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Bainbridge Island School District, 2011 1 Bainbridge High School End of Course Geometry Study Guide: Prepared April 27, 2011 Bainbridge High School Mathematics End of Course “Classroom” Exam Week: May 23-27

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Bainbridge Island School District, 2011 1

Bainbridge High School

End of Course Geometry Study Guide: Prepared April 27, 2011

Bainbridge High School Mathematics

End of Course “Classroom” Exam Week: May 23-27

Bainbridge Island School District, 2011 2

End of Course Geometry Study Guide April 27, 2011

Bainbridge High School

End of Course Geometry “Classroom” Exam Week: May 23-27

Students in all sections of Geometry will take a 3 day End-of-Course High School Proficiency Exam. The

exam is not timed and will be administered in the classroom during regular class sessions.

End-of-Course Exam Items Estimated Number of Questions (37 questions)

Logical Arguments & Proofs 6-8

Two & Three Dimensional Figures 24-26

Coordinate Geometry & Measurement 7-9

Format of exam questions: 29 multiple choice, 5 completion items and 3 short answer questions. A

graphing calculator can be used for all questions and will be cleared before the exam. A formula sheet

and graph paper will be provided in the test booklet.

“Classroom” End of Course Geometry exam will also include questions on Lines and Angles. (6-8

questions) These questions are not required for graduation requirements, but are reported as

plus/minus. The 3 day exam will be 43 questions.

Study Guide Organization:

On the next page is the Review Topic Index. These items are key “green” standards that will be assessed

for graduation. The number of test questions for each topic heading is shown to help guide your study

priorities. Following the Index are Sample Questions for each topic. After each set of questions, there

are STUDY NOTES that explain the answers.

You may want to take these questions first as a practice test, then check your answers and review the

study notes for any items that you miss. There is another .pdf file that just includes the questions & the

answer key. For additional study resources, discuss any questions with your current math teacher.

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Review Topic Index:

(G1) Logical Arguments and Proofs (6-8 points)

G1C Use deductive reasoning to prove that a valid geometric statement is true.

G1D Write the converse, inverse, and contrapositive of a valid proposition and determine their validity.

G1E Identify errors or gaps in a mathematical argument and develop counterexamples to refute invalid

statements about geometric relationships.

G1F Distinguish between definitions and undefined geometric terms and explain the role of definitions,

undefined terms, postulates (axioms), and theorems.

(G3) Two & three dimensional figures (24-26 points)

G3A Know and apply basic postulates and theorems about triangles and the special lines, line segments,

and rays associated with a triangle.

G3B Determine and prove triangle congruence and other properties of triangles.

G3C Use the properties of special right triangles (30°–60°–90° and 45°–45°–90°) to solve problems.

G3D Know, prove, and apply the Pythagorean Theorem and its converse.

G3E Solve problems involving the basic trigonometric ratios of sine, cosine, and tangent.

G3G Know, prove, and apply theorems about properties of quadrilaterals and other polygons.

(G4)(G6) Coordinate Geometry & Measurement (7-9 points)

G4B Determine the coordinates of a point that is described geometrically. Example: Determine the

coordinates for the midpoint of a given line segment.

G4C Verify and apply properties of triangles and quadrilaterals in the coordinate plane.

G6E Use different degrees of precision in measurement, explain the reason for using a certain degree of

precision, apply estimation strategies to obtain reasonable measurements for a given purpose.

G6F Solve problems involving measurement conversions within and between systems, including those

involving derived units, analyze solutions for reasonableness and appropriate units.

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(G1) Logical Arguments and Proofs G1C Use deductive reasoning to prove that a valid geometric statement is true Sample Question G1C1: Fill in the blanks to complete the two-column proof.

Given: and are supplementary. m = 135

1 2

Prove: m = 45

Proof:

Statements Reasons

1. and are supplementary. 1. Given

2. [1] 2. Given

3. m + m = 180 3. [2]

4. 135 + m = 180 4. Substitution Property

5. m = 45 5. [3]

a. [1] m = 135

[2] Definition of supplementary angles

[3] Subtraction Property of Equality

b. [1] m = 135

[2] Definition of supplementary angles

[3] Substitution Property

c. [1] m = 135

[2] Definition of supplementary angles

[3] Subtraction Property of Equality

d. [1] m = 135

[2] Definition of complementary angles

[3] Subtraction Property of Equality

Answer: C

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Sample Question G1C2: fill in the missing steps #3 and #5 in the given two-column proof.

