Lesson 24: Surface Area
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 24
7β’6
Lesson 24: Surface Area Student Outcomes ο§
Students determine the surface area of three-dimensional figures, those that are composite figures and those that have missing sections.
Lesson Notes This lesson is a continuation of Lesson 23. Students will continue to work on surface area advancing to figures with missing sections.
Classwork Example 1 (8 minutes)
Scaffolding:
Students should solve this problem on their own. Example 1 Determine the surface area of the image. Surface area of top and bottom prisms Lateral sides = π(ππ in. Γ π in.) = πππ in 2
MP.7 & MP.8
Base face = ππ in. Γ ππ in. = πππ in2
ο§ As in Lesson 23, students can draw nets of the figures to help them visualize the area of the faces. They could determine the area of these without the holes first and subtract the surface area of the holes.
Base face with hole = ππ in. Γ ππ in. β π in. Γ π in. = πππ in 2 There are two of these, making up πππ in2. Surface area of middle prism Lateral Sides = π(π in. Γ π in.) = πππ in2 Surface area: πππ in2 + πππ in2 = πππ in2
ο§
Describe the method you used to determine the surface area. οΊ
ο§
Answers will vary: I determined the surface area of each prism separately and added them together. Then I subtracted the area of the sections that were covered by another prism.
If all three prisms were separate, would the sum of their surface areas be the same as the surface area you determined in this example? οΊ
No, if the prisms were separate, there would be more surfaces shown. The three separate prisms would have a greater surface area than this example. The area would be greater by the area of four 4 in. Γ 4 in. squares (64 in2).
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
254 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
Example 2 (5 minutes)
Scaffolding:
Example 2 a.
Determine the surface area of the cube.
πΊππππππ ππππ = πππ πΊπ¨ = π(ππ ππ)π πΊπ¨ = π(πππ πππ ) πΊπ¨ = πππ πππ
ο§
7β’6
ο§ As in Lesson 23, students can draw nets of the figures to help them visualize the area of the faces. They could determine the area of these without the holes first and subtract the surface area of the holes.
Explain how 6(12 in.)2 represents the surface area of the cube. οΊ
The area of one face, one square with side length of 12 in., is (12 in. )2 , and so a total area of all six faces is 6(12 in. )2 . A square hole with a side length of π inches is drilled through the cube. Determine the new surface area.
b.
Area of interior lateral sides = π(ππ in. Γ π in.) = πππ in2 Surface Area of cube with holes π
= π(ππ in.) β π(π in. Γ π in.) + π(ππ in. Γ π in. ) = πππ in2 β ππ in2 + πππ in2 = π, πππ in2
ο§
How does drilling a hole in the cube change the surface area? οΊ
ο§
What happens to the surfaces that now show inside the cube? οΊ
ο§
A rectangular prism was drilled out of the cube with the following dimensions: 4 in. Γ 4 in. Γ 12 in.
How can we use this to help us determine the new total surface area? οΊ
ο§
These are now part of the surface area.
What is the shape of the piece that was removed from the cube? οΊ
ο§
We have to subtract the area of the square at the surface from each end.
We can find the surface area of the cube and the surface area of the rectangular prism, but we will have to subtract the area of the square bases from the cube and also exclude these bases in the area of the rectangular prism.
Why is the surface area larger when holes have been cut into the cube? οΊ
There are more surfaces showing now. All of the surfaces need to be included in the surface area.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
255 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
ο§
7β’6
Explain how the expression 6(12 in.)2 β 2(4 in. Γ 4 in.) + 4(12 in. Γ 4 in.) represents the surface area of the cube with the hole. οΊ
From the total surface area of a whole (uncut) cube, 6(12 in.)2 , the area of the bases (the cuts made to the surface of the cube) are subtracted: 6(12 in.)2 β 2(4 in. Γ 4 in.). To this expression we add the area of the four lateral faces of the cut out prism, 4(12 in. Γ 4 in.). The complete expression then is 6(12 in. )2 β 2(4 in. Γ 4 in.) + 4(12 in. Γ 4 in.).
