A triangle has sides of length three and eight-ninths, seven and four-ninths, and nine and seven-ninths. Find the perimeter of the triangle.
First, we can sketch our triangle. Now we can label each side. We know to find the perimeter of a triangle, we have to add all the side lengths. The perimeter would be the side plus side plus side. For us, that means seven and four-ninths plus three and eight-ninths plus nine and seven-ninths.
To solve this problem, let’s add all the whole numbers together and then add all the fractions together. Seven plus three equals 10, plus nine equals 19. Now we need to add the fractional parts, four-ninths plus eight-ninths plus seven-ninths. The first step when we add fractions is to check and see if they have a common denominator. These fractions all have a denominator of nine. That means that all we do now is add their numerators. We add four plus eight plus seven. Four plus eight plus seven equals 19, and our denominator would stay the same. What we have now is a perimeter equal to 19 and nineteen-ninths.
But this fraction is an improper fraction; its numerator is larger than its denominator. This means we have some simplifying to do. Nine-ninths would be equal to one whole part. So we need to ask the question how many nine-ninths are in the fraction nineteen-ninths. We want to take out the whole number pieces. If I took nine-ninths away from nineteen-ninths, we would be left with ten-ninths, and our nine over nine turns into one whole piece. This means that another way to write nineteen-ninths would be to say one and 10 over nine.
But we have the same problem; 10 over nine is still an improper fraction. What would happen if we took nine over nine away from ten-ninths? Nine over nine would be one whole, and there would only be one out of nine left over. What we’re saying is that 19 over nine is equal to two whole parts and one over nine, one-ninth of a part. Remember that we had 19 plus 19 over nine. Instead of saying 19 plus 19 over nine, we’re going to say 19 plus two and one-ninth. 19 plus two equals 21, and we bring down our fraction piece that cannot be simplified any further.
So the perimeter of this triangle would be 21 and one-ninth units.