Given: 1 and 2 are supplementary. . .

Prove: 3 is a right angle.

1

2 3

Two-column proof:

Statements Reasons

1. 1 and 2 are supplementary.

. .

1. Given

2. 1 and 2 are right angles. 2. Congruent supplementary angles form

right angles.

3. m 3. _______

4. 4. Definition of congruent angles

5. m 5. _______

6. 3 is a right angle. 6. Definition of a right angle

Answer: 3. Definition of a right angle. 5. By Substitution

Sample Question G1D1:

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Sample Question G1D2:

Write the conditional statement and converse within the biconditional.

A rectangle is a square if and only if all four sides of the rectangle have equal lengths.

a. Conditional: If all four sides of the rectangle have equal lengths, then it is a square.

Converse: If a rectangle is a square, then its four sides have equal lengths.

b. Conditional: If a rectangle is a square, then it is also a rhombus.

Converse: If a rectangle is a rhombus, then it is also a square.

c. Conditional: If all four sides have equal lengths, then all four angles are 90 .

Converse: If all four angles are 90 , then all four sides have equal lengths.

d. Conditional: If a rectangle is not a square, then its sides are of different lengths.

Converse: If the sides are of different lengths, then the rectangle is not a square.

Answer: A

STUDY NOTES:

Let p and q represent the following.

p: A rectangle is a square.

q: All four sides of the rectangle have equal lengths.

The two parts of the biconditional are and .

Conditional: If all four sides of the rectangle have equal lengths, then it is a square.

Converse: If a rectangle is a square, then its four sides have equal lengths.

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Sample Question G1D3:

Answer:

Inverse: If the table top is not rectangular, then the diagonals are not congruent.

Contrapositive: If the diagonals are not congruent, then the table top is not rectangular.

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Sample Question G1E1:

What is the truth value of the biconditional formed from the conditional, “If B is the midpoint of A and C,

then .” Explain.

a. The conditional is true.

The converse, “If then B is the midpoint of ” is false.

Since the conditional is true but the converse is false, the biconditional is false.

b. The conditional is true.

The converse, “If then B is the midpoint of ” is true.

Since the conditional is true and the converse is true, the biconditional is true.

c. The conditional is false.

The converse, “If then B is the midpoint of ” is false.

Since the conditional is false and the converse is false, the biconditional is true.

d. The conditional is false.

The converse, “If then B is the midpoint of ” is true.

Since the conditional is false and the converse is true, the biconditional is false.

Answer: A

STUDY NOTES:

The conditional statement is the definition of a midpoint, and is a true statement.

The converse is false. The picture displays a counterexample.

B

A

C

, but B is not on . Therefore, B is not the midpoint of . If either the conditional or the

converse is false, the biconditional is false.

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Sample Question G1E2:

Determine if the biconditional is true. If false, give a counterexample.

A figure is a square if and only if it is a rectangle.

a. The biconditional is true.

b. The biconditional is false. A rectangle does not necessarily have four congruent sides.

c. The biconditional is false. All squares are parallelograms with four angles.

d. The biconditional is false. A rectangle does not necessarily have four angles.

Answer: B

STUDY NOTES:

Conditional: If a figure is a square, then it is a rectangle.

True.

Converse: If a figure is a rectangle, then it is a square.

False. A rectangle does not necessarily have four congruent sides.

Because the converse is false, the biconditional is false.

Sample Question G1F1:

Write the definition as a biconditional.

An acute angle is an angle whose measure is less than .

a. An angle is acute if its measure is less than .

b. An angle is acute if and only if its measure is less than .

c. An angle’s measure is less than if it is acute.

d. An angle is acute if and only if it is not obtuse.

Answer: B

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Sample Question G3A1:

Answer: B

Sample Question G3A2:

Answer: B

STUDY NOTES

Midsegment of a triangle is half the third side.

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Sample Question G3A3:

Explain your answer below:

Answer: Show algebraic work and answer x=10

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Sample Question G3A4:

Answer: B

STUDY NOTES:

Intersection point of perpendicular bisectors is equidistant to vertices, which is the circumcenter.

