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Geometry Lesson 5 – 6 Inequalities in Two Triangles Objective: Apply the Hinge Theorem or its converse to make comparisons in two triangles. Prove triangle.

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Presentation on theme: "Geometry Lesson 5 – 6 Inequalities in Two Triangles Objective: Apply the Hinge Theorem or its converse to make comparisons in two triangles. Prove triangle."— Presentation transcript:

1 Geometry Lesson 5 – 6 Inequalities in Two Triangles Objective: Apply the Hinge Theorem or its converse to make comparisons in two triangles. Prove triangle relationships using the Hinge Theorem or its converse.

2 Inequalities in Two Triangles Hinge Theorem If two sides of a triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second triangle, then the third side of the first triangle is longer than the third side of the second triangle.

3 Converse of Hinge Theorem If two sides of a triangle are congruent to two sides of another triangle, and the third side in the first triangle is longer than the third side in the second triangle, then the include angle measure of the first triangle is greater than the included angle measure in the second.

4 Compare the given measures WX and XY WX < XY

5 Compare the given measures JK and MQ JK > MQ

6 AD and BD Compare the given measures AD > BD

7 Real World Two groups of snowmobilers leave from the same base camp. Group A goes 7.5 miles due west and then turns 35 degrees north of west and goes 5 miles. Group B goes 7.5 miles due east then turns 40 degrees north of east and goes 5 miles. At this point, which group is farther from the base camp? Explain. Draw a picture: Group A is farther from camp since the included angle is larger than Group B.

8 Find the range of possible values for x. 6x + 15 > 65 6x + 15 > 0 6x + 15 < 180 Angle has to be greater than 0, but less than x > 50 6x < 165 Don’t have to solve since we already said has to be greater than 65. Double check each time!

9 Find the range of possible values for x. 9a + 15 < 141 9a + 15 > 09a + 15 < 180 9a < 126 a < 14 9a > -15 Don’t have to solve since we already said has to be less than 141. Double check each time!

10 Find the range of possible values for x. 5x + 2 < 47 5x + 2 > 0 The length of a side must be positive. Do not need < 180 since 180 is for an angle not a side, and side has no limit on length. 5x < 45 5x > -2 x < 9

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12 Homework Pg – 8 all, 10 – 22 E, 38, 44 – 58 E


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