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Name ______________________
Important Note: the funny-looking symbol appearing after the 180 (and many other numbers) below is
supposed to be a degree symbol!!
PROPERTIES OF ANGLES, LINES, AND TRIANGLES
Parallel lines
Triangles
9
1
11
2
3 4
8
7
6
10
• corresponding angles are equal:
m1  m3
• m7  m8  m9  180Þ
• m6  m8  m9
m2  m3
• m2  m4  180Þ
(exterior angle = sum remote interior angles)
• m10  m11  90Þ
(complementary angles)
• alternate interior angles are equal:
Also shown in the above figures:
• vertical angles are equal: m1  m2
• linear pairs are supplementary: m3  m4  180Þ
and m6  m7  180Þ
In addition, an isosceles triangle, ∆ABC, has
BA=BC and mA  mC . An equilateral
triangle, ∆GFH, has GF = FH = HG and
mG  mF  mH  60Þ.
F
B
A
C
G
H
Example 1
Solve for x.
There are a number of relationships in this diagram. First, 1 and the 127˚ angle are
supplementary, so we know that m1  127Þ 180Þ so m1 = 53˚. Using the same
idea, m2 = 47˚. Next, m3  53Þ 47Þ 180Þ, so m3 = 80˚. Because angle 3 forms
a vertical pair with the angle marked
7x + 3˚, 80˚ = 7x + 3˚, so x = 11˚.
7x+3Þ
3
1
127Þ
2 133Þ
Example 2
Find the measure of the acute alternate interior angles.
Parallel lines mean that alternate interior angles are equal, so
5x  28Þ 2x  46Þ  3x  18Þ x  6Þ. Use either algebraic angle measure:
2 6Þ  46Þ 58Þ for the measure of the acute angle.
GEOMETRY Connections
5x+28Þ
2x + 46Þ
1
Use the geometric properties and theorems you have learned to solve for x in each diagram and write the
property or theorem you use in each case. You must show the equation used.
1.
2.
3.
125Þ
60Þ
68Þ
5xÞ
5x + 12Þ
xÞ
75Þ
4.
5.
142Þ
6.
xÞ
38Þ
20x + 2Þ
30Þ
23Þ
19x+3Þ
7.
8.
9.
5x + 36Þ
8x–
in.
– 18
12x
8x +
12 in
.
18Þ
5x
9xÞ
10.
11.
3x + 20Þ
EC:
45Þ
2x + 5Þ
+3
cm
cm
12.
40Þ
5x – 18Þ
18
13x + 2Þ
15x – 2Þ
58Þ
Name ______________________
AREA
AREA is the number of square units in a flat region. The formulas to calculate the area of these polygons are:
RECTANGLE PARALLELOGRAM
TRAPEZOID
b1
h
h
b
A  bh
h
h
A
1
2
h
b
b2
b
A  bh
TRIANGLE
b
1
 b2 h
b
1
2
A  bh
Note that the legs of any right triangle form a base and a height for the triangle.
The area of a more complicated figure may be found by breaking it into smaller regions of the types shown
above, calculating each area, and finding the sum.
Example 1
Find the area of each figure. All lengths are centimeters.
a)
b)
23
4
81
12
A
A  bh  8123  1863cm 2
c)
1
2
124   24 cm2
d)
42
10
108
A
1
2
10842  2268cm
8
21
2
A  218  168 cm 2
Note that 10 is a side of the parallelogram, not the height.
e)
14
12
34
A
1
2
14  34 12  12 4812  288cm2
Example 2 Find the area of the shaded region.
The area of the shaded region is the area of the
triangle minus the area of the rectangle.
triangle:
A  12 710  35cm2
rectangle:
A  5 4   20 cm 2
shaded region:
A  35  20  15 cm2
GEOMETRY Connections
7 cm.
4 cm.
5 cm.
10 cm.
3
Find the area of the following triangles, parallelograms and trapezoids. Pictures are not drawn to scale. You must
write the formulas used (eg: A=bh) and then show what numbers you plugged in (A=2x3=6).
1.
2.
3.
26
4.
10
13
10
26
20
3
9
8
6
Find the area of the shaded region. Choose 3; you may complete the fourth for extra credit.
1.
2.
18
10
48
6
24
32
24
36
3.
18
4.
16
10
6 3
34
12
16
4
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