Geometry Sample Test Items PDF
Geometry Sample Test Items PDF
Geometry Sample Test Items PDF
however, at this time, there is no plan to transition to PARCC. The sample items and student work reflect the
current EOC Geometry assessment; therefore, teachers are encouraged to use the samples provided in this
document as additional resources, but should use the current Assessment Guidance Geometry document for
up-to-date EOC testing information.
Geometry
President
Member-at-Large
Member-at-Large
Secretary/Treasurer
Member-at-Large
The mission of the Louisiana Department of Education (LDOE) is to ensure equal access to education and to promote
equal excellence throughout the state. The LDOE is committed to providing Equal Employment Opportunities and is
committed to ensuring that all its programs and facilities are accessible to all members of the public. The LDOE does
not discriminate on the basis of age, color, disability, national origin, race, religion, sex, or genetic information. Inquiries
concerning the LDOEs compliance with Title IX and other civil rights laws may be directed to the Attorney, LDOE, Office
of the General Counsel, P.O. Box 94064, Baton Rouge, LA 70804-9064; 877-453-2721 or customerservice@la.gov.
Information about the federal civil rights laws that apply to the LDOE and other educational institutions is available on
the website for the Office of Civil Rights, USDOE, at http://www.ed.gov/about/offices/list/ocr/.
This project is made possible through a grant awarded by the State Board of Elementary and Secondary Education
from the Louisiana Quality Education Support Fund8(g).
This public document was published at a total cost of $6,000.00. This web-only document was published for
the Louisiana Department of Education, Office of Assessments, P.O. Box 94064, Baton Rouge, LA 70804-9064,
by Pacific Metrics Corporation, 1 Lower Ragsdale Drive, Building 1, Suite 150, Monterey, CA 93940. This material
was published in accordance with the standards for printing by state agencies established pursuant to R.S. 43:31
and in accordance with the provisions of Title 43 of the Louisiana Revised Statutes.
Table of Contents
Introduction .......................................................................................... 1
Purpose of This Document ....................................................................... 1
Geometry Administration ......................................................................... 1
EOC Achievement Levels ......................................................................... 2
EOC Achievement-Level Definitions ......................................................... 2
Geometry .............................................................................................. 3
Constructed-Response Rubric ................................................................. 3
Testing Materials and Online Tools .......................................................... 4
Multiple-Choice Items ........................................................................... 5
Constructed-Response Item ................................................................. 31
Constructed-Response Item ................................................................... 31
Scoring Information ............................................................................... 34
Sample Student Responses .................................................................... 35
Geometry Typing Help ........................................................................... 44
Geometry Reference Sheet ..................................................................... 46
Introduction
Louisiana Believes embraces the principle that all children can achieve at high
levels, as evidenced in Louisianas recent adoption of the Common Core State
Standards (CCSS). Louisiana Believes also promotes the idea that Louisianas
educators should be empowered to make decisions to support the success
of their students. In keeping with these values, the Department has created
documents with sample test items to help prepare teachers and students as
they transition to the CCSS. The documents reflect the States commitment to
consistent and rigorous assessments and provide educators and families with
clear information about expectations for student performance.
Geometry Administration
The Geometry EOC test is administered to students who have completed one
of the following courses:
Geometry
The Geometry EOC test contains forty-six multiple-choice items and one
constructed-response item. In addition, some field test items are embedded.
Multiple-choice items assess knowledge, conceptual understanding, and
application of skills. They consist of an interrogatory stem followed by four
answer options and are scored as correct or incorrect.
Constructed-response items require students to compose an answer, and these
items generally require higher-order thinking. A typical constructed-response
item may require students to develop an idea, demonstrate a problem-solving
strategy, or justify an answer based on reasoning or evidence. The Geometry
constructed-response item is scored on a scale of 0 to 4 points. The general
constructed-response rubric, shown below, provides descriptors for each
score point.
Constructed-Response Rubric
Score
Description
1
0
protractor
inch ruler
centimeter ruler
Note: Students are NOT allowed to use calculators during session 1 unless
they have the approved accommodation Assistive Technology and are
allowed the use of a calculator.