Example 3 (5 minutes) Example 3 A right rectangular pyramid has a square base with a side length of ππ inches. The surface area of the pyramid is πππ in2. Find the height of the four lateral triangular faces. Area of base = ππ in. Γ ππ in. = πππ in 2 Area of the four faces = πππ in2 β πππ in2 = πππ in2 The total area of the four faces is πππ in2. Therefore, the area of each triangular faces is ππ in 2. π
Area of lateral side = ππ π π π ππ ππ = (ππ ππ. )π π ππ πππ = (π ππ. )π π = π ππ. The height of each lateral triangular face is π inches.
ο§
What strategies could you use to help you solve this problem? οΊ
ο§
ο§
I could draw a picture of the pyramid and label the sides so that I can visualize what the problem is asking me to do.
What information have we been given? How can we use the information? οΊ
We know the total surface area, and we know the length of the sides of the square.
οΊ
We can use the length of the sides of the square to give us the area of the square base.
How will the area of the base help us determine the slant height? οΊ
First, we can subtract the area of the base from the total surface area in order to determine what is left for the lateral sides.
οΊ
Now we can divide the remaining area by 4 to get the area of just one triangular face.
οΊ
Finally, we can work backwards. We have the area of the triangle, and we know the base is 10 in., so we can solve for the height.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
256 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exercises 1β8 (20 minutes) Students work in pairs to complete the exercises. Exercises 1β8 Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise. 1.
Top and bottom = π(ππ π Γ π π) = πππ ππ Extra interior sides = π(π π Γ π π) = ππ ππ Left and right sides = π(π π Γ ππ π) = πππ ππ Front and back sides = π((ππ π Γ ππ π) β (π π Γ π π)) = π(πππ ππ β ππ ππ ) = π(πππ ππ ) = πππ ππ Surface area = πππ ππ + ππ ππ + πππ ππ + πππ ππ = πππ ππ
2.
In addition to your calculation, explain how the surface area was determined. The surface area of the prism is found by taking the sum of the areas of the trapezoidal front and back areas of the four different sized rectangles that make up the lateral faces.
Area top = ππ ππ Γ π ππ = πππ πππ Area bottom = ππ ππ Γ π ππ = πππ πππ Area sides = (ππ ππ Γ π ππ) + (ππ ππ π ππ) = πππ πππ π Area front and back = π ( (ππ ππ + ππ ππ)(ππ ππ)) π = π(πππ πππ ) = πππ πππ Surface Area = πππ πππ + πππ πππ + πππ πππ + πππ πππ = π, πππ πππ
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
257 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
3. Surface Area of Prisms on the Sides: Area of front and back = π(π in. Γ ππ in.) = ππ in2 Area of top and bottom = π(π in. Γ ππ in.) = ππ in2 Area of side = ππ in. Γ ππ in. = πππ in2 Area of side with hole = ππ in. Γ ππ in. β π in. Γ π in. = πππ in2 There are two such rectangular prisms, so the surface area of both is πππ in2. Surface Area of Middle Prism: Area of front and back = π(π in. Γ ππ in.) = ππ in2 Area of sides = π(π in. Γ ππ in.) = ππ in2 Surface area = πππ in2 + πππ in2 = πππ in2 4.
In addition to your calculation, explain how the surface area was determined. The surface area of the prism is found by taking the area of the base of the rectangular prism and the area of its four lateral faces and adding it to the area of the four lateral faces of the pyramid. Area of base = π ππ.Γ π ππ. = ππ πππ Area of rectangular sides = π(π ππ.Γ π ππ. ) = πππ πππ π Area of triangular sides = π ( (π ππ. )(π ππ. )) π = πππ πππ Surface Area = ππ πππ + πππ πππ + πππ πππ = πππ πππ
5.
A hexagonal prism has the following base and has a height of π units. Determine the surface area of the prism. Area of bases = π(ππ + π + ππ + ππ) units2 = πππ units2 Area of π unit sides = π(π Γ π) units2 = πππ units2 Area of other sides = (π Γ π) units2 + (ππ Γ π) units2 = πππ units2 Surface Area = πππ units2 + πππ units2 + πππ units2 = πππ units2
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
258 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
6.
