Lesson 13.3Similar Right Triangles

pp. 548-553

Lesson 13.3Similar Right Triangles

pp. 548-553

Objectives:1. To prove that the altitude to the

hypotenuse of a right triangle divides it into two right triangles, each similar to the original.

2. To define and apply geometric means.

3. To compute lengths of sides and related segments for right triangles by using proportions.

Objectives:1. To prove that the altitude to the

hypotenuse of a right triangle divides it into two right triangles, each similar to the original.

2. To define and apply geometric means.

3. To compute lengths of sides and related segments for right triangles by using proportions.

Theorem 13.4An altitude drawn from the right angle to the hypotenuse of a right triangle separates the original triangle into two similar triangles, each of which is similar to the original triangle.

Theorem 13.4An altitude drawn from the right angle to the hypotenuse of a right triangle separates the original triangle into two similar triangles, each of which is similar to the original triangle.

A

C

D

B

ADB ~ ADB ~ BDCBDCADB ~ ADB ~ ABCABCBDC ~ BDC ~ ABCABC

In the proportion , notice

that the denominator of one ratio is the same as the numerator of

the other ratio. When this happens, x is called the

geometric mean.

In the proportion , notice

that the denominator of one ratio is the same as the numerator of

the other ratio. When this happens, x is called the

geometric mean.

bb

xx

xx

aa==

For example, 8 is the geometric mean between 16 and 4 because

For example, 8 is the geometric mean between 16 and 4 because

4488

881616

== ..

2727xx

xx33

==

x2 = 81

x = 81

x = 9

x2 = 81

x = 81

x = 9

EXAMPLE 1 Find the geometric mean between 3 and 27.EXAMPLE 1 Find the geometric mean between 3 and 27.

EXAMPLE 2 Find the geometric mean between 6 and 9.EXAMPLE 2 Find the geometric mean between 6 and 9.

99xx

xx66

==

x2 = 54

x = 54

x = 3 6

x2 = 54

x = 54

x = 3 6

Practice: Find the geometric mean between 5 and 25.Practice: Find the geometric mean between 5 and 25.

2525xx

xx55

==

x2 = 125x2 = 125

5555xx ==125125xx ±±==

≈ 11.2≈ 11.2

Practice: Find the geometric mean between 12 and 20.Practice: Find the geometric mean between 12 and 20.

Theorem 13.5In a right triangle, the altitude to the hypotenuse cuts the hypotenuse into two segments. The length of the altitude is the geometric mean between the lengths of the two segments of the hypotenuse.

Theorem 13.5In a right triangle, the altitude to the hypotenuse cuts the hypotenuse into two segments. The length of the altitude is the geometric mean between the lengths of the two segments of the hypotenuse.

D

CBA a b

x

BCBCDBDB

DBDBABAB

oror,,bbxx

xxaa

====

Theorem 13.6In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that the length of a leg is the geometric mean between the hypotenuse and the segment of the hypotenuse adjacent to the leg.

Theorem 13.6In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that the length of a leg is the geometric mean between the hypotenuse and the segment of the hypotenuse adjacent to the leg.

D

CBA

ACACADAD

ADADABAB

==ACACDCDC

DCDCBCBC

==

EXAMPLE 3 Given the measurements in HIJ, find x, y, and z.EXAMPLE 3 Given the measurements in HIJ, find x, y, and z.

HH

JJII

yy

zz

xx

44

16161616xx

xx44

==

x2 = 64

x = 8

x2 = 64

x = 8

EXAMPLE 3 Given the measurements in HIJ, find x, y, and z.EXAMPLE 3 Given the measurements in HIJ, find x, y, and z.

2020yy

yy44

==

y2 = 80

y = 4 5

y2 = 80

y = 4 5

HH

JJII

yy

zz

xx

44

1616

2020zz

zz1616

==

z2 = 320

z = 8 5

z2 = 320

z = 8 5

EXAMPLE 3 Given the measurements in HIJ, find x, y, and z.EXAMPLE 3 Given the measurements in HIJ, find x, y, and z.

