Download Introduction As an honors student you are expected to complete the

Introduction
As an honors student you are expected to complete the following
packet the summer before beginning the Geometry Honors course.
This packet consists of topics and problem sets from Chapter 1 of the
Geometry textbook as well as important prerequisite Algebra 1
knowledge. This packet will be assigned to students on move-up day
and extra copies will be provided to students during the first week of
school if they were unable to attend move-up day.
All packets must be completed and handed in the second week of
school. The date is to be determined by individual teachers. This
summer assignment will count as a test. A short review of the packet
will take place and then an in-class test will be given on all the
material in this packet.
Algebra Topic 1- Solving Equations
An equation is a statement that two expressions are equal.
Examples of Equations –
2x  3y 2  7 x  4 y 2
2x  y   8
4 w  13
3w  5w  6w  8
Addition and Subtraction Properties of Equality
If a  b then a  c  b  c and a  c  b  c .
We can add or subtract the same quantity from both sides of an equation without
changing the truth of the statement.
Multiplication and Division Properties of Equality
a b
If a  b , then ac  bc and (if c  0)
 .
c c
We can multiply or divide each side of an equation by the same quantity (as long as it
isn’t zero) without changing the truth of the statement.
Zero Product Property
If ab  0 , then a  0 or b  0 .
If the product of two or more numbers is zero, at least one of the numbers must be zero.
Example:
Solve the equation 3x  7  2 x  4 for x.
Subtract 2x from each side
Add 7 to each side
3x  7  2 x  4
 2x
 2x
x7
4
7
7
x 
11
Ordered Pair – An ordered pair is written as  x, y  . They are frequently listed in tables and are solutions to
equations.
Examples:
(2, 3)
 5,
 7
 3, 4
 1

  , 15 
2


Problem Set Topic 1
1. Refer to the figure.
a) Write a formula for the
perimeter P of the figure.
b) Find x if P  64.
x  10
B
C
x2
x  11
A
2. The numbers between – 3 and 5 have been shaded on the number line.
5
-3
a) Is
30 shaded?
b) Is 22 shaded?
c) Is  22 shaded?
18
d) Is  shaded?
5
14
e) Is
shaded?
3
f) Is – 3 shaded?
In problems 4 and 5, solve each equation.
3. 53x  2  50
4.
3x  6
9
4
5. Billie’s bank contains nickels, dimes and quarters.
a) If Billie has eight nickels, five dimes and seven quarters, what is the total value of the coins in the
bank?
b) Suppose Billie has n nickels, d dimes, and q quarters. Write a formula that can be used to find V, the
total value of the coins.
6. If w  2 x  y , what is
a) The value of w for which  3, 4  is a solution of the equation?
b) The value of x for which w  8 and x, 5 is a solution of the equation?
7. The perimeter of the rectangle is 62 centimeters. Find the area of the rectangle. (Hint: First find the value of
x).
2x  3
x5
8. Two hot air balloons are flying over a field. The height h, in feet, of the first balloon t seconds from now can
be found with the formula ht   50  34t . The height g, in feet, of the second balloon t seconds from now
can be found with the formula g t   400  16t .
a. In how many seconds will the balloons be at the same height?
b. How high will they be at that time?
9. Solve 3x  9 y  12 z for x.
10. Solve 2 x  3 y  10 y  6 x for y.
11. Express a in terms of x.
4x
a
11
2x
12. If y  9  4 x and z  6 x  11 , for what value of x is y = z?
In problems 13 – 14, solve each equation for x and simplify each expression.
2