7β’6
Determine the surface area of each figure. a. πΊπ¨ = πππ = π(π m)π = π(ππ m2) = πππ m2
b.
A cube with a square hole with π m side lengths has been drilled through the cube.
Lateral sides of the hole = π(π m Γ π m) = πππ m2 Surface area of cube with holes = πππ m2 β π(π m Γ π m) + πππ m2 = πππ m2
c.
A second square hole with π m side lengths has been drilled through the cube. Surface Area = πππ m2 β π(π m Γ π m) + π(π(π m Γ π m)) = πππ m2
7.
The figure below shows ππ cubes with an edge length of π unit. Determine the surface area. Area top and bottom = ππ units2 Area sides = ππ units2 Area front and back = ππ units2 Surface Area = ππ + ππ + ππ = ππ units2
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
259 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
8.
7β’6
The base rectangle of a right rectangular prism is π ft. Γ π ft. The surface area is πππ ft2. Find the height. Let π be the height in feet. Area of one base: π ft. Γ π ft. = ππ ft2 Area of two bases: π(ππ ft2 ) = ππ ft2 Numeric area of four lateral faces: πππ ft2 β ππ ft2 = πππ ft2 Algebraic area of four lateral faces: π(ππ + ππ) Solve for π π(ππ + ππ) = πππ πππ = πππ π = ππ The height is ππ feet.
Closing (2 minutes) ο§
Write down three tips that you would give a friend that is trying to calculate surface area.
Exit Ticket (5 minutes)
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
260 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
Name
7β’6
Date
Lesson 24: Surface Area Exit Ticket Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
261 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exit Ticket Sample Solutions Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area.
Area of top and bottom = π(ππ ππ Γ π ππ) = πππ πππ Area of sides = π(π ππ Γ π ππ) = ππ πππ Area of front and back = π(ππ ππ Γ π ππ) β π(π ππ Γ π ππ) = πππ πππ Area inside = π(π ππ Γ π ππ) = πππ πππ Surface Area = πππ πππ + ππ πππ + πππ πππ + πππ πππ = πππ πππ Take the sum of the areas of the four lateral faces of the main rectangular prism, and subtract the areas of the four square cuts from the area of the bases of the main rectangular prism. Finally, add the lateral faces of the prisms that were cut out of the main prism.
Problem Set Sample Solutions Determine the surface area of each figure. 1.
In addition to the calculation of the surface area, describe how you found the surface area. Area of top = ππ ππ Γ ππ ππ = πππ πππ Area of bottom = ππ ππ Γ ππ ππ = πππ πππ Area of left and right sides = ππ ππ Γ ππ ππ + ππ ππ Γ ππ ππ = πππ πππ π Area of front and back sides = π ((ππ ππ Γ ππ ππ) + (π ππ Γ ππ ππ)) π = π(πππ πππ + ππ πππ ) = π(πππ πππ ) = πππ πππ Surface area = πππ πππ + πππ πππ + πππ πππ + πππ πππ = π, πππ πππ Split the area of the two trapezoidal bases, take the sum of the areas, and then add the areas of the four different sized rectangles that make up the lateral faces.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
262 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
2.
7β’6
Area of front and back = π(ππ. π in. Γ π. π in.) = πππ. ππ in2 Area of sides = π(π. π ππ.Γ ππ ππ. ) = πππ. π πππ Area of top and bottom = = = =
π((ππ. π in. Γ ππ in.) β π. π in. Γ π. π in.) π(πππ. π in2 β ππ. ππ in2) π(πππ. ππ in2) πππ. ππ in2
Surface Area = πππ. ππ in2 + πππ. π in2 + πππ. ππ in2 = π, πππ. ππ in2
3.
