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Lesson 6-5 explores the concepts of similar triangles and their corresponding proportional relationships, including perimeters, special segments, and angles.
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Transparency 6-5 5-Minute Check on Lesson 6-4 R • Refer to the figure • If QT = 5, TR = 4, and US = 6, find QU. • If TQ = x + 1, TR = x – 1, QU = 10 and QS = 15, find x. • Refer to the figure • If AB = 5, ED = 8, BC = 11, and DC = x – 2, find x so that BD // AE. • If AB = 4, BC = 7, ED = 5, and EC = 13.75, determine whetherBD // AE. • 5. Find the value of x + y in the figure? T 7.5 S Q U 3 B A C D 19.6 E Yes Standardized Test Practice: 5y – 6 3x – 2 2x + 1 2y + 3 B A B C D 10 4 6 8 Click the mouse button or press the Space Bar to display the answers.
Lesson 6-5 Parts of Similar Triangles
Objectives • Recognize and use proportional relationships of corresponding perimeters of similar triangles • Recognize and use proportional relationships of corresponding angle bisectors, altitudes, and medians of similar triangles
Vocabulary • None New
Theorems • If two triangles are similar then • The perimeters are proportional to the measures of corresponding sides • The measures of the corresponding altitudes are proportional to the measures of the corresponding sides • The measures of the corresponding angle bisectors of the triangles are proportional to the measures of the corresponding sides • The measures of the corresponding medians are proportional to the measures of the corresponding sides • Theorem 6.11: Angle Bisector Theorem: An angle bisector in a triangle separates the opposite side into segments that have the same ratio as the other two sides
Special Segments of Similar Triangles P If ∆PMN ~ ∆PRQ, then PM PN MN special segment ----- = ----- = ----- = ------------------------- AB AC BC special segment ratios of corresponding special segments = scaling factor (just like the sides) in similar triangles Example: PM 1 median PQ 1 ----- = --- ----------------- = --- AB 3 median AD 3 Special segments are altitudes, medians, angle and perpendicular bisectors M N Q A C B D
Similar Triangles -- Perimeters If ∆PMN ~ ∆PRQ, then Perimeter of ∆PMN PM PN MN ------------------------- = ----- = ----- = ----- Perimeter of ∆PRQ PR PQ RQ ratios of perimeters = scaling factor (just like the sides) P M N Q R
Angle Bisector Theorem - Ratios P If PN is an angle bisector of P, then the ratio of the divided opposite side, RQ, is the same as the ratio of the sides of P, PR and PQ PR RN ----- = ----- PQ NQ R N Q
C Let x represent the perimeter of The perimeter of Example 1a If ∆ABC~∆XYZ, AC=32, AB=16, BC=165, and XY=24, find the perimeter of ∆XYZ
Answer: The perimeter of Example 1a cont Proportional Perimeter Theorem Substitution Cross products Multiply. Divide each side by 16.
R Answer: Example 1b If ∆PNO~∆XQR, PN=6, XQ=20, QR=202, and RX = 20, find the perimeter of ∆PNO
Example 2a ∆ ABC ~∆ MNO and 3BC = NO. Find the ratio of the length of an altitude of ∆ABC to the length of an altitude of ∆MNO ∆ ABC and ∆ MNO are similar with a ratio of 1:3. (reverse of the numbers). According to Theorem 6.8, if two triangles are similar, then the measures of the corresponding altitudes are proportional to the measures of the corresponding sides. Answer: The ratio of the lengths of the altitudes is 1:3 or ⅓
Answer: Example 2b ∆EFG~∆MSY and 4EF = 5MS. Find the ratio of the length of a median of ∆EFG to the length of a median of ∆MSY.
K Example 3a In the figure, ∆EFG~ ∆JKL, ED is an altitude of ∆EFGand JI is an altitude of ∆JKL. Find x if EF=36, ED=18, and JK=56. Write a proportion. Cross products Divide each side by 36. Answer: Thus, JI = 28.
N Example 3b In the figure, ∆ABD ~ ∆MNP and AC is an altitude of ∆ABD and MO is an altitude of ∆MNP. Find x if AC=5, AB=7 and MO=12.5 Answer: 17.5
are medians of since and If two triangles are similar, then the measures of the corresponding medians are proportional to the measures of the corresponding sides. This leads to the proportion Example 4 The drawing below illustrates two poles supported by wires with ∆ABC~∆GED , AFCF, and FGGC DC. Find the height of the pole EC.
measures 40 ft. Also, since both measure 20 ft. Therefore, Example 4 cont Write a proportion. Cross products Simplify. Divide each side by 80. Answer: The height of the pole is 15 feet.
Transparency 6-6 5-Minute Check on Lesson 6-5 • Find the perimeter of the given triangle. • ∆UVW, if ∆UVW ~ ∆UVW, MN = 6, NP = 8, MP = 12, and UW = 15.6 • ∆ABC, if ∆ABC ~ ∆DEF, BC = 4.5, EF = 9.9, and the perimeter of ∆DEF is 40.04. • Find x. • 3. 4. • 5. Find NO, if ∆MNO ~ ∆RSQ. 33.8 18.2 x = 6 x = 7.375 x – 1 2x 8 9 9 8.5 x 12 Standardized Test Practice: N S 5.5 4.5 3 R M P T Q O A B C D 8.25 3.67 7 D 6.75 Click the mouse button or press the Space Bar to display the answers.
Find x and the perimeter of DEF, if ∆DEF ~ ∆ABC A B D C P E C x + 2 8 A R P x D T 6 S B 6 15 9 4 y M L N Find x D A A J x 8 3x + 1 B K x - 2 16 F 12 12 5x - 1 C L 6 E B C 12 Find x if PT is an angle bisector Find x, ED, and DB if ED = x – 3, CA = 20, EC = 16, and DB = x + 5 Is CD a midsegment (connects two midpoints)? Find y, if ∆ABC ~ ∆PNM
Find x 3x + 1 8 ------- = ---- 5x – 1 12 36x + 12 = 40x – 8 20 = 4x 5 = x Find x and the perimeter of DEF, if ∆DEF ~ ∆ABC A J 8 3x + 1 A B K 12 5x - 1 C L B D C P E C x + 2 8 A R P x D T 6 S B 6 15 9 4 y M L N x 6 --- = ---- 16 12 12x = 96 x = 8 D A x x - 2 16 F 12 6 E B P = (x – 2) + x + 6 = 2x + 4 = 2(8) + 4 = 20 C 12 Find x, ED, and DB if ED = x – 3, CA = 20, EC = 16, and DB = x + 5 Find x if PT is an angle bisector Is CD a midsegment (connects two midpoints)? 8 6 ------- = ---- x + 2 x 8x = 6x + 12 2x = 12 x = 6 Since AC ≠ EC, then NO ! Find y, if ∆ABC ~ ∆PNM 6 4 ------- = ---- 9 y 6y = 36 y = 6 x – 3 16 ------- = ---- x + 5 20 20x - 60 = 16x + 80 4x = 140 x = 35 ED = 32 DB = 40
Summary & Homework • Summary: • Similar triangles have perimeters proportional to the corresponding sides • Corresponding angle bisectors, medians, and altitudes of similar triangles have lengths in the same ratio as corresponding sides • Homework: • Day 1: pg 319-20: 3-7, 11-15 • Day 2: pg 320-21: 17-19, 22-24,