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Sample Question G3B1:

STUDY NOTES:

Note: the vertical angles would ordinarily be written as angle ABD is congruent to EBC, but since the

angle is the same as ABD is congruent to CBE, either answer would still score as a 2-point response.

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Sample Question G3B2:

Answer: B

STUDY NOTES:

Only have one angle.

Sample Question G3B3:

Answer: A

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Sample Question G3B4:

Answer: B:

Sample Question G3B5:

Answer: B

Sample Question G3C1:

Find the measures and .

X CA

B

6.4

2.3

a. c. b. d.

Answer: A

STUDY NOTES:

Perpendicular Bisector Theorem

Substitute 6.4 for .

Given

Substitute.

Segment Addition Postulate

Substitute.

Simplify.

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Sample Question G3C2:

Given that bisects and , find .

Y

Z

X

W

a. c. b. d.

Answer: A

STUDY NOTES: WX = WZ Angle Bisector

Sample Question G3C3:

Answer: D

STUDY NOTES:

Use 30-60-90 triangle ratios to find height. Area of one triangle times 6 is area.

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Sample Question G3C4:

Answer: A

STUDY NOTES:

This is an isosceles right triangle. Divide by root two, and then rationalize denominator.

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Sample Question G3D1:

STUDY NOTES:

For this problem, you do the Pythagorean Theorem two times. First, you find the diagonal (hypotenuse)

for the base of the box. After you have calculated this, then use this leg as the base for the second right

triangle that has the height as the second leg. Calculate this diagonal (hypotenuse) for length AB.

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Sample Question G3D2: The Yield sign has a shape of an equilateral triangle with side length of 36

inches. What is the height of the sign? Will a rectangular metal sheet of 36 32 inches be big enough to

make one sign?

a. The Yield sign is about 33.7 inches tall. So the rectangular metal sheet will not be big

enough to make one sign.

b. The Yield sign is about 31.2 inches tall. So the rectangular metal sheet will be big enough

to make one sign.

c. The Yield sign is about 25.5 inches tall. So the rectangular metal sheet will be big enough

to make one sign.

d. The Yield sign is about 50.9 inches tall. So the rectangular metal sheet will not be big

enough to make one sign.

Answer: B

STUDY NOTES:

Step 1 Divide the equilateral triangle into two 30°-60°-90° triangles. The height of the Yield sign is the

length of the longer leg.

h36 in.

60º

30º

x

Step 2 Find the length x of the shorter leg.

Hypotenuse

Divide both sides by 2.

Step 3 Find the length h of the longer leg.

The Yield sign is about 31.2 inches tall. So the rectangular metal sheet will be big enough to make one

sign.

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Sample Question G3D3:

Answer: C

STUDY NOTES:

Use Pythagorean twice.

Sample Question G3E1:

Find x, y, and z.

x

y

z

10

5

a. x = 5, y = , z =

c. x = 3, y = 4, z =

b.

x = 2.5, y = , z =

d. x = , y = , z =

Answer: B

STUDY NOTES:

Solve using similar triangles.

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Sample Question G3E2:

Answer: B

STUDY NOTES:

The tangent of an angle is the ratio of the opposite side divided by its adjacent side.

Sample Question G3E3:

Answer: C

STUDY NOTES:

Use sine ratio.

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Sample Question G3G1:

Answer: D

STUDY NOTES:

All angles sum to 360 degrees. Since two angles are congruent, 180-146= 34 degrees

Sample Question G3G2:

Given isosceles trapezoid ABCD with the sides , Point Y is the intersection

of the diagonals. , and . Sketch the figure and find YD.

a. YD = 6.9 c. YD = 10.3

b. YD = 17.2 d. YD = 8.6

Answer: A

STUDY NOTES:

Diagonals of an isosceles trapezoid are congruent.

Definition of congruent segments

Substitute 17.2 for AC.

Segment Addition Postulate

Substitute the given values.

Subtract 10.3 from both sides.

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Sample Question G3G3

Answer: A

STUDY NOTES:

The midsegment of a trapezoid is the average of the bases.

Sample Question G3G4

Answer: D

STUDY NOTES:

The diagonals of a rhombus are perpendicular bisectors of each other.

