NATIONAL
MATH + SCIENCE
INITIATIVE
Mathematics
a2 b2 c2
a2 b2 b2 c2 b2
a2 c2 b2
a c b
c2 b2
Since a 0,
a
I
ABOUT THIS LESSON
2
c2 b2
LEVEL
Algebra 1 or Math 1 in a unit on solving multi-step
equations
n this lesson, students practice the essential skill
of solving literal equations for a specific variable
by showing the algebraic steps of their solution.
In addition, students also must explain the meaning
of the “new” equation.
OBJECTIVES
Students will
● solve literal equations.
● explain the meaning of geometric formulas.
MODULE/CONNECTIONS TO AP*
Rate of Change: Related Rates
*Advanced Placement and AP are registered trademarks of the
College Entrance Examination Board. The College Board was not
involved in the production of this product.
MODALITY
NMSI emphasizes using multiple representations
to connect various approaches to a situation in
order to increase student understanding. The lesson
provides multiple strategies and models for using
those representations indicated by the darkened
points of the star to introduce, explore, and reinforce
mathematical concepts and to enhance conceptual
understanding.
P
G
V
N
A
P A G E S
a 2
T E A C H E R
2
Literal Equations –
Geometric Formulas
P – Physical
V – Verbal
A – Analytical
N – Numerical
G – Graphical
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
i
Mathematics—Literal Equations – Geometric Formulas
COMMON CORE STATE STANDARDS FOR
MATHEMATICAL CONTENT
COMMON CORE STATE STANDARDS FOR
MATHEMATICAL PRACTICE
This lesson addresses the following Common
Core Standards for Mathematical Content. The
lesson requires that students recall and apply each
of these standards rather than providing the initial
introduction to the specific skill.
These standards describe a variety of instructional
practices based on processes and proficiencies
that are critical for mathematics instruction.
NMSI incorporates these important processes and
proficiencies to help students develop knowledge
and understanding and to assist them in making
important connections across grade levels. This
lesson allows teachers to address the following
Common Core State Standards for Mathematical
Practice.
MP.6: Attend to Precision.
Students use the defined variables to explain
the meaning of each re-expressed formula.
T E A C H E R
P A G E S
Targeted Standards
A-CED.4: Rearrange formulas to highlight a
quantity of interest, using the same
reasoning as in solving equations.
For example, rearrange Ohm’s law
V = IR to highlight resistance R.
See questions 1-2
ii
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
Mathematics—Literal Equations – Geometric Formulas
FOUNDATIONAL SKILLS
The following skills lay the foundation for concepts
included in this lesson:
● Solve multi-step equations
● Solve equations involving radicals
ASSESSMENTS
The following formative assessment is embedded in
this lesson:
● Students engage in independent practice.
MATERIALS AND RESOURCES
Student Activity Pages
T E A C H E R
●
P A G E S
The following additional assessments are located on
our website:
● Rate of Change: Related Rates –
Algebra 1 Free Response Questions
● Rate of Change: Related Rates –
Algebra 1 Multiple Choice Questions
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
iii
Mathematics—Literal Equations – Geometric Formulas
S
TEACHING SUGGESTIONS
T E A C H E R
P A G E S
tudents in mathematics and science must
be able to manipulate literal equations to
solve for a particular variable. Most students
seem to manage reasonably well when solving
for one variable in an equation that involves only
that variable along with other “numbers.” When
the equation involves several variables, however,
students often struggle to isolate the required
variable.
One strategy for helping students decide on the
appropriate solution steps is to create a second,
similar equation by replacing all of the variables
except the variable to be isolated with numbers.
Tell students that they may not simplify the numbers
by performing the indicated operations (addition,
multiplication, etc.). Next have students solve the
“numerical” equation and the “literal” equation side
by side. Other possible strategies include:
(1) write the variable to be isolated in a different
color or draw a circle or box around it to set it apart
from the other variables, and/or (2) have students
write in any implied operations or groupings.
Experience and practice with this skill will benefit
students at all levels.
This lesson can be used as a review or as an
assessment. In order to reach the lesson’s full
potential, students need to not only perform the
algebraic manipulations, but also to consider what
each revised formula means.
Suggested modifications for additional scaffolding
include the following:
1
Provide a template explaining the meaning of
the “new” equation. For example:
1a
The ________________ of the rectangle
is the ________________ divided by the
________________ of the rectangle.
1b
The ________________ of the triangle is
________________ the ________________
divided by the height of the ___________.
1c
The value of pi is the ________________
of a circle ________________ by
________________ the radius.
1f
Provide a template which presents the
formula, solving for the required variable, and
simplifying the new formula. Provide space
for the student to explain the reasoning. For
example,
S = Ph + 2 B Subtract Ph from each
− Ph − Ph
side of the equation.
S − Ph = 2 B Simplify.
S − Ph 2 B Divide each side of the
=
2
2
equation by 2.
S − Ph
= B Simplify.
2
B=
S − Ph
2
If preferred, write
equation with variable
solved for on
the left.
The area of the base of the prism is one-half of
the difference between the total surface area of
a prism and the product of the perimeter of its
base and the height of the prism.
iv
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
Mathematics—Literal Equations – Geometric Formulas
NMSI CONTENT PROGRESSION CHART
In the spirit of NMSI’s goal to connect mathematics across grade levels, a Content Progression Chart for
each module demonstrates how specific skills build and develop from sixth grade through pre-calculus in an
accelerated program that enables students to take college-level courses in high school, using a faster pace to
compress content. In this sequence, Grades 6, 7, 8, and Algebra 1 are compacted into three courses. Grade 6
includes all of the Grade 6 content and some of the content from Grade 7, Grade 7 contains the remainder
of the Grade 7 content and some of the content from Grade 8, and Algebra 1 includes the remainder of the
content from Grade 8 and all of the Algebra 1 content.
The complete Content Progression Chart for this module is provided on our website and at the beginning
of the training manual. This portion of the chart illustrates how the skills included in this particular lesson
develop as students advance through this accelerated course sequence.
Solve literal
equations
(perimeter, area,
and volume).
Algebra 1
Skills/Objectives
Solve literal
equations
(perimeter, area,
and volume).
Geometry
Skills/Objectives
Solve literal
equations
(perimeter, area,
and volume).
Algebra 2
Skills/Objectives
Solve literal
equations
(perimeter, area,
and volume).
Pre-Calculus
Skills/Objectives
Solve literal
equations
(perimeter, area,
and volume).
T E A C H E R
Solve literal
equations
(perimeter, area,
and volume).
7th Grade
Skills/Objectives
P A G E S
6th Grade
Skills/Objectives
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
v
Mathematics—Literal Equations – Geometric Formulas
T E A C H E R
P A G E S
vi
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
NATIONAL
MATH + SCIENCE
INITIATIVE
Mathematics
Literal Equations – Geometric Formulas
c.
f.
S = Ph + 2 B
− Ph − Ph
S − Ph = 2 B
S − Ph 2 B
=
2
2
S − Ph
=B
2
S − Ph
B=
2
The area of the base of the prism is one-half
of the difference between the total surface
area of a prism and the product of the
perimeter of its base and the height of
the prism.
P A G E S
bh
⋅2
2
2A bh
=
h
h
2A
=b
h
2A
b=
h
The base of the triangle is twice the area
divided by the height of the triangle.
b. 2 ⋅ A =
S Ph
=
h h
S
=P
h
S
P=
h
The perimeter of the base of the prism is the
lateral surface area of a prism divided by its
height.
e.
T E A C H E R
Answers
A lw
1. a. =
l
l
A
=w
l
A
w=
l
The width of the rectangle is the area divided
by the length of the rectangle.
(b1 + b2 )h
⋅2
2
2 A (b1 + b2 )h
=
h
h
2A
= b1 + b2
h
− b2
− b2
g. 2 ⋅ A =
The value of pi is the circumference of a
circle divided by twice the radius.
d.
V Bh
=
h
h
V
=B
h
V
B=
h
The area of the base of the prism is the
volume of a prism divided by the height of the
prism.
2A
− b2 = b1
h
2A
b1 =
− b2
h
The length of one of the bases of a trapezoid
is twice the area of the trapezoid divided
by its height minus the second base of the
trapezoid.
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
vii
Mathematics—Mathematics
P 2(l + w)
=
2
2
h.
k.
P
= l+w
2
−l −l
P
−l = w
2
P
w = −l
2
The width of a rectangle is one-half the
perimeter of the rectangle minus its length.
T E A C H E R
P A G E S
1
i. 3 ⋅ V = 3 ⋅ Bh
3
3V Bh
=
h
h
3V
=B
h
3V
B=
h
The area of the base of the pyramid is three
times the volume of the pyramid divided by
its height.
The height of the cylinder is the difference
between the total surface area of the cylinder
and twice the product of pi and the radius
of the cylinder squared divided by twice the
product of pi and its radius.
2.
a.
S 6s 2
=
6
6
S
= s2
6
S
= s2
6
S
= s
6
S
s > 0 because it is a length
6
The length of the side of a cube is the square
root of the quotient of the surface area of the
cube and six.
j.
s=
The value of pi is the area of a circle divided
by the radius squared.
viii
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
Mathematics—Mathematics
b. a 2 + b 2 = c 2
− b2
− b2
a 2 = c2 − b2
a 2 = c2 − b2
a = c2 − b2
But a > 0 so a = c 2 − b 2
The leg of a right triangle is the square root
of the difference between the square of the
hypotenuse and the square of the other leg of
the same right triangle.
3V
= r3
4π
3
r3 =
3
3V
4π
3V
4π
The radius of the sphere is the cube root of
the quotient of three times the volume of the
sphere and four times pi.
r=
3
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
T E A C H E R
1
1 3V
⋅
= ⋅ πr 3
π 4
π
P A G E S
3
3 4
c. ⋅V = ⋅ πr 3
4
4 3
ix
T E A C H E R
P A G E S
Mathematics—Mathematics
x
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
NATIONAL
MATH + SCIENCE
INITIATIVE
Mathematics
Literal Equations – Geometric Formulas
1. For each equation, solve for the given variable. Explain the meaning of the “new” equation.
Example:
Solve P = 4s, for s, where P = the perimeter of a square and s = the side of a square. To receive full
credit for this section, first show the steps needed to solve for the given variable.
P
. Second, explain the meaning
4
of the answer. In this case, an acceptable answer would be: The side of a square is the perimeter of the
square divided by 4.
In order to solve P = 4s for s, divide both sides by 4. The result is s =
a. Solve A = lw for w, where A = area of a rectangle, l = length of the rectangle, and w = width
of the rectangle.
bh
for b, where A = area of a triangle, b = base of the triangle, and h = height
2
of the triangle.
b. Solve A =
c. Solve C = 2πr for π, where C = circumference of a circle and r = the radius of the circle.
d. Solve V = Bh for B, where V = volume of a prism, B = the area of the base of the prism,
and h = the height of the prism.
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
1
Mathematics—Literal Equations – Geometric Formulas
e. Solve S = Ph for P, where S = the lateral surface area of a prism, P = the perimeter of the base
of the prism, and h = the height of the prism.
f. Solve S = Ph + 2B for B, where S = the total surface area of a prism, P = the perimeter of the base
of the prism, h = the height of the prism, and B = the area of the base of the prism.
(b1 + b2 )h
for b1 , where A = area of a trapezoid, h = height of the trapezoid, and b 1 and
2
b 2 are the lengths of the two bases of the trapezoid.
g. Solve A =
h. Solve P = 2(l + w) for w, where P is a rectangle’s perimeter, l is the length, and w is the width.
1
i. Solve V = Bh for B, where V = the volume of a pyramid, B = the area of the base of the pyramid,
3
and h = the height of the pyramid.
j. Solve A = πr² for π where A = the area of a circle and r = the radius of the circle.
2
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
Mathematics—Literal Equations – Geometric Formulas
k. Solve S = 2πrh + 2πr² for h where S = the total surface area of a cylinder, r = the radius of the
cylinder, and h = the height of the cylinder.
2. The formula to determine the area of a square is A = s 2 where A represents the area of the square and s
is the side of a square.
A = s2
A = s2
A =| s|
Since s is the side of the square, s > 0 , s = A .
When appropriate, use this process to solve the following questions. Remember to show the steps
needed to solve for the given variable and to explain the meaning of the answer.
a. Solve S = 6s² for s, where S = the surface area of a cube and s = the length of the side of a cube.
b. Solve a 2 + b 2 = c 2 for a, where a and b are the legs of a right triangle and c is the hypotenuse of the
same right triangle.
4
c. Solve V = πr 3 for r, where V is the volume of a sphere and r is the radius of the sphere.
3
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.
3
Mathematics—Literal Equations – Geometric Formulas
4
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org.