Download A Guide for Parents Chapter 11

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CHAPTER
11
Introduction to Geometry
Content Summary
Chapter 11 is a preview of geometry. Even so, in some places it uses linear equations.
Students also focus on radical expressions and operations with radicals.
Synthetic Geometry
The original study of geometry is now called synthetic to distinguish it from the
analytic geometry of coordinate systems, which developed much later. In the area of
synthetic geometry, the book focuses on the Pythagorean Theorem and on similar
figures—figures that are stretches or shrinks of each other by the same factor in
every direction.
Analytic Geometry
Analytic geometry is the geometry of coordinate graphs, which use algebraic
equations to represent geometric figures. Students have been working with analytic
geometry all through this course. They’ve already seen how parallel lines have the
same slope, and how that slope appears in the equations of the lines. In this chapter
they see how the slopes and equations of perpendicular lines relate. They also see
how to find the coordinates of midpoints of line segments. In considering how the
Pythagorean Theorem translates into coordinate geometry, students learn how to
work with square roots.
18
3
3
20
2
4
Trigonometry
When a geometric figure is stretched or shrunk uniformly, all angles retain their
measures and all side lengths are multiplied by the same amount; therefore, the ratio
of one side length to another remains the same. This stretching and shrinking
produces similar figures, which have equal corresponding angles and proportional
corresponding side lengths. The study of relationships between sides and angles of
similar right triangles is part of trigonometry. In particular, in similar right triangles,
the ratio of, for example, the length of the side opposite a particular acute angle to
the length of the hypotenuse is the same, no matter what the enlargement. Each
acute angle of a right triangle has several of these ratios associated with it. They’re
called trigonometric ratios.
The book considers three such ratios: the sine, the cosine, and the tangent. Students
use calculators to find values of these ratios for various angles and, in reverse, angles
that correspond to various given ratios. They thus preview the deeper consideration
of trigonometry found in Discovering Advanced Algebra.
(continued)
©2007 Key Curriculum Press
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Chapter 11 • Introduction to Geometry (continued)
Trigonometric Functions
For acute angle A in a right angle, the trigonometric functions are
length of opposite leg
sine of angle A length of hypotenuse
or
o
sin A h
length of adjacent leg
cosine of angle A length of hypotenuse
or
a
cos A h
length of opposite leg
tangent of angle A length of adjacent leg or
o
tan A a
h
A
o
a
Summary Problem
You and your student might revisit this problem, adapted from Exercise 7 in Lesson
11.1, several times while working through the chapter.
What can you say about the quadrilateral with vertices (5, 0), (1, 4), (6, 3), and
(3, 3)?
Questions you might ask in your role as student to your student include:
●
●
●
●
●
●
●
Do any sides look parallel?
Do any sides appear perpendicular?
Can you confirm your conjectures?
What are the equations of the four lines containing the sides of the
quadrilateral?
Do the diagonals meet at their midpoints?
Can you find the lengths of the diagonals?
Can you find the angle measures?
Sample Answers
Two of the sides are parallel y 1 23x and 3y 10 2x both have slope 23,
and a third side is perpendicular to them 2y 15 3x with slope 32. The
fourth side is contained in the line with equation 5y 21 x. The diagonals lie on
the lines with equations y 74(x 1) 4 and y 131 (x 5). Graphing and tracing
indicates that the diagonals meet at about (0.6, 1.2), which is not either of the
midpoints, (1, 0.5) or (0.5, 1.5). The diagonals have lengths 65 and 130 . If
you make a right triangle by dropping a perpendicular from the vertex at (1, 4) to
the parallel side opposite, you can find that the sine of the angle at vertex (6, 3)
1
1
13
x to find that the angle
is . Solve the equation sin1 2
2
26
is 45°. Therefore, the angle at (1, 4) will be 135°.
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Chapter 11 • Review Exercises
Name
Period
Date
1. (Lessons 11.1–11.3, 11.6) Plot the points A(0, 2), B(3, 1), and C(3, 4).
