Saturday, February 28, 2015

Perimeter, Fractions, and Variables. How Do They Mix?

In my 7th grade math class, my classmates and I started a project. Based on our recent test scores, we each received a problem that was a bit higher than our level. In this blogpost I'm going to show you what my problem was, and show you how I solved it. 

My problem:

A triangle has a perimeter of 25 ½ meters. The shortest side is ⅔ the length of the middle side, and the longest side is 7/6 of the length of the middle side.



I had to solve the problem. The first thing I did, was draw a scaled diagram and I wrote and equation. Then I started to solve the problem. Down below you will see my process. Also, here is my final project. I made an explain-everything video.



Equation: (7/6 x X) + (⅔ x X) + X = 25 ½


= (7/6 x X) + (4/6 x X) + X = 25 ½
combining like terms:
7/6 + 4/6 + 6/6 (which we infer is X) = 17/6
= 17/6 = 25 ½ / 1
= 17/6 = 51/2
cross multiply
= 17X x 2 = 51 x 6
= 34X =306
= X = 306/34
x = 9

Small side = ⅔ of 9 = 6
Longer side = 7/6 of 9 = 7x9 = 63/6 = 10.5

Here is my explanation:
While I was trying to solve my problem, I realized that I should find a common denominator and transform X into a fraction. Since the common denominator is 6, and X is the ‘whole’ number, X in a fraction would 6/6. So 7/6 + 4/6 (⅔) + 6/6 = 17/6. This fraction has to equal to 25 ½. 25 ½ would be 51/2 in fraction. The next thing to do is to cross multiply. 17 x 2 = 34x and 6 x 51= 306. Lastly, to find x, you need to divide 306 by 34x, which is 9. X = 9.
Of course, now you need to find the unknown lengths. Simple. Tje shortest side is ⅔ of 9, which is 6, and then 7/6 of 9, for which you multiply 9 by 7, which is 63, and then divide that by 6, which is 10.5.
This simple procedure can be used for multiple questions. Though throughout my explanation I used my problem as an example.

Part of my problem was to come up with my own, similar to this one. Here is my problem. I used the same structure, proving my method works.
A triangle has a perimeter of 15 cm. The shortest side is 8/10 the length of the middle side, and the longest side is 6/5 the length of the middle side. What are the different lengths?


Equation: (6/5 • x) + (8/10 • x) + x = 15 cm


= (12/10 • x) + (8/10 • x) + 6/6 (x) = 25 ½
combining like terms:
12/10 + 8/10 + 10/10 = 30/10
= 30/10 = 15/1
cross multiply
= 30 x 1 = 10 x 15
= 30x = 150
= x = 150/30
x = 5


Small side = ⅘ of 5 = 4
Longer side = 6/5 of 5 = 6

As you can see, the method clearly worked. Additionally, with these smaller numbers you can see more easier what I did.

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