Townhouse Tribulation

Task

1 townhouse
2 townhouses
3 townhouses
10 townhouses
N townhouses

This task was given to sixth grade students who had just finished a unit on functions and algebra. Students previously had studied the area of polygons.

What This Task Accomplishes

This task allows students to find a rule for determining the number of sides connecting townhouses will have. It also provides them with a problem-solving situation to apply concepts and skills for finding areas of polygons.

Time Required for Task

1-2 hours

Interdisciplinary Links

This task brought up an interesting discussion on why townhouses or condominiums tend to be less expensive than building single-family homes. Students who live in townhouses shared that they know that only one wall exists between two townhouses in that they can usually hear everything that happens next door. So as you can see, this task can link to construction and the pros and cons of cutting corners, or in this case, walls.

Teaching Tips

Students had a prior week of instruction on looking at patterns and creating function tables and rules based on the patterns. This provided students with some problem-solving skills with which to approach the task. Students who were not able to find an algebraic rule were still able to use logic to solve the problem. To make the task less complicated, you can provide the areas of one, two and three townhouses, but I wanted students to apply their prior knowledge of geometry so I could recheck for mastery.

Suggested Materials

• Pencil


• Paper


• Calculators


• Townhouse worksheet (see page 4)


• Formula sheet (Some students needed to refer to this for finding the area of triangles.)

Possible Solutions

• Area of three sides of townhouse = (25' x 20') = 500 x 3 = 1500 square feet


• Area of front of townhouse = 500 - 19.5 (area of door and windows) = 480.5 square feet


• Area of roof peeks: (.5) (25') (20') = 250 x 2 peeks = 500 square feet


• Total area of one townhouse = 2480.5 square feet


• Two townhouses = 2480.5 - 500 for shared wall = 1980.5 square feet + first townhouse 2480.5 = 4461 square feet


• Three townhouses = 2480 + 1950.5 + 1950.5 = 6441.5 square feet


• Rule for any number of townhouses is 1980.5N - 500 = square feet of bricks needed for any townhouses where N = number of townhouses.

Benchmark Descriptors

Novice
Few or no parts of the task are correct. Little math reasoning is used. Attempts at using math language may be present, but rudimentary.

Apprentice
Some parts of the solution are correct, but the student is unable to correctly solve for N and 10. Representations are accurate for the student's solution. Some work is missing. Some correct math language is used.

Practitioner
All work is shown. Math representations and language communicate the student's solution. Answers are correct for all parts. The approach used is clearly explained.

Expert
All work is shown, labeled and organized. Math representations are used to solve and communicate the solution. The student shows/explains the derivation of the rule for N townhouses. Algebraic notation is used correctly. All parts of the solution are correct.