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Comparing Measures of a Triangle There is a relationship between the positions of the longest and shortest sides of a triangle and the positions of its angles. largest angle shortest side longest side smallest angle Comparing Measures of a Triangle There is a relationship between the positions of the longest and shortest sides of a triangle and the positions of its angles. SIDE / ANGLE COMPARISON THEOREMS Theorem 5.10 If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side. Theorem 5.11 If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle. B 5 3 A C m A > m C D 60˚ F 40˚ E EF > DF Writing Measurements in Order from Least to Greatest Write the measurements of the triangles in order from least to greatest. J 100˚ H 45˚ 35˚ G SOLUTION mG < m H < mJ JH < JG < GH Writing Measurements in Order from Least to Greatest Write the measurements of the triangles in order from least to greatest. J 8 100˚ H 45˚ 5 35˚ G SOLUTION mG < m H < mJ JH < JG < GH R Q 7 P SOLUTION QP < PR < QR m R < mQ < m P Comparing Measurements of a Triangle Theorem 5.10 can be proved using the diagram shown below. Given AC > AB Prove mABC > mC A 2 Paragraph Proof B D 1 3 Use the Ruler Postulate to locate a point D on AC such that DA = BA. Then draw the segment BD. In the isosceles triangle ABD, 1 2. Because mABC = m1 + m3, it follows that mABC > m1. Substituting m2 for m1 produces mABC > m 2. Because m2 = m3 + mC, m2 > mC. Finally, because mABC > m2 and m2 > mC, you can conclude that mABC > mC. C Comparing Measurements of a Triangle The proof of Theorem 5.10 used an exterior angle and showed that its measure is the sum of the measures of two nonadjacent interior angles. Then the measure of the exterior angle must be greater than the measure of either nonadjacent interior angle. THEOREM Theorem 5.12 Exterior Angle Inequality A The measure of an exterior angle of a triangle is greater than the measure of either of the two nonadjacent interior angles. m1 > mA and m1 > mB 1 C B Using the Triangle Inequality Not every group of three segments can be used to form a triangle. The lengths of the segments must fit a certain relationship. Construct a triangle with the given group of side lengths, if possible. 2 cm, 2 cm, 5 cm SOLUTION Try drawing triangles with the given side lengths. Impossible Using the Triangle Inequality Not every group of three segments can be used to form a triangle. The lengths of the segments must fit a certain relationship. Construct a triangle with the given group of side lengths, if possible. 2 cm, 2 cm, 5 cm SOLUTION 3 cm, 2 cm, 5 cm Try drawing triangles with the given side lengths. Impossible Impossible Using the Triangle Inequality Not every group of three segments can be used to form a triangle. The lengths of the segments must fit a certain relationship. Construct a triangle with the given group of side lengths, if possible. 2 cm, 2 cm, 5 cm SOLUTION 3 cm, 2 cm, 5 cm 4 cm, 2 cm, 5 cm Try drawing triangles with the given side lengths. Impossible Impossible Only the last group is possible. The sum of the first and second lengths must be greater than the third length. Using the Triangle Inequality The result of the previous example is summarized as Theorem 5.13. THEOREM Theorem 5.13 Triangle Inequality A The sum of the lengths of any two sides of a triangle is greater than the length of the third side. AB + BC > AC AC + BC > AB AB + BC > BC C B Finding Possible Side Lengths A triangle has one side of 10 centimeters and another of 14 centimeters. Describe the possible lengths of the third side. SOLUTION Let x represent the length of the third side. Using the Triangle Inequality, you can write and solve inequalities. x + 10 > 14 x > 4 10 + 14 > x 24 > x So, the length of the third side must be greater than 4 centimeters and less than 24 centimeters.

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