9.1 similar right triangles

9.1 Similar Right TrianglesGeometry

Mrs. Olaniran

Spring 2013

Harlan Academy

Standard 3: Students will understand geometric concepts

and applications.

Objectives/Assignment 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.

Today’s assignment: Postulates and Theorems, and Note book/Binder Check. I need to see notes from Friday before the quiz

Algebra Review for extra credit Go Khan Academy .org (yembal@aol.com is your coach week.

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.

P

R Q

S

U T

Activity: Investigating similar right triangles. Do in pairs or threes/fours1. Cut an index card along one of

its diagonals. 2. On one of the right triangles,

draw an altitude from the right angle to the hypotenuse. Cut along the altitude to form two right triangles.

3. You should now have three right triangles. Compare the triangles. What special property do they share? Explain.

4. Tape your group’s triangles to a piece of paper and place in labwork.

What did you discover? In the activity, you may have discovered the

next theorem. A plan for proving the theorem appears on page 460 and 461. You are asked to prove the theorem in Exercise 32 on page 466. It is important you defend these in class as that will be you open ended assessement.

Theorem 9.1 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. A B

C

D

∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD

A plan for proving thm. 9.1 is shown below: Given: ∆ABC is a right triangle; altitude CD is

drawn to hypotenuse AB. Prove: ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD Plan for proof: First prove that ∆CBD ~ ∆ABC. Each

triangle has a right triangle and each includes B. The triangles are similar by the AA Similarity Postulate. You can use similar reasoning to show that ∆ACD ~ ∆ABC. To show that ∆CBD ~ ∆ACD, begin by showing that ACD B because they are both complementary to DCB. Then you can use the AA Similarity Postulate.

A B

C

D

Ex. 1: 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.

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.

h3.1 m

Y

X W

h

5.5 m

Z

Y W

5.5 m

3.1 m

6.3 m

Z

X Y

∆XYW ~ ∆YZW ~ ∆XZY.

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.

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.

A

C

BD

C D

B

A D

C

A C

B

Write this down!

A

C

BD

C D

B

A D

C

A C

B

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= CD

AD

Shorter leg of ∆CBD.

Shorter leg of ∆ACD

Longer leg of ∆CBD.

Longer leg of ∆ACD.

Copy this down!

A

C

BD

C D

B

A D

C

A C

B

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= CD

AD

Hypotenuse of ∆ABC.

Hypotenuse of ∆CBD

Shorter leg of ∆ABC.

Shorter leg of ∆CBD.

Copy this down!

A

C

BD

C D

B

A D

C

A C

B

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= CD

AD

Hypotenuse of ∆ABC.

Hypotenuse of ∆ACD

Longer leg of ∆ABC.

Longer leg of ∆ACD.

Geometric Mean Theorems Theorem 9.2: 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

Theorem 9.3: 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.

A

C

BD

BD

CD= CD

AD

AB

CB= CB

DB

AB

AC= AC

AD

What does that mean?

6

x= x

3

5 + 2

y= y

2

36

x y 5

2

18 = x2

√18 = x

√9 ∙ √2 = x

3 √2 = x

14 = y2

7

y= y

2

√14 = y

Ex. 3: Using Indirect Measurement. MONORAIL TRACK. To

estimate the height of a monorail track, your friend holds a cardboard square at eye level. Your friend lines up the top edge of the square with the track and the bottom edge with the ground. You measure the distance from the ground to your friend’s eye and the distance from your friend to the track.

In the diagram, XY = h – 5.75 is the difference between the track height h and your friend’s eye level. Use Theorem 9.2 to write a proportion involving XY. Then you can solve for h.

Upcoming: There are quizzes every week to check for

understanding Be Prepared for activities as you cannot use

the formula Test should be Tuesday or Wednesday before

you go on Spring Break – don’t miss it. If you are leaving early somewhere, take it before you leave.