.
a. Find the equation of the line through B that is parallel to AC
.
b. Find the midpoint of AB
c. Find the equation of the line containing the median of the triangle
.
through C, and show that it is the perpendicular bisector of AB
d. Find the length of each side.
e. What kind of triangle is ABC?
f. Find the area of triangle ABC.
2. (Lesson 11.5) Rewrite each expression with as few square root symbols as
possible, and no parentheses. Your final result should not have any
perfect-square factors under a radical.
a. 26
32
8
21
2
3
c. b. 53
212
3. (Lesson 11.7) The two triangles shown below are similar. Write a
proportion and solve it to find w.
w
5
4
3
4. (Lessons 11.4, 11.8) Use the Pythagorean Theorem to find x. Then find
the following ratios:
A
5
C
13
x
a. tan A
©2007 Key Curriculum Press
B
b. sin B
c. cos B
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SOLUTIONS TO CHAPTER 11 REVIEW EXERCISES
1.
5
2. a. 2
6 3
2 2 3 6 2
C
B
5
2 3
6•2
5
A
612
643
5
has slope 2, so write the equation
a. The segment AC
of the line with slope 2 that passes through the
point (3, 1). In point-slope form, this is
y 1 2(x 3), or y 2(x 3) 1.
is
b. The x-coordinate of the midpoint of AB
0 3
1.5, and the y-coordinate is
2
2 1
0.5. The midpoint is (1.5, 0.5).
2
c. Find the equation of the line through the points
0.5 4
(3, 4) and (1.5, 0.5). The slope is 1.5 3 4.5
1, so the equation is y 4 1(x 3), or
4.5
is 1 and the slope of
y x 1. The slope of AB
the median is 1, so the product of the slopes is 1.
Therefore the two lines are perpendicular. The line
y x 1 is the perpendicular bisector of AB
and it passes
because it is perpendicular to AB
.
through the midpoint of AB
d. (Lesson 11.6) Use the distance formula
d x 12 y 12 to calculate each
x 2 y 2 length. For example:
AB (3
0)2 [1 (2)]
2
2
2
AB (3) (3)
AB 18 , or 3
2
For help in changing 18 to 3
2 , look ahead to
Exercise 2.
Compute the other two lengths in a similar
fashion. The lengths are
AC 45 , or 3
5 ; and BC 45 , or 3
5.
e. Triangle ABC is a isosceles because AC BC.
f. Draw a rectangle around triangle ABC as shown on
the graph for 1a. The rectangle has area 36. The
and AC
have
right triangles with hypotenuses BC
areas 0.5(3)(6), or 9 square units. The smaller right
triangle has area 0.5(3)(3), or 4.5 square units.
Subtract the areas of the triangles from the area of
the rectangle: 36 (9 + 9 + 4.5) 13.5. The area of
triangle ABC is 13.5 square units.
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6 23
123
b. 5
3 2
12 5
3 2
4 3
Multiplication
of radical
expressions.
Multiply.
4 is a perfectsquare factor
of 12.
Take the
square root.
Multiply.
4 is a perfectsquare factor
of 12.
53 43
Take the
square root
and multiply.
(5 4)3
Addition of
radical expressions.
3
2
18
292
c. 3
2
32
2 • 3
2
3
2
2
Commutative
property of
multiplication.
Subtract.
9 is a perfectsquare factor
of 18.
Take the
square root.
Reduce.
3. Proportions may vary.
w 53
Proportion to be solved.
4
3
w
8
4 4 4 3
Multiply both sides by 4.
32
2
w 3, or 103
Multiply.
4.
52 x 2 132 Equation to be solved.
25 x 2 169 Square each term.
x 2 144 Subtract 25 from both sides.
x 12
Take the square root of 144.
Only the positive square root is a solution, because x
is the length of a segment.
opposite
12
a. tan A adjacent 5
opposite
5
b. sin B hypotenuse 13
adjacent
12
c. cos B hypotenuse 13
©2007 Key Curriculum Press