Sessions 2 and 3
calculator
protractor
inch ruler
centimeter ruler
Multiple-Choice Items
This section presents ten multiple-choice items selected to illustrate the type
of skills and knowledge students need in order to demonstrate understanding
of the CCSS in the Geometry course. These items also represent the skills
students need in order to meet performance expectations for Math Practices.
Information shown for each item includes
conceptual category,
domain,
cluster,
standard,
Math Practices,
tool restrictions,
correct answer,
COCongruence
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
A.
B.
C.
D.
*correct answer
Math Practice 7 (Look for and make use of structure.): Students must use
the structure of the two congruent circles and recognize that the circles
intersect at points that are equidistant from the endpoints of the given
line segment. These points may be connected to construct a perpendicular
bisector.
Option D: This is the correct answer. The figure shows two circles that appear
to be congruent centered at the endpoints of the given line segment.
Connecting the intersection points of these two circles guarantees the bisecting
line created is perpendicular (at a 90 degree angle) to segment AB.
Option A: The student incorrectly draws two circles that appear to be congruent
and intersect at what appears to be the midpoint of segment AB; however, the
figure does not show how a perpendicular bisector could be constructed, as it
does not guarantee that a line drawn would intersect at a 90 degree angle.
Option B: The student incorrectly draws one circle that appears to be centered
at the midpoint of segment AB. The midpoint is where a bisector will intersect,
but this figure does not show how to ensure the bisector will be perpendicular.
Option C: The student incorrectly draws two circles with different radii centered
at the endpoints of the given line segment. Connecting the intersection points
of these two circles will result in a line perpendicular to segment AB that does
not bisect it.
COCongruence
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
Reason
1. Quadrilateral ABCD is a
parallelogram.
2. AB || CD; BC || DA
1. given
3.
3.
4.
AC AC
4. reflexive property
5.
ABC CDA
6.
AB CD and BC DA
5. angle-side-angle
congruence postulate
6. Corresponding parts of
congruent triangles are
congruent (CPCTC).
2. definition of parallelogram
Which statement and reason should Missy insert into the chart as step 3
to complete the proof?
A.
B.
C.
ABD CDB and ADB CBD; When parallel lines are cut
by a transversal, alternate interior angles are congruent.
BAC DCA and BCA DAC; When parallel lines are cut
by a transversal, alternate interior angles are congruent.
D.
*correct answer
10
This item requires students to provide the missing statement and reason
in a two-column proof of the theorem that says that the opposite sides of a
parallelogram are congruent. Assessing students abilities to write proofs in a
multiple-choice item is difficult. This item allows students to demonstrate their
reasoning skills by asking for one of the middle steps in the proof, including
both the statement and reason.
This item is linked to three of the Math Practices.
Option D: This is the correct answer. Students are given a set of corresponding
congruent sides in statement and reason 4. This leads into proving triangle
congruence using the angle-side-angle congruence postulate in statement
and reason 5. In order to reach statement and reason 5, students need to
provide two pairs of corresponding congruent angles, which include the given
corresponding congruent sides in statement and reason 4.
Option A: This answer is a true statement with a correct reason for the
statement based on the given information; however, it is not relevant to
completing the proof because it does not support statement and reason 5.
Option B: The student uses the final result of the proof as the statement and
provides an incorrect reason for the statement.
Option C: This response is also a true statement with a correct reason, but it
is not relevant to the given proof because it does not support statement and
reason 5.
11
Cluster:
Standard:
Math Practices:
Calculator:
Tool Restrictions:
None
12
y
10
8
6
2
10
10
2
4
6
8
10
A.
1
The result is a different line the size of the
5
original line.
B.
C.
1
The result is a different line with a slope of .
3
D.
*correct answer
This item requires students to evaluate the effect of a dilation of a line not
passing through the center of the dilation. Specifically, students must recognize
that the dilated line will have the same slope as the original line.
This item is linked to one of the Math Practices.
Math Practice 7 (Look for and make use of structure.): Students must
recognize that the effect of the described dilation would not change the
slope of the line, and that it would not be the same line either.
13
Option B: This is the correct answer. The result of a dilation of a line not
passing through the center of the dilation is a line parallel to the original line.
Therefore, the slope will remain the same.
Option A: The student confuses the line with line segment. A dilation of a line
segment with a scale factor of 1/5 would result in a smaller line segment;
however, a dilation of a line results in a line.
Option C: The student incorrectly assumes that the result of the dilation is a
line that is perpendicular to the original line.
Option D: The student incorrectly assumes that the result of the dilation is the
original line; however, this is only true of lines that pass through the center of
the dilation.
14
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
15
B'
6
4
C'
A'
10
10
2
4
6
10
Kelly dilates triangle ABC using point P as the center of dilation and creates
triangle A'B'C'.
By comparing the slopes of AC and CB and A'C' and C'B', Kelly found that
ACB and A'C'B' are right angles.
Which set of calculations could Kelly use to prove ABC is similar to A'B'C'?
A.
slope AB =
7 (7) 14
=
=2
2 (5)
7
slope A B =
B.
73
4
= =2
3 (5) 2
AB2 = 72 + 142
AB2 = 22 + 42
C.
tan ABC =
AC 7
=
BC 14
tan A B C =
D.
A C 2
=
B C 4
*correct answer
Geometry EOC Sample Test ItemsOctober 2013
16
This item requires students to decide which evidence is sufficient for proving
that the given triangles are similar. Providing the calculations in the answer
choices eliminates the possibility of a student missing the item due to an
arithmetic error. This allows the item to focus on assessing the students
reasoning in determining the AA criterion for triangle similarity.
This item is linked to two of the Math Practices.
17
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
B.
C.
D.
*correct answer
This item requires students to determine which information is needed to
complete an argument that two triangles are similar. Specifically, students
must recognize that two pairs of corresponding congruent angles are sufficient
for proving two triangles are similar.
This item is linked to two of the Math Practices.
18
19
CCircles
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
B.
C.
D.
*correct answer
Geometry EOC Sample Test ItemsOctober 2013
20
Option C: This is the correct answer. Tangent segments drawn from the same
fixed point outside of the circle to points of tangency on the circumference of
the circle are congruent. Therefore, the triangle is isosceles.
Option A: The student incorrectly assumes that a triangle could be created if
side TR passed through the center of the circle; however, if this were possible,
then the lines tangent to the circle at T and R would be perpendicular to
segment TR. Thus, a triangle could not be created.
Option B: The student incorrectly assumes that angle S must always be smaller
than angle T and angle R; however, as points T and R move closer to each other
on the circumference of the circle, point S moves closer to the circle. This
eventually allows angle S to grow larger than angle T and angle R.
Option D: The student incorrectly assumes that triangle STR cannot be a
right triangle. It is true that angles T and R can never be right angles, but it is
possible for angle S to be equal to 90.
21
Cluster:
Standard:
Math Practices:
Calculator:
Tool Restrictions:
None
center: (2, 5)
radius: 6
B.
center: ( 2, 5)
radius: 6
C.
center: (2, 5)
radius: 36
D.
center: ( 2, 5)
radius: 36
*correct answer
This item requires students to use the method of completing the square in
order to determine the center and radius of a circle given the equation.
This item is linked to one of the Math Practices.
Math Practice 7 (Look for and make use of structure.): Students must
complete the square for the equation to create an equivalent equation of the
form (x h )2 + (y k )2 = r 2. Then, they must use this structure to determine
that the center is (h, k ) and the radius is r.
22
Option B: This is the correct answer. Completing the square in the original
equation results in (x + 2)2 + (y 5)2 = 62. Therefore, the center of the circle is
the point (2, 5) and the radius is 6 units.
Given
Equation
x 2 + y 2 + 4x 10y = 7
Step 1
x + 4x + y 10y = 7
2
(x 2 + 4x +
Step 2
)= 7
+ 4
+4
+25 +25
2
2
(x + 4x + 4) + (y 10y + 25) = 36
Step 3
(x + 2)2 + (y 5)2 = 36
Step 4
Step 5
) + (y 2 10y +
(x h )2 + (y k )2 = r 2
(x (2))2 + (y 5)2 = 62
h = 2
k =5 r= 6
36 = 6
Rearrange x-terms
together and y-terms
together.
Complete the square
for the expression
x 2 + 4x by adding 4 to
both sides because
4 2
= 4.
2
Complete the square
for the expression
y 2 10y by adding 25
to both sides because
10 2
= 25.
2
Convert to factored
form.
Find center at (h, k ).