13.  x  3  7  6 x  2
3


14. abc  ac  ab   bab  bc  ac 
15. Refer to the figure.
a. Write an expression for the surface area of the box.
b. Find x if the surface area is 261.
12
x
6
16. Refer to the diagram.
a. Find the coordinates of point R.
b. Find the lengths of PR and RQ .
c. Find the area of PQR.
y
P 0, 3
d. Use the Pythagorean Theorem to find the length of PQ .
x
R
17. If s  5 x  7 and t  x  10 , for what value of x is s  2t  4 ?
Q12,2 
Algebra Topic 2 - Exponents and Radicals
The Product Rule – If a is a real number and m and
n are integers, then a m * a n  a m n
Rule – If x is a real number and n is a positive
integer, then
Example 3 * 3  (3 * 3 * 3)(3 * 3)  3 * 3 * 3 * 3 * 3  35
3
2
The Power Rule – If a is a real number and m and n
 
are integers, then a m
5 
3 5
n
 a mn
Example -
 53 * 53 * 53 * 53 * 53  533333  515
 x
n
n
 x whenever
 x  is defined.
n
n
Example - (4 16 ) 4  16
Rule – If x and y are nonnegative real numbers, then
x
x
x y  xy ,
where y  0

y
y
Example -
5 2  10 and
5
2

5
2
The Quotient Rule – If a is a real number and m and
am
n are integers, then n  a mn
a
Standard Radical Form – Radicals written in the
form a b where b is an integer have no perfect
square factors.
Example 35 3 * 3 * 3 * 3 * 3 3 * 3


* 3 * 3 * 3  1* 33  33 or
2
3*3
3*3
3
5
3
 352  33
2
3
Example -
Rule – If a is a real number, and a  0 , and n is an
1
integer, then a  n  n .
a
1
Example - 5 3  3
5
The Distributive Property of Exponentiation over
Multiplication – If x and y are real numbers and n is
an integer, then x n * y n  ( x * y) n
Example - 6 5 * 2 5  (6 * 2) 5
The Distributive Property of Exponentiation over
Division – If x and y are real numbers y  0 ,
and n is an integer, then
xn  x 
 
y n  y 
n
5
Example -
65  6 
    35
5
2
2
Rule – If a is a real number and a  0 then a 0  1
Example - 5 0  1
252  9 * 28  3 4 7  6 7
Conjugates – Expressions in the form a  b c and
a b c
Example - 2  7 and 2  7 are conjugates.
Problem Set –Topic 2
In problems 1-2 simplify each expression.
   
3
2
1. 2b 3 * 3 b 4
2.
12 w 4 15 w12
*
8w8 (6 w) 2
In problems 3-5, indicate whether each statement is True or False. ( x  0)
4. x a * x b  x ab
3. x 2a  ( x a ) 2
5.
xa
 x1
a
x
6. Evaluate the following
 
a. 5 2
 
3
b. 53
2
 
c. 5 5
d. 5 2.5
7. Find an expression for the sum of the volumes of the cube and the box.
x2
x2
x
x
x2
x2
3
8. Evaluate each expression.

a. 6 2

b. 32  1

d. 2 1  31
1
1 1
c.   
 2 3
1

1
In problems 9-10 rewrite each expression without parenthesis using only negative
exponents.
 x 3 y 2
9.   4
 w




5
 x  4 y 3 

10. 
4

 w 
5
2
11. Write an expression, in simplified form, for the area of the trapezoid.
x4 y9
4x 2 y 5
3x 4 y 9
12. Simplify x * x 2  2 y * 4 y 2
1
13. Solve 5  5 x  5 3
xy 
14. Evaluate
x y 
2 18
19
2
*y
1
1
 (5 z ) 3  w 3 for w, x, y, and z.
 x4
*  4
y
5

 for (x, y) = (5.2, -2).


