SOLVING APPLIED PROBLEMS

Example: Working with Fractions

PROBLEMAdam wants to fence a triangular shaped yard. He wants to determine how many feet of fencing he will need. The sides of his yard measure as shown in the diagram below. How many feet of fencing should Adam buy?

7 ½ ft10 3/4 ft

13 1/8 ft

SOLUTION

Note: To find the perimeter of a triangle, we need to add the lengths of all sides of the triangle.

In order to find out how many feet of fencing should be bought, we need to find out the distance around the backyard. This is the same as finding the distance around

the triangle, or finding the perimeter of the triangle.

7 ½ ft

13 1/8 ft

10 3/4 ft

The perimeter of the triangle = sum of all sides of the triangle

Based on the diagram we have the following: Length of first side = 7 ½ ftLength of second side = 13 1/8 ft Length of third side = 10 ¾ ft.

4

310

8

313

2

17

Notice that we have fractions with different denominators

Since the denominators for the fractions are different, we need to first find the LCD. The denominators are 2, 4, and 8. The least common denominator is 8. So we can rewrite the fractions so that they will all have a denominator of 8.

8

6

4

3

8

3

8

3

8

4

2

1

So to find the sum of 7 ½ ft ,13 1/8 ft , and 10 ¾ ft, we add the whole number part and then the fractional part.

8

610

8

313

8

47

4

310

8

313

2

17

8

6

8

3

8

4)10137(

8

531

8

5130

8

1330

Therefore, the perimeter of the triangle is 31 5/8 feet. And so Adam will have to purchase 31 5/8 ft of fencing for the yard.

Solving applied problems involving fractions Open University Malaysia is licensed under a Creative Commons Attribution-Noncommercial 3.0 Unported License. Based on a work at http://oer.oum.edu.my. Permissions beyond the scope of this license may be available at http://www.oum.edu.my.