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Key Learning  Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle.  Use a geometric mean.

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Presentation on theme: "Key Learning  Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle.  Use a geometric mean."— Presentation transcript:

1 Key Learning  Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle.  Use a geometric mean to solve problems such as estimating a climbing distance.

2 Proportions in right triangles  In Lesson 8.4, you learned that two triangles are similar if two of their corresponding angles are congruent. For example ∆PQR ~ ∆STU. Recall that the corresponding side lengths of similar triangles are in proportion.

3  If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD

4 Finding the Height of a Roof  Roof Height. A roof has a cross section that is a right angle. The diagram shows the approximate dimensions of this cross section.  A. Identify the similar triangles.  B. Find the height h of the roof.

5 Solution:  You may find it helpful to sketch the three similar triangles so that the corresponding angles and sides have the same orientation. Mark the congruent angles. Notice that some sides appear in more than one triangle. For instance XY is the hypotenuse in ∆XYW and the shorter leg in ∆XZY.  ∆XYW ~ ∆YZW ~ ∆XZY.

6 Solution for b.  Use the fact that ∆XYW ~ ∆XZY to write a proportion. YW ZY = XY XZ h 5.5 = 3.1 6.3 6.3h = 5.5(3.1) h ≈ 2.7  The height of the roof is about 2.7 meters. Corresponding side lengths are in proportion. Substitute values. Cross Product property Solve for unknown h.

7 Using a geometric mean to solve problems  In right ∆ABC, altitude CD is drawn to the hypotenuse, forming two smaller right triangles that are similar to ∆ABC From Theorem 9.1, you know that ∆CBD ~ ∆ACD ~ ∆ABC.

8 Notice that CD is the longer leg of ∆CBD and the shorter leg of ∆ACD. When you write a proportion comparing the legs lengths of ∆CBD and ∆ACD, you can see that CD is the geometric mean of BD and AD. BD CD = AD Shorter leg of ∆CBD. Shorter leg of ∆ACD Longer leg of ∆CBD. Longer leg of ∆ACD.

9 Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below: BD CD = AD Hypotenuse of ∆ABC. Hypotenuse of ∆CBD Shorter leg of ∆ABC. Shorter leg of ∆CBD.

10 Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below: BD CD = AD Hypotenuse of ∆ABC. Hypotenuse of ∆ACD Longer leg of ∆ABC. Longer leg of ∆ACD.

11 Geometric Mean Theorems  In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments  In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. BD CD = AD AB CB = DB AB AC = AD

12 What does that mean? 6 x = x 3 5 + 2 y = y 2 18 = x 2 √18 = x √9 ∙ √2 = x 3 √2 = x 14 = y 2 7 y = y 2 √14 = y


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