4.1.1 Classifying Triangles (continued)
Next, classify △ SPR . In the given figure, PS is divided into two parts, PT and TS , where PT = 5 and TS = 4. So, by the Segment Addition Postulate, PS = PT + TS = 5 + 4 = 9. The length of PR can be found in a similar way, and PR = 9 as well. So, since PS = 9 and PR = 9, the two sides of △ SPR are congruent. A triangle with at least two congruent sides is classified as isosceles. Therefore, △ SPR is an isosceles triangle. This figure has tick marks indicating congruency on AB and BC . When two segments are congruent, their lengths are equal. Therefore, AB = BC by the definition of congruent segments. In the figure, expressions are given for AB and BC . So, by the Substitution Property of Equality, 4 x − 3 = 3 x + 2. Solve this equation for x . Once the value of x is found, substitute that value into the expression for the length of each side of △ ABC and simplify. The result is the length of each side. AB = 4 x − 3 BC = 3 x + 2. In this example, a musical triangle is described as an equilateral triangle. Therefore, the musical triangle must have three congruent sides, or three sides with equal length. It is given that each side of the musical triangle is 3 inches long. Since a triangle has 3 sides, the length of steel used to make one musical triangle is 3 inches + 3 inches + 3 inches, or 9 inches. The full length of steel is 80 inches. So, divide 80 by 9 to find the maximum number of musical triangles that can be formed from the 80 inches of steel. The quotient of 80 and 9 is not a whole number. However, the number of musical triangles formed should be a whole number. Even though 8 and 8/9 is closer to 9 than it is to 8, the answer is 8 musical triangles because the extra 8/9 inches of steel cannot be used to form an additional musical triangle.
Example 3 Using Triangle Classification
Example 4 Music Application
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