, we have the tan of 30 degrees is equal to π over 12. And multiplying through by 12, π is equal to 12 multiplied by the tan of 30 degrees. As tan 30 is equal to root three over three, then π is equal to four root three. This confirms the π¦-coordinate of the center of gravity of our triangle.

Now that we have the centers of gravity of the mass and the triangle, we can use these to find the center of gravity of the whole system. The π₯-coordinate will be equal to π one, π₯ one plus π two, π₯ two divided by π one plus π two, where π one and π two are the two masses and π₯ one and π₯ two are the π₯-coordinates of the corresponding centers of gravity. We can use a similar formula to calculate the π¦-coordinate of the center of gravity.

Substituting the values into the right-hand side of our first equation, we have 149 multiplied by eight plus 298 multiplied by 12 all divided by 149 plus 298. Typing this into our calculator gives us 32 over three. Since π¦ sub one equals zero, the π¦-coordinate is 298 multiplied by four root three divided by 149 plus 298. This is equal to eight root three over three. We can therefore conclude that the coordinates of the systemβs center of gravity are 32 over three, eight root three over three.