π Area of front and back = π ( (ππ π + ππ π)ππ π) π = πππ ππ Area of top = ππ π Γ ππ π = πππ ππ Area of left and right sides = π(ππ π Γ ππ π) = π(πππ ππ ) = π, πππ ππ
32 m
Area of bottom = ππ π Γ ππ π = π, πππ ππ Surface Area = πππ ππ + π, πππ ππ + π, πππ ππ + πππ ππ = π, πππ ππ
4.
Determine the surface area after two square holes with a side length of π m are drilled through the solid figure composed of two rectangular prisms. Surface Area of Top Prism Before the Hole is Drilled: Area of top = π π Γ π π = ππ ππ Area of front and back = π(π π Γ π π) = ππ ππ Area of sides = π(π π Γ π π) = ππ ππ Surface Area of Bottom Prism Before the Hole is Drilled: Area of top = ππ π Γ ππ π β ππ ππ = ππ ππ Area of bottom = ππ π Γ ππ π = πππ ππ Area of front and back = π(ππ π Γ π π) = ππ ππ
Surface Area of Interiors: Area of Interiors = π(π π Γ π π) + π(π π Γ π π) = ππ ππ
Area of sides = π(ππ π Γ π π) = ππ ππ Surface area = πππ ππ + πππ ππ + ππ ππ β ππ ππ = πππ ππ
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
263 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
Lesson 24
7β’6
The base of a right prism is shown below. Determine the surface area if the height of the prism is ππ cm. Explain how you determined the surface area. Take the sum of the areas of the two bases made up of two right triangles, and add to it the sum of the areas of the lateral faces made up by rectangles of different sizes.
Area of sides
= (ππ ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ) + (π ππ Γ ππ ππ) = πππ πππ + πππ πππ + πππ πππ + ππ πππ = πππ πππ
Area of bases
π π = π ( (π ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ)) π π = (π ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ) = πππ πππ + πππ πππ = πππ πππ
Surface area
= πππ πππ + πππ πππ = π, πππ πππ
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
264 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
7β’6
Lesson 24: Surface Area Student Outcomes ο§
Students determine the surface area of three-dimensional figures, those that are composite figures and those that have missing sections.
Lesson Notes This lesson is a continuation of Lesson 23. Students will continue to work on surface area advancing to figures with missing sections.
Classwork Example 1 (8 minutes)
Scaffolding:
Students should solve this problem on their own. Example 1 Determine the surface area of the image. Surface area of top and bottom prisms Lateral sides = π(ππ in. Γ π in.) = πππ in 2
MP.7 & MP.8
Base face = ππ in. Γ ππ in. = πππ in2
ο§ As in Lesson 23, students can draw nets of the figures to help them visualize the area of the faces. They could determine the area of these without the holes first and subtract the surface area of the holes.
Base face with hole = ππ in. Γ ππ in. β π in. Γ π in. = πππ in 2 There are two of these, making up πππ in2. Surface area of middle prism Lateral Sides = π(π in. Γ π in.) = πππ in2 Surface area: πππ in2 + πππ in2 = πππ in2
ο§
Describe the method you used to determine the surface area. οΊ
ο§
Answers will vary: I determined the surface area of each prism separately and added them together. Then I subtracted the area of the sections that were covered by another prism.
If all three prisms were separate, would the sum of their surface areas be the same as the surface area you determined in this example? οΊ
No, if the prisms were separate, there would be more surfaces shown. The three separate prisms would have a greater surface area than this example. The area would be greater by the area of four 4 in. Γ 4 in. squares (64 in2).
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
254 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
Example 2 (5 minutes)
Scaffolding:
Example 2 a.
Determine the surface area of the cube.
πΊππππππ ππππ = πππ πΊπ¨ = π(ππ ππ)π πΊπ¨ = π(πππ πππ ) πΊπ¨ = πππ πππ
ο§
7β’6
ο§ As in Lesson 23, students can draw nets of the figures to help them visualize the area of the faces. They could determine the area of these without the holes first and subtract the surface area of the holes.
