50 likes | 191 Views
2 and 5 are corresponding angles formed by AB and XY cut by transversal BC . Because AB || XY , you can conclude that 2 5 by the Corresponding Angles Postulate. Thus, m 2 = m 5. Inequalities in Triangles. LESSON 5-5. Additional Examples. Quick Check.
E N D
2 and 5 are corresponding angles formed by AB and XY cut by transversal BC. Because AB || XY, you can conclude that 2 5 by the Corresponding Angles Postulate. Thus, m2 = m5. Inequalities in Triangles LESSON 5-5 Additional Examples Quick Check Explain why m4 > m5. 4 is an exterior angle of XYC. The Corollary to the Exterior Angle Theorem states that the measure of an exterior angle of a triangle is greater than the measure of each of its remote interior angles. The remote interior angles are 2 and 3, so m4 > m2. Substituting m5 for m2 in the inequality m4 > m2 produces the inequality m4 > m5.
In RGY, RG = 14, GY = 12, and RY = 20. List the angles from largest to smallest. Theorem 5-10 states If two sides of a triangle are not congruent, then the larger angle lies opposite the longer side. No two sides of RGY are congruent, so the larger angle lies opposite the longer side. Find the angle opposite each side. The longest side is 20. The opposite angle, G, is largest. The shortest side is 12. The opposite angle, R, is smallest. From largest to smallest, the angles are G, Y, R. Inequalities in Triangles LESSON 5-5 Additional Examples Quick Check
In ABC, C is a right angle. Which is the longest side? Theorem 5-11 states If two angles of a triangle are not congruent, then the longer side lies opposite the larger angle. Find the largest angle. m A + m B + m C = 180 by the Triangle Angle-Sum Theorem. m A + m B + 90 = 180 because the measure of a right angle is 90, so m A + m B = 90. m A < m C and m B < m C by the Comparison Property of Inequality, so C is the largest angle. Because AB is opposite C, the longest side in ABC is AB . Inequalities in Triangles LESSON 5-5 Additional Examples Quick Check
a. 2 + 2 > 4 No. The sum of 2 and 2 is not greater than 4, contradicting Theorem 5-12. Inequalities in Triangles LESSON 5-5 Additional Examples Can a triangle have sides with the given lengths? Explain. a. 2 cm, 2 cm, 4 cm b. 8 in., 15 in., 12 in. According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. b. 8 + 15 > 12 8 + 12 > 15 15 + 12 > 8 Yes. The sum of any two lengths is greater than the third length. Quick Check
FH + 9 > 17 FH + 17 > 9 9 + 17 > FH The length of FH must be longer than 8 m and shorter than 26 m. Inequalities in Triangles LESSON 5-5 Additional Examples In FGH, FG = 9 m and GH = 17 m. Describe the possible lengths of FH. The Triangle Inequality Theorem states the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Solve three inequalities. 26 > FH This can be written “FH < 26.” FH > 8 This can be written “8 < FH.” FH > –8 Quick Check