In problems 15-17 rewrite each expression without parentheses, using only positive
exponents.
 x 6
15.  4
y




5
 x3 
16.  2 
x 
5

17. x 3 y 4

5
In problems 18-20 simplify each expression.
18.
 20 xy 3 z 2
24 x 4 y 2 z
19. x 2 ( x 2  x  2)  x 3 ( x  1)
20.  xy ( x 2 y 2 ) 2
4
In problems 21-24 simplify each expression.
21.
xy
2
x y  xy
23.
3x y  * 5x
15 x y 
2
22.
4
4
xy 
5 18
3
5 3
10 x 3  20 x 2
15 x 4  30 x 3
24.
xy 17
Algebra Topic 3–Factoring & Solving Quadratic Equations
Multiplication of a monomial and polynomial example –
Find the product 2 x 3 ( x 3  3x 2  2 x  5)
2 x 3 ( x 3  3x 2  2 x  5
Write product
 2 x 3 ( x 3 )  2 x 3 (3x 2 )  2 x 3 (2 x)  2 x 3 (5)
Distributive Property
 2 x 6  6 x 5  4 x 4  10 x 3
Product of powers property
Multiplication of binomials using FOIL example Find the product 3a  4a  2
 (3a)( a)  (3a)( 2)  (4)( a)  (4)( 2)
Write product of terms
 3a 2  (6a)  4a  (8)
Multiply
 3a 2  2a  8
Combine like terms
Factoring Trinomials ( x 2  bx  c )
Rule - x 2  bx  c =(x + p) (x + q) provided p + q = b and p*q=c.
Example when b and c are positive –
Factor x 2  11x  18
Find two positive factors of 18 whose sum is 11.
18, 1
9, 2
6, 3
The factors 9 & 2 have a product of 18 and a sum of 11, so they are the correct value of p and q.
You can check your solution by distribution. Does ( x  9)( x  2)  x 2  11x  18 ?
Example when b is negative and c is positive –
Factor n 2  6n  8
Because b is negative and c is positive, p and q must both be negative.
Make a list of the negative factors of 18
-8, -1
-4, -2
The factors -4 & -2 have a product of 18 and a sum of -6, so they are the correct value of p and q.
You can check your solution by distribution. Does (n  4)( n  2)  n 2  6n  8 ?
Example when b is positive and c is negative –
Factor y 2  2 y  15
Because c is negative, p and q must have different signs.
Make a list of the factors of -15
-15, 1
15, -1
-5, 3
5, -3
The factors 5 & -3 have a product of -15 and a sum of 2, so they are the correct value of p and q.
You can check your solution by distribution. Does ( y  5)( y  3)  y 2  2 y  15
Example when b is negative and c is negative –
Factor s 2  2s  24
Because both b and c are negative, p and q must have different signs.
Make a list of the factors of -24
1, -24
-1, 24
2, -12
-2, 12
3, - 8
-3, 8
4, -6
-4, 6
The factors 4 & -6 have a product of -24 and a sum of -2, so they are the correct value of p and q.
You can check your solution by distribution. Does ( s  4)( s  6)  s 2  2s  24
Solving Quadratic Equations
Solving by Factoring – Once you have factored the equation you set each parenthesis equal to
zero to determine the values of x.
Example –
Solve the quadratic equation for x using factoring
m 2  5m  6  0
Step 1 – Factor (m + 2)(m + 3) = 0
Step 2 – Set each parenthesis equal to zero (m + 2) = 0 and (m + 3) = 0
Step 3 – Solve each equation for x.
m+2=0
m = -2
m+3=0
m = -3
Solving by the Quadratic Formula - Sometimes quadratic equations can not be factored so we
must use the quadratic formula.
If ax 2  bx  c  0 then x 
 b  b 2  4ac
2a
Example – Solve the quadratic equation 3 x 2  5 x  8  0 for x.
Step 1 – Determine the value of a, b & c.
a = 3 b = 5 c = -8
Step 2 – Substitute the values for a, b, & c into the quadratic formula.
x
 5  (5) 2  4(3)( 8)
2*3
Step 3 – Simplify.
 5  (5) 2  4(3)( 8)  5  25  96  5  121  5  11
x



2*3
6
6
6
Step 4 – Break expression into two parts.
x
 5  11 6
  1 and
6
6
x
 5  11  16