Explain how 6(12 in.)2 represents the surface area of the cube. οΊ
The area of one face, one square with side length of 12 in., is (12 in. )2 , and so a total area of all six faces is 6(12 in. )2 . A square hole with a side length of π inches is drilled through the cube. Determine the new surface area.
b.
Area of interior lateral sides = π(ππ in. Γ π in.) = πππ in2 Surface Area of cube with holes π
= π(ππ in.) β π(π in. Γ π in.) + π(ππ in. Γ π in. ) = πππ in2 β ππ in2 + πππ in2 = π, πππ in2
ο§
How does drilling a hole in the cube change the surface area? οΊ
ο§
What happens to the surfaces that now show inside the cube? οΊ
ο§
A rectangular prism was drilled out of the cube with the following dimensions: 4 in. Γ 4 in. Γ 12 in.
How can we use this to help us determine the new total surface area? οΊ
ο§
These are now part of the surface area.
What is the shape of the piece that was removed from the cube? οΊ
ο§
We have to subtract the area of the square at the surface from each end.
We can find the surface area of the cube and the surface area of the rectangular prism, but we will have to subtract the area of the square bases from the cube and also exclude these bases in the area of the rectangular prism.
Why is the surface area larger when holes have been cut into the cube? οΊ
There are more surfaces showing now. All of the surfaces need to be included in the surface area.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
255 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
ο§
7β’6
Explain how the expression 6(12 in.)2 β 2(4 in. Γ 4 in.) + 4(12 in. Γ 4 in.) represents the surface area of the cube with the hole. οΊ
From the total surface area of a whole (uncut) cube, 6(12 in.)2 , the area of the bases (the cuts made to the surface of the cube) are subtracted: 6(12 in.)2 β 2(4 in. Γ 4 in.). To this expression we add the area of the four lateral faces of the cut out prism, 4(12 in. Γ 4 in.). The complete expression then is 6(12 in. )2 β 2(4 in. Γ 4 in.) + 4(12 in. Γ 4 in.).
Example 3 (5 minutes) Example 3 A right rectangular pyramid has a square base with a side length of ππ inches. The surface area of the pyramid is πππ in2. Find the height of the four lateral triangular faces. Area of base = ππ in. Γ ππ in. = πππ in 2 Area of the four faces = πππ in2 β πππ in2 = πππ in2 The total area of the four faces is πππ in2. Therefore, the area of each triangular faces is ππ in 2. π
Area of lateral side = ππ π π π ππ ππ = (ππ ππ. )π π ππ πππ = (π ππ. )π π = π ππ. The height of each lateral triangular face is π inches.
ο§
What strategies could you use to help you solve this problem? οΊ
ο§
ο§
I could draw a picture of the pyramid and label the sides so that I can visualize what the problem is asking me to do.
What information have we been given? How can we use the information? οΊ
We know the total surface area, and we know the length of the sides of the square.
οΊ
We can use the length of the sides of the square to give us the area of the square base.
How will the area of the base help us determine the slant height? οΊ
First, we can subtract the area of the base from the total surface area in order to determine what is left for the lateral sides.
οΊ
Now we can divide the remaining area by 4 to get the area of just one triangular face.
οΊ
Finally, we can work backwards. We have the area of the triangle, and we know the base is 10 in., so we can solve for the height.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
256 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exercises 1β8 (20 minutes) Students work in pairs to complete the exercises. Exercises 1β8 Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise. 1.
Top and bottom = π(ππ π Γ π π) = πππ ππ Extra interior sides = π(π π Γ π π) = ππ ππ Left and right sides = π(π π Γ ππ π) = πππ ππ Front and back sides = π((ππ π Γ ππ π) β (π π Γ π π)) = π(πππ ππ β ππ ππ ) = π(πππ ππ ) = πππ ππ Surface area = πππ ππ + ππ ππ + πππ ππ + πππ ππ = πππ ππ
2.
In addition to your calculation, explain how the surface area was determined. The surface area of the prism is found by taking the sum of the areas of the trapezoidal front and back areas of the four different sized rectangles that make up the lateral faces.
