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8.4 Similar Triangles

8.4 Similar Triangles. Geometry Mr. Parrish. Identify similar triangles. Use similar triangles in real-life problems such as using shadows to determine the height of the Great Pyramid. pp. 483-485 1, 4-7, 10, 12-26, 33-48, 55 not proof, 57 a-c. Objectives/Assignment.

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8.4 Similar Triangles

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  1. 8.4 Similar Triangles Geometry Mr. Parrish

  2. Identify similar triangles. Use similar triangles in real-life problems such as using shadows to determine the height of the Great Pyramid pp. 483-485 1, 4-7, 10, 12-26, 33-48, 55 not proof, 57 a-c Objectives/Assignment

  3. Identifying Similar Triangles • In this lesson, you will continue the study of similar polygons by looking at the properties of similar triangles.

  4. In the diagram, ∆BTW ~ ∆ETC. Write the statement of proportionality. Find mTEC. Find ET and BE. Ex. 1: Writing Proportionality Statements 34° 79°

  5. In the diagram, ∆BTW ~ ∆ETC. Write the statement of proportionality. Ex. 1: Writing Proportionality Statements 34° ET TC CE = = BT TW WB 79°

  6. In the diagram, ∆BTW ~ ∆ETC. Find mTEC. B  TEC, SO mTEC = 79° Ex. 1: Writing Proportionality Statements 34° 79°

  7. In the diagram, ∆BTW ~ ∆ETC. Find ET and BE. Ex. 1: Writing Proportionality Statements 34° CE ET Write proportion. = WB BT 3 ET Substitute values. = 12 20 3(20) ET Multiply each side by 20. = 79° 12 5 = ET Simplify. Because BE = BT – ET, BE = 20 – 5 = 15. So, ET is 5 units and BE is 15 units.

  8. If two angles of one triangle are congruent to the two angles of another triangle, then the two triangles are similar. If JKL  XYZ and KJL  YXZ, then ∆JKL ~ ∆XYZ. Postulate 25 Angle-Angle Similarity Postulate

  9. Color variations in the tourmaline crystal shown lie along the sides of isosceles triangles. In the triangles, each vertex measures 52°. Explain why the triangles are similar. Ex. 2: Proving that two triangles are similar

  10. Solution. Because the triangles are isosceles, you can determine that each base angle is 64°. Using the AA Similarity Postulate, you can conclude the triangles are similar. Ex. 2: Proving that two triangles are similar

  11. Use the properties of similar triangles to explain why any two points on a line can be used to calculate slope. Find the slope of the line using both pairs of points shown. Ex. 3: Why a Line Has Only One Slope

  12. By the AA Similarity Postulate, ∆BEC ~ ∆AFD, so the ratios of corresponding sides are the same. In particular, Ex. 3: Why a Line Has Only One Slope CE BE By a property of proportions, = DF AF CE DF = BE AF

  13. The slope of a line is the ratio of the change in y to the corresponding change in x. The ratios Ex. 3: Why a Line Has Only One Slope Represent the slopes of BC and AD, respectively. and CE BE DF AF

  14. Because the two slopes are equal, any two points on a line can be used to calculate its slope. You can verify this with specific values from the diagram. Ex. 3: Why a Line Has Only One Slope 3-0 3 = Slope of BC 4-2 2 6-(-3) 9 3 = = Slope of AD 6-0 6 2

  15. Aerial Photography. Low-level photos can be taken using a remote-controlled camera suspended from a blimp. You want to take an aerial photo that covers a ground of g of 50 meters. Use the proportion Ex. 4: Using Similar Triangles n f h f n = h g g To estimate the altitude h that the blimp should fly at to take the photo. In the proportion, use f = 8 cm and n = 3 cm. These two variables are determined by the type of camera used.

  16. Ex. 4: Using Similar Triangles f n = Write proportion. h g n f 8cm 3cm = Substitute values. h 50 m 3h = 400 h ≈ 133 Cross product property. h Divide each side by 3. The blimp should fly at an altitude of about 133 meters to take a photo that covers a ground distance of 50 meters. g

  17. Note: • In Lesson 8.3, you learned that the perimeters of similar polygons are in the same ratio as the lengths of the corresponding sides. This concept can be generalized as follows: If two polygons are similar, then the ratio of any two corresponding lengths (such as altitudes, medians, angle bisector segments, and diagonals) is equal to the scale factor of the similar polygons.

  18. Find the length of the altitude QS. Solution: Find the scale factor of ∆NQP to ∆TQR. Ex. 5: Using Scale Factors NP 12+12 24 3 = = = TR 8 + 8 16 2 Now, because the ratio of the lengths of the altitudes is equal to the scale factor, you can write the following equation: QM 3 = QS 2 Substitute 6 for QM and solve for QS to show that QS = 4

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