lesson 22: surface area - engageny
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NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
308
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Lesson 22: Surface Area
Student Outcomes
Students find the surface area of three-dimensional objects whose surface area is composed of triangles and
quadrilaterals, specifically focusing on pyramids. They use polyhedron nets to understand that surface area is
simply the sum of the area of the lateral faces and the area of the base(s).
Lesson Notes
Before class, teachers need to make copies of the composite figure for the Opening Exercise on cardstock. To save class
time, they could also cut these nets out for students.
Classwork
Opening Exercise (5 minutes)
Make copies of the composite figure on cardstock and have students cut and fold the net to form the three-dimensional
object.
Opening Exercise
What is the area of the composite figure in the diagram? Is the diagram a net for a three-dimensional image? If so,
sketch the image. If not, explain why.
There are four unit squares in each square of the
figure. There are ๐๐ total squares that make up the
figure, so the total area of the composite figure is
๐จ = ๐๐ โ ๐ ๐ฎ๐ง๐ข๐ญ๐ฌ๐ = ๐๐ ๐ฎ๐ง๐ข๐ญ๐ฌ๐.
The composite figure does represent the net of a
three-dimensional figure. The figure is shown below.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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Example 1 (5 minutes)
Pyramids are formally defined and explored in more depth in Module 6. Here, we simply introduce finding the surface
area of a pyramid and ask questions designed to elicit the formulas from students. For example, ask how many lateral
faces there are on the pyramid; then, ask for the triangle area formula. Continue leading students toward stating the
formula for total surface area on their own.
Example 1
The pyramid in the picture has a square base, and its lateral faces are triangles that are exact copies of one another. Find
the surface area of the pyramid.
The surface area of the pyramid consists of one square base and four lateral triangular faces.
๐ณ๐จ = ๐ (๐๐
๐๐) ๐ฉ = ๐๐
๐ณ๐จ = ๐ โ๐๐
(๐ ๐๐ฆ โ ๐ ๐๐ฆ) ๐ฉ = (๐ ๐๐ฆ)๐
๐ณ๐จ = ๐(๐ ๐๐ฆ โ ๐ ๐๐ฆ) ๐ฉ = ๐๐ ๐๐ฆ๐
๐ณ๐จ = ๐(๐๐ ๐๐ฆ๐) The pyramidโs base area is ๐๐ ๐๐ฆ๐.
๐ณ๐จ = ๐๐ ๐๐ฆ๐
The pyramidโs lateral area is ๐๐ ๐๐ฆ๐. ๐บ๐จ = ๐ณ๐จ + ๐ฉ
๐บ๐จ = ๐๐ ๐๐ฆ๐ + ๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐๐ ๐๐ฆ๐
The surface area of the pyramid is ๐๐๐ ๐๐ฆ๐.
Example 2 (4 minutes): Using Cubes
Consider providing 13 interlocking cubes to small groups of students so they may construct a model of the diagram
shown. Remind students to count faces systematically. For example, first consider only the bottom 9 cubes. This
structure has a surface area of 30 (9 at the top, 9 at the bottom, and 3 on each of the four sides). Now consider the four
cubes added at the top. Since we have already counted the tops of these cubes, we just need to add the four sides of
each. 30 + 16 = 46 total square faces, each with side length 14
inch.
Example 2: Using Cubes
There are ๐๐ cubes glued together forming the solid in the diagram. The edges of each cube are ๐๐
inch in length. Find the
surface area of the solid.
The surface area of the solid consists of ๐๐ square faces, all having side lengths of ๐๐
inch. The area of a square having
sides of length ๐๐
inch is ๐
๐๐ ๐ข๐ง๐ .
๐บ๐จ = ๐๐ โ ๐จ๐ฌ๐ช๐ฎ๐๐ซ๐
๐บ๐จ = ๐๐ โ๐
๐๐ ๐ข๐ง๐
๐บ๐จ =๐๐
๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐
๐ข๐ง๐ The surface are of the solid is ๐๐๐
๐ข๐ง๐.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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Compare and contrast the methods for finding surface area for pyramids and prisms. How are the methods
similar?
When finding the surface area of a pyramid and a prism, we find the area of each face and then add
these areas together.
How are they different?
Calculating the surface area of pyramids and prisms is different because of the shape of the faces. A
prism has two bases and lateral faces that are rectangles. A pyramid has only one base and lateral
faces that are triangles.
Example 3 (15 minutes)
Example 3
Find the total surface area of the wooden jewelry box. The sides and bottom of the box are all ๐๐ inch thick.
What are the faces that make up this box?
The box has a rectangular bottom, rectangular lateral faces,
and a rectangular top that has a smaller rectangle removed
from it. There are also rectangular faces that make up the
inner lateral faces and the inner bottom of the box.
How does this box compare to other objects that you have
found the surface area of?
The box is a rectangular prism with a smaller rectangular prism
removed from its inside. The total surface area is equal to the
surface area of the larger right rectangular prism plus the
lateral area of the smaller right rectangular prism.
Large Prism: The surface area of the large right rectangular
prism makes up the outside faces of the box, the rim of the box,
and the inside bottom face of the box.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ท โ ๐ ๐ฉ = ๐๐
๐ณ๐จ = ๐๐ ๐ข๐ง.โ ๐ ๐ข๐ง. ๐ฉ = ๐๐ ๐ข๐ง.โ ๐ ๐ข๐ง.
๐ณ๐จ = ๐๐๐ ๐ข๐ง๐ ๐ฉ = ๐๐ ๐ข๐ง๐
The lateral area is ๐๐๐ ๐ข๐ง๐.
The base area is ๐๐ ๐ข๐ง๐.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐บ๐จ = ๐๐๐ ๐ข๐ง๐ + ๐(๐๐ ๐ข๐ง๐)
๐บ๐จ = ๐๐๐ ๐ข๐ง๐ + ๐๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐ ๐ข๐ง๐
The surface area of the larger prism is ๐๐๐ ๐ข๐ง๐.
Small Prism: The smaller prism is ๐๐
๐ข๐ง. smaller in
length and width and ๐๐
๐ข๐ง. smaller in height due to
the thickness of the sides of the box.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ท โ ๐
๐ณ๐จ = ๐ (๐๐
๐ ๐ข๐ง. +๐
๐
๐ ๐ข๐ง. ) โ ๐
๐
๐ ๐ข๐ง.
๐ณ๐จ = ๐(๐๐ ๐ข๐ง. +๐ ๐ข๐ง. ) โ ๐๐
๐ ๐ข๐ง.
๐ณ๐จ = ๐(๐๐ ๐ข๐ง. ) โ ๐๐
๐ ๐ข๐ง.
๐ณ๐จ = ๐๐ ๐ข๐ง.โ ๐๐
๐ ๐ข๐ง.
๐ณ๐จ = ๐๐ ๐ข๐ง๐ +๐๐
๐ ๐ข๐ง๐
๐ณ๐จ = ๐๐ ๐ข๐ง๐ + ๐๐๐
๐ ๐ข๐ง๐
๐ณ๐จ = ๐๐๐๐
๐ ๐ข๐ง๐
The lateral area is ๐๐๐๐๐
๐ข๐ง๐.
Surface Area of the Box
๐บ๐จ๐๐จ๐ฑ = ๐บ๐จ + ๐ณ๐จ
๐บ๐จ๐๐๐ = ๐๐๐ ๐ข๐ง๐ + ๐๐๐๐๐
๐๐๐
๐บ๐จ๐๐๐ = ๐๐๐๐๐
๐ข๐ง๐
The total surface area of the box is ๐๐๐๐๐
๐ข๐ง๐.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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Discussion (5 minutes): Strategies and Observations from Example 3
Call on students to provide their answers to each of the following questions. To
encourage a discussion about strategy, patterns, arguments, or observations, call on more
than one student for each of these questions.
What ideas did you have to solve this problem? Explain.
Answers will vary.
Did you make any mistakes in your solution? Explain.
Answers will vary; examples include the following:
I subtracted 12
inch from the depth of the box instead of 14
inch;
I subtracted only 14
inch from the length and width because I
didnโt account for both sides.
Describe how you found the surface area of the box and what that surface area is.
Answers will vary.
Closing (2 minutes)
What are some strategies for finding the surface area of solids?
Answers will vary but might include creating nets, adding the areas of polygonal faces, and counting
square faces and adding their areas.
Exit Ticket (9 minutes)
Scaffolding:
To help students visualize the various faces involved on this object, consider constructing a similar object by placing a smaller shoe box inside a slightly larger shoe box. This will also help students visualize the inner surfaces of the box as the lateral faces of the smaller prism that is removed from the larger prism.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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Name Date
Lesson 22: Surface Area
Exit Ticket
1. The right hexagonal pyramid has a hexagon base with equal-length sides. The
lateral faces of the pyramid are all triangles (that are exact copies of one
another) with heights of 15 ft. Find the surface area of the pyramid.
2. Six cubes are glued together to form the solid shown in the
diagram. If the edges of each cube measure 112
inches in length,
what is the surface area of the solid?
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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Exit Ticket Sample Solutions
1. The right hexagonal pyramid has a hexagon base with equal-length sides. The lateral faces of the pyramid are all
triangles (that are exact copies of one another) with heights of ๐๐ ๐๐ญ. Find the surface area of the pyramid.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ โ๐๐
(๐๐) ๐ฉ = ๐จrectangle + ๐๐จtriangle
๐ณ๐จ = ๐ โ๐๐
(๐ ๐๐ญ.โ ๐๐ ๐๐ญ. ) ๐ฉ = (๐ ๐๐.โ ๐ ๐๐. ) + ๐ โ๐๐
(๐ ๐๐.โ ๐ ๐๐. )
๐ณ๐จ = ๐ โ ๐๐ ๐๐ญ๐ ๐ฉ = ๐๐ ๐๐ญ๐ + (๐ ๐๐ญ.โ ๐ ๐๐ญ. )
๐ณ๐จ = ๐๐๐ ๐๐ญ๐ ๐ฉ = ๐๐ ๐๐ญ๐ + ๐๐ ๐๐ญ๐
๐ฉ = ๐๐ ๐๐ญ๐
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐บ๐จ = ๐๐๐ ๐๐ญ๐ + ๐๐ ๐๐ญ๐
๐บ๐จ = ๐๐๐ ๐๐ญ๐
The surface area of the pyramid is ๐๐๐ ๐๐ญ๐.
2. Six cubes are glued together to form the solid shown in the diagram. If the edges of each cube measure ๐๐๐
inches in
length, what is the surface area of the solid?
There are ๐๐ square cube faces showing on the surface area of the solid (๐ each from the top and bottom view, ๐
each from the front and back view, ๐ each from the left and right side views, and ๐ from the โinsideโ of the front).
๐จ = ๐๐ ๐บ๐จ = ๐๐ โ (๐๐๐
๐ข๐ง๐)
๐จ = (๐๐๐
๐ข๐ง. )๐
๐บ๐จ = ๐๐ ๐ข๐ง๐ +๐๐๐
๐ข๐ง๐
๐จ = (๐๐๐
๐ข๐ง. ) (๐๐๐
๐ข๐ง. ) ๐บ๐จ = ๐๐ ๐ข๐ง๐ + ๐ ๐ข๐ง๐ +๐๐
๐ข๐ง๐
๐จ = ๐๐๐
๐ข๐ง. (๐ ๐ข๐ง. +๐๐
๐ข๐ง. ) ๐บ๐จ = ๐๐๐๐
๐ข๐ง๐
๐จ = (๐๐๐
๐ข๐ง.โ ๐ ๐ข๐ง. ) + (๐๐๐
๐ข๐ง.โ๐๐
๐ข๐ง. )
๐จ = ๐๐๐
๐ข๐ง๐ +๐๐
๐ข๐ง๐
๐จ = ๐๐๐
๐ข๐ง๐ +๐๐
๐ข๐ง๐
๐จ = ๐๐๐
๐ข๐ง๐ = ๐๐๐
๐ข๐ง๐ The surface area of the solid is ๐๐๐๐
๐ข๐ง๐.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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Problem Set Sample Solutions
1. For each of the following nets, draw (or describe) the solid represented by the net and find its surface area.
a. The equilateral triangles are exact copies.
The net represents a triangular pyramid where the three lateral
faces are identical to each other and the triangular base.
๐บ๐จ = ๐๐ฉ since the faces are all the same size and shape.
๐ฉ =๐๐
๐๐ ๐บ๐จ = ๐๐ฉ
๐ฉ =๐๐
โ ๐ ๐ฆ๐ฆ โ ๐๐๐
๐ฆ๐ฆ ๐บ๐จ = ๐ (๐๐๐
๐๐ ๐ฆ๐ฆ๐)
๐ฉ =๐๐
๐ฆ๐ฆ โ ๐๐๐
๐ฆ๐ฆ ๐บ๐จ = ๐๐๐ ๐ฆ๐ฆ๐ +๐
๐๐ ๐ฆ๐ฆ๐
๐ฉ =๐๐๐
๐ฆ๐ฆ๐ +๐๐๐๐
๐ฆ๐ฆ๐ ๐บ๐จ = ๐๐๐๐๐
๐ฆ๐ฆ๐
๐ฉ =๐๐๐๐๐
๐ฆ๐ฆ๐ +๐๐๐๐
๐ฆ๐ฆ๐
๐ฉ =๐๐๐๐๐
๐ฆ๐ฆ๐ The surface area of the triangular pyramid is ๐๐๐๐๐
๐ฆ๐ฆ๐.
๐ฉ = ๐๐๐
๐๐ ๐ฆ๐ฆ๐
b. The net represents a square pyramid that has four identical
lateral faces that are triangles. The base is a square.
๐บ๐จ = ๐ณ๐จ + ๐ฉ
๐ณ๐จ = ๐ โ๐๐
(๐๐) ๐ฉ = ๐๐
๐ณ๐จ = ๐ โ๐๐
(๐๐ ๐ข๐ง.โ ๐๐๐๐
๐ข๐ง. ) ๐ฉ = (๐๐ ๐ข๐ง.)๐
๐ณ๐จ = ๐ (๐๐ ๐ข๐ง.โ ๐๐๐๐
๐ข๐ง.) ๐ฉ = ๐๐๐ ๐ข๐ง๐
๐ณ๐จ = ๐(๐๐๐ ๐ข๐ง๐ + ๐ ๐ข๐ง๐)
๐ณ๐จ = ๐๐๐ ๐ข๐ง๐ + ๐๐ ๐ข๐ง๐
๐ณ๐จ = ๐๐๐ ๐ข๐ง๐
๐บ๐จ = ๐ณ๐จ + ๐ฉ
๐บ๐จ = ๐๐๐ ๐ข๐ง๐ + ๐๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐ ๐ข๐ง๐
The surface area of the square pyramid is ๐๐๐ ๐ข๐ง๐.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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2. Find the surface area of the following prism.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ท โ ๐
๐ณ๐จ = (๐ ๐๐ฆ + ๐๐๐
๐๐ฆ + ๐๐๐
๐๐ฆ + ๐๐๐
๐๐ฆ) โ ๐ ๐๐ฆ
๐ณ๐จ = (๐๐ ๐๐ฆ +๐๐
๐๐ฆ +๐๐
๐๐ฆ +๐๐
๐๐ฆ) โ ๐ ๐๐ฆ
๐ณ๐จ = (๐๐ ๐๐ฆ +๐๐๐๐
๐๐ฆ +๐
๐๐ ๐๐ฆ +
๐๐๐
๐๐ฆ) โ ๐ ๐๐ฆ
๐ณ๐จ = (๐๐ ๐๐ฆ +๐๐๐๐
๐๐ฆ) โ ๐ ๐๐ฆ
๐ณ๐จ = ๐๐๐ ๐๐ฆ๐ +๐๐๐๐๐
๐๐ฆ๐
๐ณ๐จ = ๐๐๐ ๐๐ฆ๐ + ๐๐๐๐๐
๐๐ฆ๐
๐ณ๐จ = ๐๐๐๐๐๐๐
๐๐ฆ๐
๐ฉ = ๐จ๐ซ๐๐๐ญ๐๐ง๐ ๐ฅ๐ + ๐จ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ ๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ฉ = (๐๐๐
๐๐ฆ โ ๐ ๐๐ฆ) +๐๐
(๐ ๐๐ฆ โ ๐๐๐
๐๐ฆ) ๐บ๐จ = ๐๐๐๐๐๐๐
๐๐ฆ๐ + ๐ (๐๐๐๐
๐๐ฆ๐)
๐ฉ = (๐๐ ๐๐ฆ๐ + ๐ ๐๐ฆ๐) + (๐ ๐๐ฆ โ ๐๐๐
๐๐ฆ) ๐บ๐จ = ๐๐๐๐๐๐๐
๐๐ฆ๐ + ๐๐ ๐๐ฆ๐
๐ฉ = ๐๐ ๐๐ฆ๐ + ๐๐๐
๐๐ฆ๐ ๐๐ = ๐๐๐๐๐๐๐
๐๐ฆ๐
๐ฉ = ๐๐๐๐
๐๐ฆ๐
The surface area of the prism is ๐๐๐๐๐๐๐
๐๐ฆ๐.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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3. The net below is for a specific object. The measurements shown are in meters. Sketch (or describe) the object, and
then find its surface area.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ท โ ๐ ๐ฉ = (๐๐
๐๐ฆ โ๐๐
๐๐ฆ) + (๐๐
๐๐ฆ โ ๐ ๐๐ฆ) + (๐๐
๐๐ฆ โ ๐๐๐
๐๐ฆ) ๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ ๐๐ฆ โ๐๐
๐๐ฆ ๐ฉ = (๐๐
๐๐ฆ๐) + (๐๐
๐๐ฆ๐) + (๐๐
๐๐ฆ๐) ๐บ๐จ = ๐ ๐๐ฆ๐ + ๐ (๐๐๐
๐๐ฆ๐)
๐ณ๐จ = ๐ ๐๐ฆ๐ ๐ฉ = (๐๐
๐๐ฆ๐) + (๐๐
๐๐ฆ๐) + (๐๐
๐๐ฆ๐) ๐บ๐จ = ๐ ๐๐ฆ๐ + ๐ ๐๐ฆ๐
๐ฉ =๐๐
๐๐ฆ๐ ๐บ๐จ = ๐ ๐๐ฆ๐
๐ฉ = ๐๐๐
๐๐ฆ๐
The surface area of the object is ๐ ๐๐ฆ๐.
4. In the diagram, there are ๐๐ cubes glued together to form a solid. Each cube has a volume of ๐๐
๐ข๐ง๐. Find the
surface area of the solid.
The volume of a cube is ๐๐, and ๐๐
๐ข๐ง๐ is the same as (๐๐
๐ข๐ง. )๐
, so the
cubes have edges that are ๐
๐๐ข๐ง. long. The cube faces have area ๐๐, or
(๐๐
๐ข๐ง. )๐
, or ๐
๐ ๐ข๐ง๐. There are ๐๐ cube faces that make up the surface
of the solid.
๐บ๐จ =๐๐
๐ข๐ง๐ โ ๐๐
๐บ๐จ = ๐๐๐๐
๐ข๐ง๐
The surface area of the solid is ๐๐๐๐
๐ข๐ง๐.
(3-Dimensional Form)
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
317
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c. a. b.
5. The nets below represent three solids. Sketch (or describe) each solid, and find its surface area.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ท โ ๐
๐ณ๐จ = ๐๐ โ ๐
๐ณ๐จ = ๐๐ ๐๐๐
๐ฉ = ๐๐
๐ฉ = (๐ ๐๐ฆ)๐
๐ฉ = ๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ + ๐(๐ ๐๐ฆ๐)
๐บ๐จ = ๐๐ ๐๐ฆ๐ + ๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐จ๐ฌ๐ช๐ฎ๐๐ซ๐ + ๐๐จ๐ซ๐ญ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ + ๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐
๐จ๐ฌ๐ช๐ฎ๐๐ซ๐ = ๐๐
๐จ๐ฌ๐ช๐ฎ๐๐ซ๐ = (๐ ๐๐ฆ)๐
๐จ๐ฌ๐ช๐ฎ๐๐ซ๐ = ๐ ๐๐ฆ๐
๐จ๐ซ๐ญ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐
๐๐๐
๐จ๐ซ๐ญ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐
๐โ ๐ ๐๐ฆ โ ๐ ๐๐ฆ
๐จ๐ซ๐ญ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐
๐
๐จ๐ซ๐ญ ๐ญ๐ซ๐ข๐๐ง๐ ๐ = ๐๐
๐ ๐๐ฆ2
๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐
๐๐๐
๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐
๐โ (๐
๐
๐ ๐๐ฆ) โ (๐
๐
๐๐ ๐๐ฆ)
๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ = ๐๐
๐๐๐๐ฆ โ ๐
๐
๐๐ ๐๐ฆ
๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐๐
๐๐ ๐๐ฆ โ
๐๐
๐๐ ๐๐ฆ
๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ =๐๐๐
๐๐๐ ๐๐ฆ๐
๐จ๐๐ช๐ฎ ๐ญ๐ซ๐ข๐๐ง๐ ๐ฅ๐ = ๐๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐(๐ ๐๐ฆ๐) + ๐ (๐๐
๐ ๐๐ฆ๐) + ๐
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ + (๐๐ +๐
๐) ๐๐ฆ๐ + ๐
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ +๐
๐ ๐๐ฆ๐ +
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ +๐๐
๐๐๐ ๐๐ฆ๐ +
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ +๐๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐จ๐๐ ๐๐๐๐๐๐๐๐ + ๐จ๐๐๐ ๐๐๐๐๐๐๐๐
๐บ๐จ = ๐ (๐๐
๐) ๐๐ฆ๐ + ๐
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ +๐
๐ ๐๐ฆ๐ + ๐ ๐๐ฆ๐ +
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ +๐
๐ ๐๐ฆ๐ +
๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐ ๐๐ฆ๐ + ๐ ๐๐ฆ๐ +๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐๐๐
๐๐๐ ๐๐ฆ2
๐บ๐จ = ๐๐ ๐๐ฆ๐ + ๐ ๐๐ฆ๐ +๐๐
๐๐๐ ๐๐ฆ๐
๐บ๐จ = ๐๐๐๐
๐๐๐ ๐๐ฆ๐
Sketch B
Sketch C
Sketch A
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
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d. How are figures (b) and (c) related to figure (a)?
If the equilateral triangular faces of figures (b) and (c) were matched together, they would form the cube in
part (a).
6. Find the surface area of the solid shown in the diagram. The solid is a right triangular prism (with right triangular
bases) with a smaller right triangular prism removed from it.
๐บ๐จ = ๐ณ๐จ + ๐๐ฉ
๐ณ๐จ = ๐ท โ ๐
๐ณ๐จ = (๐ ๐ข๐ง. +๐ ๐ข๐ง. +๐๐๐
๐๐ ๐ข๐ง. ) โ ๐ ๐ข๐ง.
๐ณ๐จ = (๐๐๐๐
๐๐ ๐ข๐ง. ) โ ๐ ๐ข๐ง.
๐ณ๐จ = ๐๐ ๐ข๐ง๐ +๐๐
๐๐ ๐ข๐ง๐
๐ณ๐จ = ๐๐ ๐ข๐ง๐ + ๐ ๐ข๐ง๐ +๐
๐๐ ๐ข๐ง๐
๐ณ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐
The ๐
๐ ๐ข๐ง. by ๐
๐๐๐๐
๐ข๐ง. rectangle has to be taken away from the lateral area:
๐จ = ๐๐ ๐๐ = ๐๐๐
๐๐ ๐ข๐ง๐ โ ๐
๐๐๐๐
๐ข๐ง๐
๐จ = ๐๐๐๐๐
๐ข๐ง โ๐๐
๐ข๐ง ๐๐ = ๐๐๐๐๐๐
๐ข๐ง๐ โ ๐๐๐๐๐
๐ข๐ง๐
๐จ = ๐ ๐ข๐ง๐ +๐๐๐๐
๐ข๐ง๐ ๐ณ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐
๐จ = ๐๐๐๐๐
๐ข๐ง๐ ๐ณ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐
Two bases of the larger and smaller triangular prisms must be added:
๐บ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐ + ๐ (๐
๐
๐ ๐ข๐ง โ
๐
๐ ๐ข๐ง) + ๐ (
๐
๐โ ๐ ๐ข๐ง โ ๐ ๐ข๐ง)
๐บ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐ + ๐ โ
๐
๐ ๐ข๐ง โ ๐
๐
๐ ๐ข๐ง + ๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐ +
๐
๐ ๐ข๐ง โ ๐
๐
๐ ๐ข๐ง + ๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐ + (
๐
๐ ๐ข๐ง๐ +
๐
๐ ๐ข๐ง๐) + ๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐
๐๐ ๐ข๐ง๐ + ๐ ๐ข๐ง๐ +
๐
๐๐ ๐ข๐ง๐ +
๐
๐๐ ๐ข๐ง๐ + ๐๐ ๐ข๐ง๐
๐บ๐จ = ๐๐๐๐
๐๐ ๐ข๐ง๐
The surface area of the solid is ๐๐๐๐๐๐
๐ข๐ง๐.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข3 Lesson 22
Lesson 22: Surface Area
319
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7. The diagram shows a cubic meter that has had three square holes punched completely through the cube on three
perpendicular axes. Find the surface area of the remaining solid.
Exterior surfaces of the cube (๐บ๐จ๐):
๐บ๐จ๐ = ๐(๐ ๐ฆ)๐ โ ๐ (๐
๐ ๐ฆ)
๐
๐บ๐จ๐ = ๐(๐ ๐ฆ๐) โ ๐ (๐
๐ ๐ฆ๐)
๐บ๐จ๐ = ๐ ๐ฆ๐ โ๐
๐ ๐ฆ๐
๐บ๐จ๐ = ๐ ๐ฆ๐ โ (๐๐
๐ ๐ฆ๐)
๐บ๐จ๐ = ๐๐
๐ ๐ฆ๐
Just inside each square hole are four intermediate surfaces that can be treated as the lateral area of a rectangular
prism. Each has a height of ๐๐
๐ฆ and perimeter of ๐๐
๐ฆ +๐๐
๐ฆ +๐๐
๐ฆ +๐๐
๐ฆ or ๐ ๐ฆ.
๐บ๐จ๐ = ๐(๐ณ๐จ)
๐บ๐จ๐ = ๐ (๐ ๐ฆ โ๐
๐ ๐ฆ)
๐บ๐จ๐ = ๐ โ๐
๐ ๐ฆ2
๐บ๐จ๐ = ๐ ๐ฆ๐
The total surface area of the remaining solid is the sum of these two areas:
๐บ๐จ๐ป = ๐บ๐จ๐ + ๐บ๐จ๐.
๐บ๐จ๐ป = ๐๐
๐ ๐ฆ2 + ๐ ๐ฆ๐
๐บ๐จ๐ป = ๐๐
๐ ๐ฆ๐
The surface area of the remaining solid is ๐๐๐
๐ฆ๐.