HH

JJII

yy

zz

xx

44

1616

Practice: Given: Right JKL with altitude to the hypotenuse, MK; LJ = 20, and MJ = 4, find KM.

Practice: Given: Right JKL with altitude to the hypotenuse, MK; LJ = 20, and MJ = 4, find KM.

K

J LM

Practice: Given: Right JKL with altitude to the hypotenuse, MK; MJ = 4, and KJ = 6, find LJ.

Practice: Given: Right JKL with altitude to the hypotenuse, MK; MJ = 4, and KJ = 6, find LJ.

K

J LM

Homeworkpp. 552-553Homeworkpp. 552-553

►A. ExercisesSolve each proportion; assume that x is positive.

3.

►A. ExercisesSolve each proportion; assume that x is positive.

3.xx44

99xx

==

►A. ExercisesSolve each proportion; assume that x is positive.

5.

►A. ExercisesSolve each proportion; assume that x is positive.

5.xx22

22x – 3x – 3

==

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.

7. AD = 15 units; DB = 5 units; find AC

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.

7. AD = 15 units; DB = 5 units; find AC

AA DD BB

CC

AA DD BB

CC

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.

9. AB = 32 units; DB = 6 units; find CD

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.

9. AB = 32 units; DB = 6 units; find CD

AA DD BB

CC

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.11. AD = 6 units; AB = 10 units; find CD

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.11. AD = 6 units; AB = 10 units; find CD

AA DD BB

CC

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.13. AD = 11 units; DB = 5 units; find AC

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.13. AD = 11 units; DB = 5 units; find AC

AA DD BB

CC

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.15. AD = 12 units; AB = 18 units; find

CB

►B. ExercisesGiven that ∆ABC is a right triangle and DC is an altitude to the hypotenuse, AB, find the length of the indicated sides.15. AD = 12 units; AB = 18 units; find

CB

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

23. Two circles with radii 3 and 6

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

23. Two circles with radii 3 and 6

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

24. Two rectangles: 6 by 9 and 8 by 12.

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

24. Two rectangles: 6 by 9 and 8 by 12.

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

25. Two rectangles: 6 by 8 and 16 by 18.

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

25. Two rectangles: 6 by 8 and 16 by 18.

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

26. Two regular tetrahedra with sides of length 9 and 6 respectively

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

26. Two regular tetrahedra with sides of length 9 and 6 respectively

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

27. Two squares with sides of length s and t respectively.

■ Cumulative Review

Which pairs of figures are similar? For each pair of similar figures, give the scale factor, k.

27. Two squares with sides of length s and t respectively.

Analytic Geometry

Slopes of Perpendicular Lines

Analytic Geometry

Slopes of Perpendicular Lines

Theorem

If two distinct nonvertical lines are perpendicular, then their slopes are negative reciprocals.

Theorem

If two distinct nonvertical lines are perpendicular, then their slopes are negative reciprocals.

l1l1 l2l2

X(x1, 0)X(x1, 0) Y(x3, 0)Y(x3, 0)

W(x2, 0)W(x2, 0)

Z(x2, y1)Z(x2, y1)

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

1. y = -4/3x + 5 with y-intercept (0, -8)

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

1. y = -4/3x + 5 with y-intercept (0, -8)

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

2. y = 2x – 1 and passing through(1, 4)

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

2. y = 2x – 1 and passing through(1, 4)

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

3. the line containing (2, 5) and (3, 4) at the first point.

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

3. the line containing (2, 5) and (3, 4) at the first point.

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

4. y = 1/2x + 5, if their point of

intersection occurs when x = 2

►Exercises

Give the equation of the line perpendicular to the line described and satisfying the given conditions.

4. y = 1/2x + 5, if their point of

intersection occurs when x = 2