Area top = ππ ππ Γ π ππ = πππ πππ Area bottom = ππ ππ Γ π ππ = πππ πππ Area sides = (ππ ππ Γ π ππ) + (ππ ππ π ππ) = πππ πππ π Area front and back = π ( (ππ ππ + ππ ππ)(ππ ππ)) π = π(πππ πππ ) = πππ πππ Surface Area = πππ πππ + πππ πππ + πππ πππ + πππ πππ = π, πππ πππ
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
257 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
3. Surface Area of Prisms on the Sides: Area of front and back = π(π in. Γ ππ in.) = ππ in2 Area of top and bottom = π(π in. Γ ππ in.) = ππ in2 Area of side = ππ in. Γ ππ in. = πππ in2 Area of side with hole = ππ in. Γ ππ in. β π in. Γ π in. = πππ in2 There are two such rectangular prisms, so the surface area of both is πππ in2. Surface Area of Middle Prism: Area of front and back = π(π in. Γ ππ in.) = ππ in2 Area of sides = π(π in. Γ ππ in.) = ππ in2 Surface area = πππ in2 + πππ in2 = πππ in2 4.
In addition to your calculation, explain how the surface area was determined. The surface area of the prism is found by taking the area of the base of the rectangular prism and the area of its four lateral faces and adding it to the area of the four lateral faces of the pyramid. Area of base = π ππ.Γ π ππ. = ππ πππ Area of rectangular sides = π(π ππ.Γ π ππ. ) = πππ πππ π Area of triangular sides = π ( (π ππ. )(π ππ. )) π = πππ πππ Surface Area = ππ πππ + πππ πππ + πππ πππ = πππ πππ
5.
A hexagonal prism has the following base and has a height of π units. Determine the surface area of the prism. Area of bases = π(ππ + π + ππ + ππ) units2 = πππ units2 Area of π unit sides = π(π Γ π) units2 = πππ units2 Area of other sides = (π Γ π) units2 + (ππ Γ π) units2 = πππ units2 Surface Area = πππ units2 + πππ units2 + πππ units2 = πππ units2
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
258 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
6.
7β’6
Determine the surface area of each figure. a. πΊπ¨ = πππ = π(π m)π = π(ππ m2) = πππ m2
b.
A cube with a square hole with π m side lengths has been drilled through the cube.
Lateral sides of the hole = π(π m Γ π m) = πππ m2 Surface area of cube with holes = πππ m2 β π(π m Γ π m) + πππ m2 = πππ m2
c.
A second square hole with π m side lengths has been drilled through the cube. Surface Area = πππ m2 β π(π m Γ π m) + π(π(π m Γ π m)) = πππ m2
7.
The figure below shows ππ cubes with an edge length of π unit. Determine the surface area. Area top and bottom = ππ units2 Area sides = ππ units2 Area front and back = ππ units2 Surface Area = ππ + ππ + ππ = ππ units2
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
259 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
8.
7β’6
The base rectangle of a right rectangular prism is π ft. Γ π ft. The surface area is πππ ft2. Find the height. Let π be the height in feet. Area of one base: π ft. Γ π ft. = ππ ft2 Area of two bases: π(ππ ft2 ) = ππ ft2 Numeric area of four lateral faces: πππ ft2 β ππ ft2 = πππ ft2 Algebraic area of four lateral faces: π(ππ + ππ) Solve for π π(ππ + ππ) = πππ πππ = πππ π = ππ The height is ππ feet.
Closing (2 minutes) ο§
Write down three tips that you would give a friend that is trying to calculate surface area.
Exit Ticket (5 minutes)
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
260 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
Name
7β’6
Date
Lesson 24: Surface Area Exit Ticket Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
261 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
7β’6
Exit Ticket Sample Solutions Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area.
Area of top and bottom = π(ππ ππ Γ π ππ) = πππ πππ Area of sides = π(π ππ Γ π ππ) = ππ πππ Area of front and back = π(ππ ππ Γ π ππ) β π(π ππ Γ π ππ) = πππ πππ Area inside = π(π ππ Γ π ππ) = πππ πππ Surface Area = πππ πππ + ππ πππ + πππ πππ + πππ πππ = πππ πππ Take the sum of the areas of the four lateral faces of the main rectangular prism, and subtract the areas of the four square cuts from the area of the bases of the main rectangular prism. Finally, add the lateral faces of the prisms that were cut out of the main prism.
Problem Set Sample Solutions Determine the surface area of each figure. 1.
In addition to the calculation of the surface area, describe how you found the surface area. Area of top = ππ ππ Γ ππ ππ = πππ πππ Area of bottom = ππ ππ Γ ππ ππ = πππ πππ Area of left and right sides = ππ ππ Γ ππ ππ + ππ ππ Γ ππ ππ = πππ πππ π Area of front and back sides = π ((ππ ππ Γ ππ ππ) + (π ππ Γ ππ ππ)) π = π(πππ πππ + ππ πππ ) = π(πππ πππ ) = πππ πππ Surface area = πππ πππ + πππ πππ + πππ πππ + πππ πππ = π, πππ πππ Split the area of the two trapezoidal bases, take the sum of the areas, and then add the areas of the four different sized rectangles that make up the lateral faces.
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
262 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 24
NYS COMMON CORE MATHEMATICS CURRICULUM
2.
7β’6
Area of front and back = π(ππ. π in. Γ π. π in.) = πππ. ππ in2 Area of sides = π(π. π ππ.Γ ππ ππ. ) = πππ. π πππ Area of top and bottom = = = =
π((ππ. π in. Γ ππ in.) β π. π in. Γ π. π in.) π(πππ. π in2 β ππ. ππ in2) π(πππ. ππ in2) πππ. ππ in2
Surface Area = πππ. ππ in2 + πππ. π in2 + πππ. ππ in2 = π, πππ. ππ in2
3.
π Area of front and back = π ( (ππ π + ππ π)ππ π) π = πππ ππ Area of top = ππ π Γ ππ π = πππ ππ Area of left and right sides = π(ππ π Γ ππ π) = π(πππ ππ ) = π, πππ ππ
32 m
Area of bottom = ππ π Γ ππ π = π, πππ ππ Surface Area = πππ ππ + π, πππ ππ + π, πππ ππ + πππ ππ = π, πππ ππ
4.
Determine the surface area after two square holes with a side length of π m are drilled through the solid figure composed of two rectangular prisms. Surface Area of Top Prism Before the Hole is Drilled: Area of top = π π Γ π π = ππ ππ Area of front and back = π(π π Γ π π) = ππ ππ Area of sides = π(π π Γ π π) = ππ ππ Surface Area of Bottom Prism Before the Hole is Drilled: Area of top = ππ π Γ ππ π β ππ ππ = ππ ππ Area of bottom = ππ π Γ ππ π = πππ ππ Area of front and back = π(ππ π Γ π π) = ππ ππ
Surface Area of Interiors: Area of Interiors = π(π π Γ π π) + π(π π Γ π π) = ππ ππ
Area of sides = π(ππ π Γ π π) = ππ ππ Surface area = πππ ππ + πππ ππ + ππ ππ β ππ ππ = πππ ππ
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
263 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
Lesson 24
7β’6
The base of a right prism is shown below. Determine the surface area if the height of the prism is ππ cm. Explain how you determined the surface area. Take the sum of the areas of the two bases made up of two right triangles, and add to it the sum of the areas of the lateral faces made up by rectangles of different sizes.
Area of sides
= (ππ ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ) + (π ππ Γ ππ ππ) = πππ πππ + πππ πππ + πππ πππ + ππ πππ = πππ πππ
Area of bases
π π = π ( (π ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ)) π π = (π ππ Γ ππ ππ) + (ππ ππ Γ ππ ππ) = πππ πππ + πππ πππ = πππ πππ
Surface area
= πππ πππ + πππ πππ = π, πππ πππ
Lesson 24: Date: Β© 2014 Common Core, Inc. Some rights reserved. commoncore.org
Surface Area 7/12/15
264 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
