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9.1 Similar Right Triangles from Taos schools

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					9.1 Similar Right Triangles
Geometry
Mrs. Spitz
Spring 2006
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.
   Assignment: pp. 531-532 #1-34
   Today’s assignment: Ch. 9 Definitions, Ch. 9
    Postulates and Theorems, and Binder Check. I need
    to see notes from 7.1, 7.2 (except 5th period), 8.1,
    8.2, 8.3, 8.4, 8.5(except 2nd period), 8.6, and 8.7.
   Algebra Review for extra credit is due by the end of
    the week.
Proportions in right triangles
   In Lesson 8.4, you learned
    that two triangles are           S

    similar if two of their
    corresponding angles are
    congruent. For example P
    ∆PQR ~ ∆STU. Recall                  U   T

    that the corresponding
    side lengths of similar
    triangles are in
    proportion.


                                 R           Q
Activity: Investigating similar right
triangles. Do in pairs or threes
1.   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 528. You are asked to prove
    the theorem in Exercise 34 on page 533.
Theorem 9.1
   If the altitude is drawn
    to the hypotenuse of a              C
    right triangle, then the
    two triangles formed
    are similar to the
    original triangle and to
    each other.              A          D     B


      ∆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.             C




                                         A          D     B
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        Y

    sketch the three similar   3.1 m                       Z
    triangles so that the              h

    corresponding angles and X         W
    sides have the same                        6.3 m
    orientation. Mark the                  Z               5.5 m
    congruent angles. Notice
    that some sides appear in
    more than one triangle.     5.5 m
    For instance XY is the                     X           Y
    hypotenuse in ∆XYW and                         3.1 m
    the shorter leg in ∆XZY.               W
                               Y     h
      ∆XYW ~ ∆YZW ~ ∆XZY.
Solution for b.
    Use the fact that ∆XYW ~ ∆XZY to write a
     proportion.
     YW           XY      Corresponding side lengths are in
              =
     ZY           XZ      proportion.

       h          3.1
              =           Substitute values.
      5.5         6.3

       6.3h = 5.5(3.1)    Cross Product property

            h ≈ 2.7       Solve for unknown h.

    The height of the roof is about 2.7 meters.
Using a geometric mean to solve
problems
   In right ∆ABC,                         C


    altitude CD is drawn
    to the hypotenuse,         A

    forming two smaller
                                           D   B

                                       B           C
    right triangles that
    are similar to ∆ABC    C           D
                                                       B
    From Theorem 9.1,              A               D
    you know that ∆CBD
    ~ ∆ACD ~ ∆ABC.

                                       A               C
        Write this down!
                C
                                Notice that CD is the longer leg of
                                ∆CBD and the shorter leg of
                                ∆ACD. When you write a
                                proportion comparing the legs
    A
                D       B
            B
                    C
                            B   lengths of ∆CBD and ∆ACD,
C           D
                                you can see that CD is the
                    D
                                geometric mean of BD and AD.
    A


        A                   C Segment of hypo              ALTITUDE
                                                            Longer leg of ∆CBD.
                                Shorter leg of ∆CBD.

                                                BD     =    CD
                                                CD          AD
                                Shorter leg of ∆ACD         Longer leg of ∆ACD.
                                   ALTITUDE                OTHER Segment
                                                           of hypo
        Copy this down!
                 C

                                 Sides CB and AC also appear in
                                 more than one triangle. Their side
    A
                 D   B
                                 lengths are also geometric means, as
             B           C       shown by the proportions below:
                                                              Orig leg
             D                                                NEAREST
C

                         D
                             B      Hypotenuse original       SEG OF HYPO
         A                       Hyopotenuse of ∆ABC.        Shorter leg of ∆ABC.
                                              AB             CB
                                                        =
                                              CB             BD
                                  hypotenuse of ∆CBD         Shorter leg of ∆CBD.
                                   Orig leg
        A                    C     NEAREST SEG
                                                            Segment of hypo
                                   OF HYPO
        Copy this down!
                 C
                                     Sides CB and AC also appear in
                                     more than one triangle. Their
    A
                                     side lengths are also geometric
                 D   B

                         C           means, as shown by the
             B
                                     proportions below:
             D                                              Orig leg NEAREST
C

                         D
                             B    Hypotenuse original       SEG OF HYPO
         A                       Hypotenuse of ∆ABC.         Longer leg of ∆ABC.
                                                AB           AC
                                                        =
                                                AC           AD
                                 Hypotenuse of ∆ACD          Longer leg of ∆ACD.


        A                    COrig leg NEAREST
                              SEG OF HYPO
                                                               SEG OF HYPO
    Geometric Mean Theorems                               C

   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          A
                                                          D    B
    altitude is the geometric mean of the        BD       CD
    lengths of the two segments                       =
                                                 CD       AD
   Theorem 9.3: In a right triangle, the
    altitude from the right angle to the         AB       CB
    hypotenuse divides the hypotenuse into            =
    two segments. The length of each leg         CB       DB
    of the right triangle is the geometric
    mean of the lengths of the hypotenuse        AB   =   AC
    and the segment of the hypotenuse that       AC       AD
    is adjacent to the leg.
What does that mean?
                               2

              x        y                 5

      6           3

 6            x        5+2                   y
       =                             =
 x            3            y                 2
   18 = x2                 7                 y
                                     =
  √18 = x                  y                 2
                                   14 = y2
√9 ∙ √2 = x
  3 √2 = x                     √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:
   March 23 is the last day of the Third Quarter.
    Grades are due soon after.
   There are quizzes
   Test should be

				
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