Pearson Unit 1 Topic 5: Relationships Within Triangles 5-8: Inequalities in Two Triangles Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007.

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Presentation transcript:

Pearson Unit 1 Topic 5: Relationships Within Triangles 5-8: Inequalities in Two Triangles Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (6)(D) Verify theorems about the relationships in triangles, including proof of the Pythagorean Theorem, the sum of interior angles, base angles of isosceles triangles, midsegments, and medians, and apply these relationships to solve problems. (1)(A) Apply mathematics to problems arising in everyday life, society, and the workplace. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (5)(D) Verify the Triangle Inequality theorem using constructions and apply the theorem to solve problems.

Example: 1 The diagram below shows a robotic arm in two different positions. In which position is the tip of the robotic arm closer to the base? Use the Hinge Theorem to justify your answer. The robotic arm is closer to the base in the diagram on the right because the length of the side opposite of the 40° angle is less than the length of the side opposite of the 60° angle.

Example: 2 Ship A and Ship B leave from the same point in the ocean. Ship A travels 150 miles due west, turns 65° towards north, and then travels another 100 miles. Ship B travels 150 miles due east, turns 70° toward south and then travels another 100 miles. Which ship is farther from the starting point? Explain. Ship B is farther away from the starting point because 70° > 65° so the side opposite of the 70° is longer than the side opposite of the 65°.

Example: 3 Compare m∡BAC and m∡DAC. Compare the side lengths in ∆ABC and ∆ADC. AB = AD AC = AC BC > DC By the Converse of the Hinge Theorem, mBAC > mDAC.

Example: 4 Compare EF and FG. Compare the sides and angles in ∆EFH angles in ∆GFH. mGHF = 180° – 82° = 98° EH = GH FH = FH mEHF > mGHF By the Hinge Theorem, EF < GF.

Example: 5 Find the range of values for k. Step 1 Compare the side lengths in ∆MLN and ∆PLN. LN = LN LM = LP MN > PN By the Converse of the Hinge Theorem, mMLN > mPLN. 5k – 12 < 38 Substitute the given values. k < 10 Add 12 to both sides and divide by 5.

Example: 5 continued Step 2 Since PLN is in a triangle, mPLN > 0°. 5k – 12 > 0 Substitute the given values. k  2.4 Add 12 to both sides and divide by 5. Step 3 Combine the inequalities k < 10 with k > 2.4: The range of values for k is 2.4 < k < 10.

Example: 6 John and Luke leave school at the same time. John rides his bike 3 blocks west and then 4 blocks north. Luke rides 4 blocks east and then 3 blocks at a bearing of N 10º E. Who is farther from school? Explain. Note: Like the example with Ship A and Ship B, a bearing of 10º E means it is 10º off of the vertical, so 90 + 10 = 100º. Luke is farther away from school because 100º is more than the right angle for John.

Example: 7 When the swing ride is at full speed, the chairs are farthest from the base of the swing tower. What can you conclude about the angles of the swings at full speed versus low speed? Explain. The ∡ of the swing at full speed is greater than the ∡ at low speed because the length of the triangle on the opposite side is the greatest at full swing.

Example: 8 A quilter cuts out and sews together two triangles. She tries to make ΔABD and ΔBDC equilateral triangles, but the actual angle measures are as shown. Assuming that the common side is the same length in both triangles, list the 5 segments in order from shortest to longest. Step 1: list the sides of ΔABD in order from shortest to longest. BD, AB, AD Step 2: repeat with sides of ΔBDC. DC, BC, BD. Step 3: Now list all sides in order from shortest to longest. DC, BC, BD (list once), AB, AD.