Current Problem of the Week
June 17-21, 1996
First we discussed the problem as a class, so everyone could voice their opinion or views on how to get an accurate answer. To get our minds working we all physically folded a piece of paper and thought about how to go about solving it. The first suggestion was to get a piece of paper with the same dimensions and fold it and measure the fold and then you would have the answer. But how accurate would this be? Well, it depends on how accurate the fold and measuring is, so we decided to work it out mathematically. Thinking about this method we figured out the length of the fold would be between 18 and 24cm.
To get us started, Mr. Vermont drew two parallel lines on either side of the fold which then left a square (or two right angle triangles) and two smaller rectangles. We knew that the longer side was 24cm and the shorter, 18cm. On a diagram (see above) one side of a rectangle was labelled 'a' and the other rectangle was 'b'. A diagonal line was drawn from corner to corner in one of the rectangles. We folded this line and Rodney, among other people, rightly discovered that this line was the same length as 'b'. This line also formed two more right angle triangles, one of which we were to use for the next part of the investigation.
For the next part we used Pythagoras' theorem; a^2 + b^2 = h^2. In this case it was 18^2 + a^2 = b^2. This gave us two rules for our investigation: b^2 - a^2 = 18^2 and a + b = 24. With these rules in mind we then began working by ourselves (or with partners) to find the solution. I began by using 'Guess, Check and Improve' to find out what 'a' and 'b' were. The following table shows the data I came up with.
The first Total was to determine whether or not the numbers I was investigating added up to 24 and the TOTAL was to see if the numbers, using the rule; b^2 - a^2 = 18^2 or 324, did in fact equal 324. By using this method I found out that 'a' = 5.25cm and 'b' = 18.75cm.
The next bit proved to be a bit easier as it involved certain steps and rules. To find the hypotenuse I needed to know 'c' and this was achieved by using the rule; 24 - 2a = c or because I knew 'a' this rule was 24 - 10.5 = 13.5cm. 'C' equals 13.5cm.
Using Pyagothoras' theorem again I was able to determine the length of the hypotenuse. 5.25^2 + 18.75^2 = c^2. This equation equalled 506.25. So c^2 = 506.25. Because this number was squared I had to find the square root of this figure. Using my calculator this was soon achieved and the answer was 22.5cm. This answer seemed reasonable because it fitted in with my first hypothesis that the answer would be between 18 and 24cm.
Because the original question was actually in inches with the rectangle having sides of 6" and 8" I had to convert my answer to inches. I noticed the numbers we used, 24 and 18, were triple the inches. That would mean the answer I got, 22.5cm, would just have to be divided by 3 to achieve the original answer. 22.5 / 3 = 7.5. The answer would make sense because it is between 6" and 8".
That means the answer to our question, using centimetres, is 22.5cm and the answer using inches is 7.5".
Erin McPartland
South Fremantle Senior High School
Perth, Western Australia