The outside walls of a new and unusal house have an area of 5060 square feet.This area includes windows and doors, but not the elevated part of the roof. The owner wants to paint the house's exterior walls (the siding),not the windows, with a high quality paint. If the owner buys 45 gallons of expensive exterior paint, and if each gallon of paint can paint up to 100 sq. ft, give examples of possible shapes , dimensions, and the number of windows and doors the owner must buy in order to use up all of the paint: that is, so that no paint will go to waste.(1)(11) This windows can be in the form of any of the following geometrical shapes: rectangular, squared, oval, circlular, triangular, etc. (Hint: Multiplicationand subtraction is necessary to figure out the area of the house not covered by the paint, which is the area the windows and doors must occupy. Also, formulas for figuring out the area of various geometrical shapes, which will be related to these windows and doors, must be employed.
In addition, describe three sets of dimensions this house may have in order to obtain the area in question. Give the names to the shapes with these dimensions. (13)
Example 1: To figure out the area not covered by the paint, we multiply 45x100=4500sq.ft. We subtract 4500 from 5060sq.ft. which leaves us with 560sq.ft.=the area needed to be filled in by the doors and windows. Then one could come up with 28 rectangular windows each having the same dimensions of 6'x3'= 18 sq.ft. The total area these 28 windows will cover is 504 sq. ft.(28x18=504) which leaves us with 56 sq. ft. to fullfill the remaining part of the house not covered by the paint(which is 560 sq. ft.). We can include two doors with dimensions of 4'x7'=28sq.ft. for a total area of (28+28=) 56sq. ft. 56sq.ft. of the doors + 504sq.ft. of the windows = 560sq.ft. that the paint will not be able to cover. (5) (7)
Example 2: We can obtain the same result for the area needed by following the first part of example 1. One could come up with 10 squared windows each having the same dimensions of 6'x6'=36 sq.ft. The total area these 10 windows will cover is 360sq. ft.(36x10=360) which leaves us with 200 sq.ft. to fullfill the remaining part of the house not covered by the paint(which is 560 squ.ft). We can then add 12 trianglular windows each having the base 5 ft and height 4 ft. with the total area of (5 x 4)/2= 10sqft. Then the total area would be 10 x 12 = 120 sq ft. With the area these squared and trianglular windows occupy(480sq.ft.), we have 80sq.ft. left which we can make two retangular doors each with dimensions of 5x8=40sq.ft. 40sq.ft.x2 doors=80sq.ft.(4) (7) Adding all gives 560sq.ft.
This generative activity is for 10th graders who are taking geometry classes and requires about three days of work.(12) Since this activity is for 10th graders, the teacher can use it to challenge their 9th graders as well.(3) The specific mathematical ideas emphasized involve the subject of areas. Before working with these activities, students need to have the knowledge of subtraction, multiplication, and addition of the areas involved along with formulas for finding the area of various geometrical shapes, which will be related to the shapes of windows. This knowledge is required so that the area these windows and doors must fully occupy can be easily manipulated.(2) In order for them to work with these activities, students need to have papers ,pencils and/or calculators to help them work with the images.(9) To start working on the problems, students can use the materials mentioned above to draw different geometrical shapes on papers.(10)
First day; before asking students to work on these activities, teachers should give them a review on formulas for some geometrical shapes and give an example of how to calculate the area of the shape she /he is giving as example by using the over head projector. Then the teacher can divide students into groups so that they may work together. It is optional for students to divide the jobs among themselves, and each of them have different responsibilities. For instance, one person is responsible for the drawings , another person is responsible for calculations, and the third person is responsible for putting their answers on a poster. Second day: continue working on the problems. Third day; each group can give an oral presentation of their solutions.(6)
In the whole-class discussion students will be able to see that generative mathematical problems which they may be dealing with can be solved in different ways with different methods.(8)
In order to see how effective these activities are, I've tested one of my friend's sister who is in 10th grade and who is taking geometry. She thought the problem was pretty straight forward. The only thing she did not like was that too many calculations are involved and too many formulas had to be memorized.
The following is the details of how this 10th grade student came up the with the solution: First, she multiplied 45 with 100 to get 4500 sq. ft. Then she subtracted 4500 sq. ft from 5060 sq.ft to get 560 sq ft which is the remaining part of the house not covered by the paint, and is also the area the windows and doors must fully occupy. Then, she drew four big retangles, which represented the four sides of the house. After that, she decided that one side had windows and one door only. The second side had windows and also a door. The rest of the two sides had only windows but no doors. Second, she started putting windows into the retangles she drew: (A)She drew one window on the first side with the dimensions of 8 x 4 = 32sq.ft. She also added a door with the dimesions 10 x 5 = 50 sq.ft. (B)She drew another window in the second rectangle with dimensions 8 x 8 = 64sq.ft. She also drew a door with dimensions 8 x 8 = 64 sq.ft. (C)She drew three windows in the third rectangle, each with dimensions of 10 x 5 = 50 sq.ft, for a total area of 50 x 3windows = 150sq.ft. (D)For the fourth rectangle, she did the same thing as to the third rectangle, which was also 150sq.ft. After she finished steps (A) throught (D), she added the total areas calculated, and came up with a total of 446 sq.ft. After she subtracted 446sq.ft. from 560sq.ft, she noticed that she had 114sq.ft. left to work with, so she added four squared windows to every rectangle , each with dimensions of 5 x 5 =25 sq.ft, for a total area of 25 x 4windows = 100sq.ft. Now, she subtracted 100sq.ft. from 114sq.ft to get 14sq.ft and she decided that she wanted to put another window in the first rectangle with dimensions of 2 x 7 = 14sq.ft.(14)
Details in the directions to keep students from collapsing the possible
solutions to a few easy ones may be performed as follows: since our tested
students only used squares to get the solution of the problem, this implies
that students have to be reminded that they must try solving this problem
with different window figures. They need to be informed that they can use
a variety of shapes (including composite geometrical figures) to solve the
problem.(15).
Shapes for the different geometrical figures: