Math Projects on Building for Middle School

Wood frame construction projects can provide a variety of math exercises for middle grade students.

Relating math to practical situations can be a valuable learning device for middle grade students. Construction projects, especially those concerning wood frame houses, provide multiple opportunities for young math students to appreciate how numerical systems work. Students can also learn how to apply basic geometric principles to real-life circumstances,

Stud Spacing

Explain how vertical walls are built with wooden studs that are placed either 16 or 24 inches apart. Then calculate the number of studs needed to cover a 20-foot length for each sized gap. This involves converting 20 feet to 240 inches. Now divide 240 by 16 and 24. Your results are 15 and 20, but you are not quite done with the math yet, for you have to have to add one stud to each total to allow for the beginning framing member. So the answers are 16 and 21.

Window Spacing

Design a wall with window openings. As an example, you might select a wall that has a width of 45 feet and height of 12 feet. Then install five windows that are 3 feet wide and 6 feet high. If you wish to place the windows halfway between the top and bottom of the wall, all you need do is to subtract the height of the wall by the height of the window and then divide by 2. In this case, the answer is 3 feet.

However, spacing is more problematic. For even spacing, you first multiply the number of windows time their width. This number will equal 15 feet. Now subtract 15 from the total width and you will have 30 feet of open space. This number gets divided by six (don't forget the extra space at the end), and equals five feet. Now have the students draw a diagram to scale, where each inch equals eight feet.

Paint coverage

Now your students can calculate the square footage of the same wall that will take paint. This is a simple calculation that involves calculating the square area of one window and then multiplying it by five. The answer should be 90 square feet. Now calculate the total area of the wall (540 feet) and subtract the smaller number from the larger. The answer should be 450 square feet.

Height of a Roof Peak

You can use the mathematical properties of an equilateral or right triangle to calculate the height of a roof. For example, say you are building a peaked roof on top of a building that is 20 feet wide. The longest available length of lumber for a roof rafter is also 20 feet. Because this building project is undertaken in an area with heavy snowfall, the architect wants to build the roof as steep as possible. The solution is achieved by constructing a drawing of an equilateral triangle, where each side is equal to 20 feet. In such a geometric form, the height of the triangle is calculated by using Pythagoras' theorem. The formula used is the square root of [S²-(S/2)²] = X, where S equals one side of the triangle (20 feet) or the square root of 400 - 100 and X is the height. The answer is 17.32. Incidentally, the pitch of such a roof will equal 45 degrees.

References

• Math is Fun: Pythagoras' Theorem