Find the radius.
Option A: This response results from a sign error when finding the center of
the circle in step 4.
Option C: This response is the result of a sign error when finding the center
of the circle in step 4, and not taking the square root of the constant term to
determine the radius in step 5.
Option D: This response results from not taking the square root of the constant
term to determine the radius in step 5.
23
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
8 in.
6 in.
B.
C.
D.
*correct answer
24
Option B: This is the correct answer. The formula for the volume of a
1
pyramid is V = Bh, where B is the area of the base and h is the height.
3
1
For this pyramid, V = (61.94)(8) = 165.17 cubic inches.
3
1
by the product of the
3
height, the given side length, and the number of sides of the base.
Option C: This response results from computing the product of the height,
the given side length, and the number of sides of the base.
Option D: This answer results from failing to multiply
area of the base and the height.
1
by the product of the
3
25
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
Aviary
5 yards
B.
C.
D.
*correct answer
Geometry EOC Sample Test ItemsOctober 2013
26
Math Practice 4 (Model with mathematics.): Students must model the given
real-world situation using the equation for volume and the process to
determine density.
Option B: This is the correct answer. The volume of the aviary is given by the
product of 5 yards, 12 yards, and 9 yards. The density of birds per cubic yard
is given by the quotient of the number of birds and the volume of the aviary.
Option A: This response is the result of incorrectly using a sum of the given
dimensions instead of calculating the volume of the aviary.
Option C: This answer results from incorrectly computing the number of birds
per square yard of floor space.
Option D: This response is the result of calculating the number of cubic yards
per bird rather than the number of birds per cubic yard.
27
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
None
28
fenced-in
area
width
house
12 feet
B.
16 feet
C.
24 feet
D.
32 feet
*correct answer
This item requires students to solve an optimization problem involving area and
perimeter. The most efficient pathway to solve the problem is to write equations
to represent the area and perimeter. Then, students may solve for one variable
in the linear perimeter formula and substitute this expression into the area
formula. This will lead to a quadratic equation representing the area in terms of
the length of one of the sides. Finding the maximum point of this parabola will
determine the side length that will maximize the area.
This item is linked to three of the Math Practices.
29
Math Practice 4 (Model with mathematics.): Students must model the given
real-world situation with equations for the area and perimeter.
Determine necessary
equations to use.
Step 3
48 x
A=x
2
Step 4
A=
Step 1
Step 2
Step 5
x=
1 2
x + 24x
2
b
=
2a
24
= 24
1
2
2
Step 4
A = 48w 2w 2
Step 5
b
48
w =
=
= 12
2a 2(2)
Simplify expression.
Find the width.
(incorrect to stop at
this point)
Option B: The student incorrectly assumes that a square will maximize the
area of the fenced-in area without doing any calculations. This is true for
situations where all four sides of the enclosed area are being used.
Option D: The student incorrectly assumes that the largest given value for x
will result in the maximum area without doing any calculations.
30
Constructed-Response Item
This section presents a constructed-response item and samples of student
responses that received scores of 4, 3, 2, 1, 1 for minimal understanding, and 0.
This section also includes information used to score this constructed-response
item: an exemplary response, an explanation of how points are assigned, and a
specific scoring rubric. In addition to the online resources available for all test
questions, students have access to the Geometry Typing Help (pages 44 and 45),
which describes how to enter special characters, symbols, and formatting into
typed responses.
Constructed-Response Item
Conceptual Category: GGeometry
Domain:
Cluster:
Standard:
Math Practices:
Calculator:
Calculator allowed
Tool Restrictions:
31
top of stairs
5 in.
12 in.
5 in.
12 in.
5 in.
Leah needs to add a wheelchair ramp over her stairs. The ramp
will start at the top of the stairs. Each stair makes a right angle
with each riser.
Part A
1
. To the nearest
12
hundredth of a foot, what is the shortest length of ramp that Leah
can build and not exceed the maximum slope? Show your work
or explain your answer.
The ramp must have a maximum slope of
Length of ramp:
Work/Explanation:
Part B
Leah decides to build a ramp that starts at the top of the stairs
and ends 18 feet from the base of the bottom stair. To the nearest
hundredth of a foot, what is the length of the ramp?
(student enters response in text box)
32
Part C
To the nearest tenth of a degree, what is the measure of the angle
created by the ground and the ramp that Leah builds in part B?
(student enters response in text box)
This item requires students to reason about how slope, angle measures,
trigonometric ratios, and the Pythagorean Theorem relate mathematically.
Students must determine a horizontal measure for a leg of a right triangle
1
that results in a hypotenuse slope of no more than
. Students then use the
12
Pythagorean Theorem to find the measure of the hypotenuse in two different
right triangles. Finally, students must use a trigonometric ratio to determine
the measure of an angle in a right triangle.
This item is linked to five of the Math Practices.
33
Scoring Information
Scoring Rubric
Score
Description
Exemplary Response
Part A
Length of ramp: 15.05 feet
Work/Explanation: First find length of base of triangle (x)
1.25ft/x = 1/12
1.25*12 = 1*x
x = 15
Then find hypotenuse/ramp (c)
1.25^2 + 15^2 = c^2
226.5625 = c^2
c =* 15.05
Part B
20.04 feet
Part C
3.6 degrees
Points Assigned
Part A: 1 point for correct length
Part A: 1 point for complete and accurate work or explanation
Part B: 1 point for correct length
Part C: 1 point for correct degree measure
Note: The student may receive credit for part C for correctly finding the angle
measure based on an incorrect value from part B.
34
35
36
Score Point 3
The following authentic student responses show the work of two students who
each earned a score of 3 for their responses. A score of 3 is received when a
student completes three of the four required components of the task. There may
be simple errors in calculations or some confusion with communicating his or
her ideas effectively.
Score Point 3, Student 1
Part A
15.05 feet
The shortest length of ramp that Leah can build and not exceed the maximum slope
is 15.05 feet. In order to get the length of the ramp, I first used the slope (1/12) as
a ratio to find the length of the stairs. By multiplying the total heights of the risers,
15 inches, by 12, I found that the total length of the stairs would be 180 inches.
By using the Pythagorean Theorum to find the length of the hypotenuse, I found that
15^2 + 180^2 =* 180.62^2. I then divided the hypotenuse by 12 to get the length of
the ramp in feet. 180.62 / 12 = 15.05 feet.
Part B
19.04 feet
Part C
3.8 degrees
This student calculates the correct length and provides a correct and
complete explanation of his or her work for part A. The student does not
provide the correct length of the ramp in part B; however, the students
response in part C is correct based on this incorrect value in part B.
37
38
Score Point 2
The following authentic student responses show the work of two students who
each earned a score of 2 for their responses. A score of 2 is received when a
student completes two of the four required components of the task. There may
be simple errors in calculations, one or two missing responses, or unclear or
incorrect communications of his or her ideas.
Score Point 2, Student 1
Part A
15.05
I used the pythagorem thereom to get find that , with a slope of 1/12, each in down
would take 12.04 inches. With this I could find the length of the ramp by multiplying
it by the height,15 inches. Which came out to be 180.6 inches. Next, I needed to
convert the answer to feet, which gave me 15.05 feet.
Part B
18.04
Part C
6.90%
This student provides the correct length of the ramp in part A and correctly
reasons that each vertical rise of 1 inch will result in a hypotenuse length of
12.04 inches and multiplies that length by the height. The student does not
provide the correct length of the ramp in part B or the correct angle measure
in part C.
Score Point 2, Student 2
Part A
15.05 feet
I used a ratio and Pythagorean Theorem. Then I found how many feet.
Part B
20.04 feet
Part C
30 degrees
The student provides the correct length for the ramp; however, the explanation
is incomplete. There is not enough detail to receive full credit for their reasoning.
The student provides the correct length of the ramp in part B, but the angle
measure provided in part C is incorrect.
39
Score Point 1
The following authentic student responses show the work of two students who
each earned a score of 1 for their responses, and two students who each earned
a score of 1 for demonstrating minimal understanding. A score of 1 is received
when a student correctly addresses one of the four required components, or
demonstrates at least minimal understanding of the key concepts.
Score Point 1, Student 1
Part A
180.62
First I started out with what I know is going to be the height of the triangle we are
going to make with the ramp and the floor. Since slope = rise/run I then figured out
that the base of our tiangle would have to be 180in for the ratio to match 1/12. After
that I simply used the pathagoream therom to figue out the hypotenuse or in this
case, our ramp.
Part B
216.52
Part C
4
This student provides the correct length of the ramp in part A, but expresses
the measurement in the wrong unit. The item specifically asks for the length
of the ramp in feet; however, the reasoning provided is correct and complete,
so the student earned 1 point. The student does not provide the correct length
of the ramp in part B or the correct angle measure in part C.
40
41
42
Score Point 0
The following samples show the work of two students who each earned a score
of 0. A score of 0 is received when a student response is incorrect, irrelevant,
too brief to evaluate, or blank.
Score Point 0, Student 1
Part A
12/13
12^2 + 5^2= 169 sqrt(169)= 13 rise/run 24/26 = 12/13
Part B
51 feet
Part C
90 degrees
This student does not provide the correct length of the ramp in part A.
The reasoning provided uses the Pythagorean Theorem, but it is applied using
the wrong numbers. There is no mention of the slope of the ramp. The student
does not provide the correct ramp length in part B or the correct angle measure
in part C.
Score Point 0, Student 2
Part A
0.0ft
If you divide 12 into 1 it comes out to be 0.083. round that to the nearest hundredth of
a foot it comes out to be 0.1ft., anything less then 0.ft is 0.0ft.
Part B
1.5ft
Part C
2ft
This student does not provide the correct length of the ramp in part A.
The reasoning provides no useful information for calculating the length of
the ramp. The student does not provide the correct ramp length in part B.
The answer provided for the missing angle in part C is incorrect in value and
is expressed in the wrong unit of measure.
43
As of July 2014, the Geometry Typing Help has been updated to include typing
complex roots and the trigonometric functions. A current version can be found
at https://www.louisianaeoc.org/Documents/GeometryTypingHelp.pdf
multiplication symbol
division symbol
mixed number
32
exponent
square root
about equal to
45
degree symbol
3 * 4 = 12
12 / 3 = 4
/
forward slash
(12 7)/(3 1)
Note: Parentheses
are required.
2 3/4
^
caret (SHIFT + 6)
3^2 = 9
(pi)
Area = 9(pi)
square inches
pi symbol
3 x 4 = 12
/
forward slash
fraction or ratio
3
2
4
3. Example:
>=
greater than sign, followed
by equals sign
<=
less than sign, followed by
equals sign
sqrt()
the letters sqrt, with the
radicand in parentheses
=*
equals sign, followed by
asterisk (SHIFT + 8)
degrees
(spell out the word)
y >= 13
y <= 13
sqrt(4) = 2
(pi) =* 3.14
The angle measures 45
degrees.
44
triangle
(spell out the word)
m q
perpendicular symbol
perpendicular
(spell out the word)
Line m is perpendicular to
line q.
m || q
parallel symbol
||
two pipes (SHIFT +
backslash)
OR parallel
(spell out the word)
m || q
OR
Line m is parallel to line q.
ABC
RST
congruence symbol
congruent
(spell out the word)
STU VWX
similarity symbol
similar
(spell out the word)
line segment
line segment
(spell out the words)
line
line
(spell out the word)
Line AB is parallel to
line CD.
ray
ray
(spell out the word)
angle symbol
ABC
triangle symbol
45
Pythagorean Theorem
a2 + b2 = c2
Sphere
3.14
Surface Area = 4 r
4
Volume = r 3
3
Cone
Surface Area
h s
rs
r 2
1
Volume = r 2 h
3
Cylinder
r
Surface Area = 2 r 2 + 2 rh
Volume = r 2 h
Regular Pyramid
B = area of base
L = area of lateral surfaces
Surface Area = B + L
Volume = 1 Bh
3
Rectangular Solid
h
l
46
Circle
Area = r 2
Circumference = 2 r
Rectangle
w
l
Area = lw
Perimeter = 2l
2w
Trapezoid
b1
h
Area =
1
h ( b1+ b 2 )
2
Area =
1
bh
2
b2
Triangle
h
b
Parallelogram
Area = bh
h
b
Cartesian Distance
Formula
Point 1: ( x1 , y1 )
Point 2 : ( x 2, y2 )
d=
( x2 x1 ) 2 + ( y2 y1 ) 2
47