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Sample Question G4B1:

Answer: (1,-2)

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Sample Question G4B2:

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Sample Question G4B3:

1) Determine the mid-point for the given line segment:

(4, -2)

(-2, -4)

a. (-3, -6)

b. (-2, -6)

c. (1, -3)

d. (1, -6)

Answer: C

STUDY NOTES:

Mid-point = (x, y)

To find x, you add x1 +x2 and divide by 2, or rather -2 + 4 = 2, then divide by 2 = 1.

To find y, you add y1 + y2 and divide by 2, or rather -4 + -2 = -6, then divide by 2 = -3.

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Sample Question G4C1:

2) Parallelogram ABCD has the following coordinates for three of its vertices:

(-1, 4) (4, 4) (3, -2)

Of the following coordinates listed below, which one cannot be the fourth vertex for parallelogram

ABCD?

a. (0, 10)

b. (8, -2)

c. (-2, -2)

d. (2, 10)

Answer: D

STUDY NOTES:

Sketch a graph of the three given vertices. For ABCD to be a parallelogram, opposite sides need to be

parallel. Therefore, opposite sides need to have the same slope, or rather the same rise/run.

All points create opposite sides with the same slope, except (2,10).

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Sample Question G4C2:

3) The vertices of a quadrilateral are given below:

(0, -4) (3, 0) (0, 4) (0, -3)

Best classify the described quadrilateral as either a rhombus, a square, a rectangle, or a parallelogram.

a. rhombus

b. square

c. rectangle

d. parallelogram

Answer: A

STUDY NOTES:

Sketch the above quadrilateral on a coordinate plane. To be a rhombus, opposite sides must be equal,

and opposite sides must be parallel. Since opposite angles are not 90 degrees, it is not a rectangle nor a

square.

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Sample Question G4C3:

Consider the points , , , and . P is on the bisector of . Write an

equation of the line in point-slope form that contains the bisector of .

a. c.

b. d.

Answer: A

STUDY NOTES:

A

B

C

P

(–2, 5)

(2, –3)

(8, 0)

(4, 3)

x

y

Step 1 Find the slope of the line that contains the bisector of .

slope of

Step 2 Use the point-slope form to write an equation.

The line that contains the bisector of has slope 3 and passes through .

Point-slope form

Substitute 3 for , 3 for m, and 4 for .

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Sample Question G4C4:

Use the following parallelogram RSTU:

R (2, 5) U (8, 5)

S (1, 1) T (7, 1)

The length of diagonal RT is:

A. 41

B. √41

C. 9

D. √9

Answer: B

STUDY NOTES:

Use the distance formula. It does not matter which point you call (x1, y1) and (x2. Y2):

Sample Question G4C5:

Use the following parallelogram RSTU:

R (2, 5) U (8, 5)

S (1, 1) T (7, 1)

Find the coordinates of the midpoint of diagonal RT:__________

Answer: (9/2, 3)

STUDY NOTES:

Use (x1 + x2) / 2 and (y1 + y2) / 2 to find the coordinates of the midpoint.

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Sample Question G4C6:

Below is a diagram of a miniature golf hole as drawn on a coordinate grid. The dimensions of

the miniature golf hole are 4 feet by 12 feet and the space is enclosed with a wall so that players can

bounce their golf shots off of the four surrounding walls. There are three tees to choose from: Player 1

must start his ball from (1, 3). The actual hole is located at (10, 3). A wall separates the tees from the

hole. Sketch the intended path of the ball so that the player banks off of the lower wall. If Player 1 hits a

hole in one, what is the shortest distance that the ball traveled?

A. 145 feet

B. √117 feet

C. 13 feet

D. √145 feet

Answer: B

STUDY NOTES:

You can reflect an imaginary construction line from Player 1’s starting point at (1, 3) over the x-axis to an

imaginary point at (1, -3). Drawing a line from this point to the hole at (10,3) is the total distance. This

imaginary path creates congruent angles for the bank shot from the tee to the wall, and from the wall

up to the hole. Use Pythagorean Theorem or distance formula.

92 + 62 = √117 or use the distance formula below:

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Sample Question G6F1:

Sample Question G6F2:

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Sample Question G6F3:

Answer: A

STUDY NOTES:

To convert from cubic yards to cubic feet, you multiply by 3 feet for each of the dimensions, so: 33 or 27.

Sample Question G6F4:

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Sample Question G6F5:

Sample Question G6F6: