final exam key concepts. vertical angles find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

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Final Exam Key Concepts

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Page 1: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Final Exam

Key Concepts

Page 2: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Vertical Angles

• Find the value of “x”.

(2x + 32)68

68 = 2x + 32

36 = 2x

18 = x

Page 3: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Segments and Lengths• In the diagram, and S is the

midpoint of . QR = 4, and ST = 5. Find the following values.

1. RS =

2. PR =

3. PQ =

PR RTRT

5

P Q R S T10

6

Page 4: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Parallel Lines

• Find the missing values.

z

y

x

110

70o

70o

110o

Page 5: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Straight Line is 180o

• Find the difference between the larger angle and the smaller angle.

(4x + 19)

(3x)3x + (4x + 19) = 180

7x + 19 = 180

7x =161

x =23

3(23) = 69 and 4(23) + 19 = 111

111 – 69 = 42

Page 6: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Angles of Triangle = 180o

• The angles of a triangle are in the ratio of 1:2:3. Find the measure of each angle.

1x + 2x + 3x = 180

6x = 180

x = 30

30o, 60o, 90o

Page 7: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Exterior Angle Theorem

• Find the missing value.

x

Note: Figure not drawn to scale.

35

25

35o + 25o = 60o

60o

Page 8: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Angles and Sides of a Triangle

• Remember, largest angle is OPPOSITE of longest side….

• And smallest angle is OPPOSITE of smallest side.

45

67

68

B

C

A

BC

AB

Page 9: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

TRIANGLE INEQUALITY THEOREM

• The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

• Is it possible for a triangle to have the following lengths?

3, 6, 8 6 + 3 = 9 > 8 YES

Page 10: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Three Theorems about Triangles

1) If c2 = a2 + b2, then the triangle is a right triangle.

2) If c2 < a2 + b2, then the triangle is an acute triangle.

3) If c2 > a2 + b2, then the triangle is an obtuse triangle.

12

Page 11: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

RIGHT TRIANGLES

Page 12: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Special Right Triangles

Page 13: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Pythagorean Theorem

62 + 72 = x2

36 + 49 = x2

85 = x2

85x

Page 14: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Pythagorean Triples

3, 4, 5

5, 12, 13

8, 15, 17

7, 24, 25

Page 15: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Quadrilaterals

Page 16: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Quadrilaterals

• Complete the worksheet.

Page 17: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

5 Ways to Prove Parallelograms

1. Both pairs of opposite sides are parallel

2. Both pairs of opposite sides are congruent

3. Both pairs of opposite angles are congruent

4. One pair of opposite sides congruent and parallel

5. Diagonals bisect each other.

Page 18: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

INTERIOR MEASURES

SUM of the INTERIOR

Measures of Any Polygon

(n – 2)180o

(4-2)180o = 360o (8-2)180o = 1080o

Page 19: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

EXTERIOR ANGLES

Sum of the measures of the EXTERIOR angles of any polygon

= 360o

e s

r

o

h

p

a

r

c

h + o + r + s + e = 360oc + r + a + p = 360o

Page 20: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

REGULAR POLYGON

A polygon that is both equilateral and equiangular.

Page 21: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Areas

2

1 2

1 2

AREA

Rectangles =

Squares =

Parallelogram

1Triangles =

21

Rhombus= 2

1Regular Polygon =

21

Trapezoids =2

bh

s

bh

bh

d d

ap

h b b

Page 22: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Circles

2

2

CIRCLES

Area of Circle =

Circumference of Circle = 2

Arc length = 2360

Area of a sector = 360

r

r

xr

xr

Page 23: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Angles in a Circle

central angles =

1inscribed angles =

21

interior angles = ( )21

exterior angles = ( )2

arc

arc

big small

big small

80o

50o

50o

30o

Page 24: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Lengths in a Circle

12(9) = 18x

108 = 18x

6 = x

Page 25: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Lengths in a Circle

3(8) = 2(12)

24 = 24

Page 26: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

Lengths in a Circle

122 = x(x+12+x)

144 = 2x2 + 12x

x = 8

Page 27: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x
Page 28: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x
Page 29: Final Exam Key Concepts. Vertical Angles Find the value of “x”. 68 = 2x + 32 36 = 2x 18 = x

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