Tyrese wants to build a three-dimensional structure for a project. He needs to choose from a cube, rectangular prism, triangular prism, or square pyramid. He knows that whatever he chooses will impact what his friend, Miguel, will need to build. If you were Tyrese, what structure would you build? Once he chooses his structure, he needs to sketch the figure and label the whole number dimensions needed to find the surface area of the figure. His friend, Miguel, needs to build the same figure, but needs to double all of the dimensions. Write an explanation of what happens to the surface area of the figures when the dimensions are doubled. Include all relevant math work to help clarify your explanation.
If I were Tyrese, I would choose to build a cube for the project. A cube has all sides equal in length, which makes it easier for Miguel to double all of the dimensions for his project.
To find the surface area of a cube, you use the formula: 6s^2, where s is the length of one side. In this case, let's say the cube has sides of length 3. So, the surface area of the cube is 6(3)^2 = 6(9) = 54 square units.
Now, if Miguel doubles all of the dimensions for his cube, the new side length would be 6. To find the surface area of Miguel's cube, we use the formula: 6(6)^2 = 6(36) = 216 square units.
When Miguel doubled all of the dimensions, the surface area of the cube increased by a factor of 4 (216/54 = 4). This is because the surface area of a three-dimensional figure is proportional to the square of its dimensions. Therefore, when the dimensions are doubled, the surface area is quadrupled.