Summer Mathematics Prep
Chesterfield County Public Schools
Department of Mathematics
Entering Geometry
SOLUTIONS
NOTE: There are many ways to solve problems in mathematics. The solution shown is just one way to solve
each problem in this document.
Set 1: Solving linear equations
1.
2.
3.
4.
2y  2  4y  4
2  2y  4
6  2y
3 y
5.
x3
 12
2
x  3  24
x  21
6.
9( p  4)  18
9 p  36  18
9 p  54
p6
7.
(9x + 8) + (4x + 4) = 90
13x + 12 = 90
13x = 78
x=6
1
2 x  4  10
2
2 x  4  20
2 x  16
x8
(19y + 8) – (7y +10) = 22
12y – 2 = 22
12y = 24
Y=2
x3 8

5
2
2  x  3  40
(13x – 14) + (8x – 15) = 180
21x – 29 = 180
21x = 209
209
x = 21
8.
2 x  6  40
2 x  34
x  17
Set 2: The Coordinate Plane
9.
y y
The slope formula is m  2 1 .
x2  x1
m
 6,  3 


 x1 , y1  ,
 1, 2 


 x2 , y 2 
2  (3)
5

 1
1 6
5
Chesterfield County Public Schools
Department of Mathematics
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May 2012
10.
Use the same procedure as in #9 or simply count the spaces, making sure you go up or down first, then
left or right second. m = -5
1
1
x  7 is . Parallel lines have the same slope.
2
2
1
1
The slope of a line parallel to y  x  7 is .
2
2
3
y  3x  2 . The slope is 3 (or ) and the y-intercept is  2.
1
Graph the y-intercept at  2 on the y-axis. Using the slope, move 3 units up and 1 unit to the right of
the y-intercept and graph another point. Draw the line.
The slope of the line y 
11.
12.
Writing the Equation of the Line
2
13.
Slope =  , y-intercept =  8.
3
14.
2
Substitute the values of m and b into y  mx  b . y   x  8
3
5
Slope =  , point (6,  9).
6
Calculate the y-intercept, b, by substituting the slope and the x and y coordinates of the given point into
y  mx  b .
5
 9   ( 6)  b
6
 9  5  b
4 b
5
Substitute the values of m and b into y  mx  b . y   x  4
6
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Department of Mathematics
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15.
Points (10, 3) and (5, 12) . Calculate the slope. m 
Calculate the y-intercept (b) using either point.
(10, 3) y  mx  b
(5, 12)
3
3  (10)  b
5
3  6  b
9 b
12  3
9 3


5  (10) 15 5
y  mx  b
3
12  (5)  b
5
12  3  b
9 b
3
Substitute the values of m and b into y  mx  b . y  x  9
5
16.
Point ( 2, 4) , and parallel to y  3x  7 . The slope of y  3x  7 is 3. Parallel lines have
the same slope. Calculate the y-intercept, b, by substituting the slope and the x and y coordinates of the
given point into y  mx  b .
4  3( 2)  b
4  6b
2b
Substitute the values of m and b into y  mx  b . y  3x  2
Set 3: Radicals
20  4  5  2 5
17.
18.
80  16  5  4 5
19.
128  64  2  8 2
Evaluating Square Roots
20.
10
21.
5
3
22.
8
23.
0.7
24. 75, you must square the 5, which is 25, then square the square root of 3, which is 3, then multiply
25 times 3
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Department of Mathematics
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May 2012
Set 4: Classifying solid figures
25. C
26. A
27. B
28. D
Volume and Surface Area
29.
Surface area of a rectangular prism: SA  2(lw  lh  wh) ; l  5, w  6, h  4
SA  2(5  6  5  4  6  4)
 2(30  20  24)
 2(74)
 148 square units
30.
Volume of a rectangular prism: V  lwh; l  5, w  4, h  9
V  5 49
 180 cubic units
31.
Surface area of a cylinder: SA  2 r (h  r ); r  7, h  8
SA  2  3.14  7(8  7)
 2  3.14  7  15
 659.4 square inches
32.
Volume of a cylinder: V   r 2 h; r  9, h  6
V  3.14(9 2 )(6)
 3.14(81)(6)
 1,526.04 cubic meters
Set 5: Angle Relationships
33.
Acute
34.
Obtuse
35.
Straight
36.
Right
37.
Supplementary angles add up to 180. Supplement = 180  75  105.
38.
Complementary angles add up to 90 . Complement = 90  66  24.
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Department of Mathematics
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39.
Vertical angles are equal so m1  m2.
8x  2  6x  20
m2  6x  20
 6(9)  20
2x  2  20
2 x  18
 54  20
x9
m2  74
40. A linear pair of angles are supplementary, which means the sum of their measure adds up to 180
(7x – 5) + (13x – 15) = 180
20x – 20 = 180
20x = 200
x = 10
You must substitute the 10 in for x in 7x – 5, to find the measure of angle 2, which is the same as the
measure of angle 1. 7(10) – 5. The answer is 65.
Set 6: Formulas
Pythagorean Theorem
Since the triangle shown is a right triangle, one may use the Pythagorean Theorem ( a 2  b 2  c 2 ) to solve
either problem.
41.
a 2  32  5 2
62  82  c 2
a 2  9  25
36  64  c 2
a 2  16
100  c 2
a 2  16
a  4 or  4
Since we are speaking of side lengths,
 4 is not a possible answer.
42.
100  c 2
c  10 or  10
Since we are speaking of side lengths,
 10 is not a possible answer.
43.
Area. A  r 2 Given that r = 4, A   (4 2 )  16  50.24 using   3.14 .
Circumference. C  2r Given that r = 4, C  2 (4)  8  25.12 using   3.14 .
44.
Area. A  lw Given l = 6mm and w = 3mm, A  6mm  3mm  18mm 2 .
Perimeter. P  2(l  w) P  2(6mm  3mm)  2(9mm)  18mm
45.
Area. A  lw or since we are talking about a rectangle that is a square, l  w , so
A  ( side length) 2 or A  s 2 . So, the area of the square is A  (8units) 2  64units 2 .
Perimeter. P  2(l  w) , so P  2(8units  8units)  32units .
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Department of Mathematics
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46.
1
1
bh . We know that the base = 12units and the height = 9units, so A   12  9  54units 2 .
2
2
Perimeter. Perimeter is the distance around an object, so add up all three side lengths, so
Area. A 
P  9  12  15  36units
47.
Area. A = πr2 . Since the area is 81π, substitute 81π for A. 81π = πr2 . The π cancels out from each side
Leaving 81 = r2 . 9 = r as we know the radius cannot be -9.
Circumference. C = 2πr. C = 2π(9), C = 18π cm.
48.
Area. A = s2 . Since the area of the square is 196, substitute 196 for A. 196 = s2 . 14 = s as we know
the side of the square cannot be -14.
Perimeter. P = 4s. P = 4(14). P = 56m.
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Department of Mathematics
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May 2012