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Lesson 7.3 Using Similar Right Triangles Students need scissors, rulers, and note cards. Today, we are going to… …use geometric mean to solve problems.

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Presentation on theme: "Lesson 7.3 Using Similar Right Triangles Students need scissors, rulers, and note cards. Today, we are going to… …use geometric mean to solve problems."— Presentation transcript:

1 Lesson 7.3 Using Similar Right Triangles Students need scissors, rulers, and note cards. Today, we are going to… …use geometric mean to solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle

2 With a straight edge, draw one diagonal of the note card. Draw an altitude from one vertex of the note card to the diagonal. Cut the note card into three triangles by cutting along the segments.

3 C B A A D B DB C hypotenuse hyp short leg short long leg long Color code all 3 sides of all 3 triangles on the front and back. Label all right angles (front and back)

4 Arrange the small and medium triangles on top of the large triangle like this?

5 If the altitude is drawn to the hypotenuse of a right triangle, then… Theorem 7.5 … the three triangles formed are similar to each other.

6 BD CD AD = BD is a side of the medium  and a side of the small  A B C D

7 x m n x x = m n BD = CD AD

8 Theorem 7.6 …the altitude is the geometric mean of the two segments of the hypotenuse. D A C B x mn x x = m n

9 x 8 3 x x = 8 3 1. Find x. x ≈ 4.9

10 4 8 x 4 4 = 8 x 2. Find x. x = 2

11 CB CDCB CA = A B C D CB is a side of the large  and a side of the medium 

12 x m h x x = m h CB = CD CA

13 AB ADAB AC = A B C D AB is a side of the large  and a side of the small 

14 x n h x x = n h AB = AD AC

15 Theorem 7.7 …the leg of the large triangle is the geometric mean of the “adjacent leg” and the hypotenuse. D A C B x mn x x = m h y h y y = n h

16 x 9 14 x x = 9 3. Find x. x ≈ 11.2

17 x 4 10 x x = 4 4. Find x. x ≈ 6.3

18 5. Find x, y, z x y z 94 x x = 9 13 x ≈ 10.8 y y = 9 4 y = 6 z z = 4 13 z ≈ 7.2

19 3 3 = x 5 6. Find h. x = 1.8 h 4 3 5 x 1.8 3.2 h h = 1.8 h = 2.4

20 Day Two

21 Today, we are going to use geometric mean to solve problems involving similar right triangles formed by … …the altitude drawn to the hypotenuse of a right triangle & … the leg of the large triangle as the geometric mean of the “adjacent leg” and the hypotenuse.

22 Right Triangles and Geometric Mean We formed 3 triangles! What do these 3 triangles have in common? Let’s review what we did yesterday… take out your triangles.

23 Right Triangles and Geometric Mean Let’s consider this diagram A C B D

24 Right Triangles and Geometric Mean We’ll put the triangles up for reference (set up your triangles like this) A C B D A A C C C B B DD

25 Right Triangles and Geometric Mean Let’s label the sides and the angles a a a b b b c c h h h d d e e A C B D A A C C C B B DD

26 Right Triangles and Geometric Mean We can use similarity properties to set up proportions: a a a b b b c c h h h d d e e A C B D A A C C C B B DD

27 Right Triangles and Geometric Mean To conclude, a, b, and h, can all be written as the Geometric Mean of two segments. a b c h de

28 Right Triangles and Geometric Mean Putting it in words: The altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. –The length of this altitude is the geometric mean between the lengths of these segments. –The length of a leg of this triangle is the geometric mean between the segment of the hypotenuse adjacent to that leg and length of the hypotenuse.

29 Example x y z 5 20

30 Let’s practice…

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