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Lesson 13.3 Similar 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. - PowerPoint PPT Presentation
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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.
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.
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.
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
►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.
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.
►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