 2.6
6
6
Problem Set - Topic 3
Directions – Simplify the following expressions.
1. x  82x  5
2.  x  6 
3. 5 x 2 (3x 2  6 x  12)
4. 3xx  4x  2


5. 3x 2  6 2 x 2  7
2

Directions – Factor the following expressions.
6. x 2  8 x  20
7. x 2  16 x  64
8. 2 x 2  8 x  24
9. x 3  11x 2  30 x
10. y 2  8 y  48
Directions – Solve the following quadratic equations using either factoring or the quadratic
formula.
11. x 2  2 x  63  0
12. 5 x 2  13x  6
13. n 2  n  6
14. 3 x 2  7 x  4  0
15. 3x 2  4  11x
16. 6 x 2  2 x  28  0
17. x 2  17 x  18
Algebra Topic 4 – System of Equations
Linear Systems by Substitution - the strategy is to obtain an equation in only one variable. The
solution is the point of intersection of two lines.
Example -
1

y  x  5
Solve the system 
2
2 x  y  7
The first equation is solved for y, so we can substitute
equation. Then 2 x  y  7 becomes 2 x 
1
x  5 for the value of y in the second
2
1
x5 7.
2
2.5 x  5  7
2.5 x  2
Now this equation has only one variable
And we can solve for x.
x  0.8
To find the y-coordinate, substitute 0.8 for the value of x in the first equation.
y
1
(0.8)  5
2
 0.4  5
 5.4
y
1
x  5 becomes
2
By this method,
the solution is , x, y   0.8,5.4
Linear Systems by Elimination - There are two types of elimination. The first is called the
addition/subtraction method. The second requires you to multiply before you can eliminate with
addition and subtraction.
Example Addition & Subtraction Method –
Step 1 – Add or Subtract the equations to eliminate one variable
Step 2 – Solve the resulting equation for the other variable
Step 3 – Substitute in either original equation to find the value of the
eliminated variable
2 x  3 y  11
 2 x  5 y  13
Solve the linear system 
Step 1 – Add the equations to eliminate the variable.
Step 2 – Solve for y.
2 x  3 y  11
 2 x  5 y  13
Step 3 – Substitute 3 for y in either equation and solve for x.
2 x  3 y  11
2 x  3(3)  11
x 1
Write Equation 1
Substitute 3 for y
Solve for x
8 y  24
y3
The solution is (1, 3)
Example Multiplication First Method Sometimes in a linear system variables can not be eliminated by using addition and
subtraction. First we must multiply one or both of the equations by a constant so that adding or
subtracting the equations will eliminate a variable.
6 x  5 y  19
2 x  3 y  5
Solve the linear system 
Equation 1
Equation 2
Solution –
Step 1 – Multiply Eqn 2 by -3 so that the coefficients of x are opposites.
Step 2 – Add the equations
6 x  5 y  19
 6 x  9 y  15
 4y  4
Step 3 – Solve for y
y  1
6 x  5 y  19
2 x  3 y  5 *(-3)
Step 4 – Substitute -1 for y in either of the original equations and solve
for x.
2x  3 y  5
2 x  3(1)  5
2 x  (3)  5
2x  8
x4
The solution is 4,1
Write Equation 2
Substitute -1 for y
Multiply
Subtract -3 from each side
Divide each side by 2.
Problem Set Topic 4
Directions – Solve each system of equations for the missing values using the indicated method.
Write your solution as a point of intersection. Check your solution in both equations. Show all
work.
4 x  3 y  27
 y  2x  1
1. 
5 x  y  13
4 x  3 y  18
2. 
Substitution Method
Add/Sub/Mult. Method
3x  4 y  26
x  2 y  2
Substitution Method
7 x  3 y  23
 x  2 y  13
Add/Sub/Mult. Method
 x  3 y  13
x  y  5
Either Method
3 x  7 y  22
2 x  8 y  2
Either Method
3. 
4. 
5. 
6. 
Chapter 1 –Introduction to Geometry
Geometry Topic 1 - 1.1 Getting Started
Points are represented by dots. We use capital letters to name points.
B
A
E
C
D
Lines are made up of points and are straight. The arrows at the end of a line show that the lines
extend infinitely far in both directions.
E
B
m
l
C
D


F
The line on the left is called line m.
We can name a line by any two points on it, so the line in the middle can be called one of
the following;
BD , BC , CD , CB , DB , DC

The line on the right can be called by any of three names;
line l, EF , FE
In Algebra 1 you learned that a number line is formed when a numerical value is assigned to each
point on a line.
B
A
The coordinate for A is -2.
-4 -3-2-1 0 1 2 3 4
The coordinate for B is 1
1
2
A line segment or segment are made up of points and are straight. A segment has a definite
beginning and end.
S
R
P
X
Q
A segment is named in terms of its endpoints.
 The segment on the left can be called either RS or SR .
 In the second figure there are two segments. The vertical one can be called
either PX or XP . The horizontal segment can be named as either XQ or QX .
A ray is like a line and segment because it is made up of points and it is straight. It is different
from a line or segment because it begins at an endpoint and then extends infinitely far in only one
direction.
B
C
D
E
A
When we name a ray, we name the endpoint first so that it is clear where the ray begins.

The ray on the left is called AB .

The ray on the right is called CD or CE . As long as we use the endpoint first any other
point on the ray can be used in its name.
An angle is made up of two rays with a common endpoint. This endpoint is called the vertex of
the angle. The rays are called the sides of the angle.
A
Y
P
B
C
3
1
O
2
R
X


The angle of the left is called 3 . The 3 placed inside the angle near the vertex names it.
The middle angle can be called by any of these three names;

In the figure on the right there are three angles.
BAC
o
o
o
CAB
A
1 can be called POY or YOP
2 can be called YOR , YOX , ROY or XOY .
The other angle can be named POR . See if you can find three other names for
this angle.
A triangle has three segments as its sides and three angles even though they do have endpoints.
Triangle ABC = ABC
A
B
C
The triangle is the union of three segments ABC  AB  BC  AC .
The intersection of any two sides is a vertex of the triangle. AB  BC  B .
Problem Set 2.1 (pg. 7 #1-6, 8-14)
l
1. What are three possible names for the line shown?
B
A
C
2. What are four possible names for the angle shown?
7
E
D
3. Can the ray shown be called XY ?
Y
X
4. Name the sides of RST .
R
S
T
5.
a. AB  BC 
C
D
b.EC  EA 
c. AC  DB 
E
d .DC  AB 
e AC  EC 
f .BA  BC 
A
g .EC  CB  BE 
6. a. Name OPR in all possible ways.
B
S
T
O
b. What is the vertex of TOS .
c. How many angles have vertex R?
d. Name TSP in all other possible ways.
P
R
e. How many triangles are there in the figure?
7a. A line is named up of
b. An angle is the union of two
.
with a common
.
8. Draw a number line and label points F, G, H and J with the coordinates  4
2
, 2, 5, and 3.5
3
respectively. One of these points is the midpoint (the halfway point) between two others. Which
is it?
8.6 cm
9. Given a rectangle with sides 2.5 cm and 8.6 cm long, find
a. The rectangle’s area
2.5 cm
b. The rectangle’s perimeter (the distance around it)
H
10a. In HJK , HJ is twice as long as JK and exactly as
long as HK . If the length of HJ is 15, find the
perimeter of HJK
J
b. If the length of HJ were 4x, the length of HK were
K
3x, the length of JK were 2x, and the perimeter of
HJK were 63, what would the length of HJ be?
11. Draw a diagram in which AB  CD  CB.
12. Draw a diagram in which the intersection of AEF and DPC is ED.
13. a What percentage of the triangles in the diagram
T
E
have CT as a side?
A
b. What percentage have AC as a side?
X
C