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NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
267
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Lesson 24: Surface Area
Student Outcomes
Students determine the surface area of three-dimensional figuresโthose that are composite figures and those
that have missing sections.
Lesson Notes
This lesson is a continuation of Lesson 23. Students continue to work on surface area advancing to figures with missing
sections.
Classwork
Example 1 (8 minutes)
Students should solve this problem on their own.
Example 1
Determine the surface area of the image.
Surface area of top or bottom prism:
Lateral sides = ๐(๐๐ ๐ข๐ง.ร ๐ ๐ข๐ง. ) = ๐๐๐ ๐ข๐ง๐
Base face = ๐๐ ๐ข๐ง.ร ๐๐ ๐ข๐ง. โ= ๐๐๐ ๐ข๐ง๐
Base face with hole = ๐๐ ๐ข๐ง.ร ๐๐ ๐ข๐ง. โ ๐ ๐ข๐ง.ร ๐ ๐ข๐ง.
= ๐๐๐ ๐ข๐ง๐
There are two of these, making up ๐๐๐ ๐ข๐ง๐.
Surface area of middle prism:
Lateral sides: = ๐(๐ ๐ข๐ง.ร ๐ ๐ข๐ง. ) = ๐๐๐ ๐ข๐ง๐
Surface area: ๐๐๐ ๐ข๐ง๐ + ๐๐๐ ๐ข๐ง๐ = ๐๐๐ ๐ข๐ง๐
Describe the method you used to determine the surface area.
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?
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.ร 4 in. squares (64 in2).
Scaffolding:
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.
MP.7 &
MP.8
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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Example 2 (5 minutes)
Example 2
a. Determine the surface area of the cube.
Explain how 6(12 in. )2 represents the surface area of the cube.
The area of one face, one square with a side length of 12 in., is (12 in. )2, and so a total area of all six
faces is 6(12 in. )2.
b. A square hole with a side length of ๐ inches is cut through the cube. Determine the new surface area.
How does cutting a hole in the cube change the surface area?
We have to subtract the area of the square at the surface from each end.
We also have to add the area of each of the interior faces to the total surface area.
What happens to the surfaces that now show inside the cube?
These are now part of the surface area.
What is the shape of the piece that was removed from the cube?
A rectangular prism was cut out of the cube with the following dimensions: 4 in.ร 4 in.ร 12 in.
How can we use this to help us determine the new total surface area?
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?
There are more surfaces showing now. All of the surfaces need to be included in the surface area.
Scaffolding:
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. ๐๐ฎ๐ซ๐๐๐๐ ๐๐ซ๐๐ = ๐๐๐
๐บ๐จ = ๐(๐๐ ๐ข๐ง. )๐ ๐บ๐จ = ๐(๐๐๐ ๐ข๐ง๐)
๐บ๐จ = ๐๐๐ ๐ข๐ง๐
Area of interior lateral sides
= ๐(๐๐ ๐ข๐ง.โร ๐ ๐ข๐ง. )
= ๐๐๐ ๐ข๐ง๐
Surface area of cube with holes
= ๐ (๐๐ ๐ข๐ง. )๐ โ ๐(๐ ๐ข๐ง.โร ๐ ๐ข๐ง. ) + ๐(๐๐ ๐ข๐ง.โร ๐ ๐ข๐ง. )
= ๐๐๐ ๐ข๐ง๐ โ ๐๐ ๐ข๐ง๐ + ๐๐๐ ๐ข๐ง๐
= ๐, ๐๐๐ ๐ข๐ง๐
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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Explain how the expression 6(12 in. )2 โ 2(4 in.ร 4 in. ) + 4(12 in.ร 4 in. ) represents the surface area of the
cube with the hole.
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 โ 2(4 in.ร 4 in. ). To this expression we add the
area of the four lateral faces of the cutout prism, 4(12 in.ร 4 in. ). The complete expression then is
6(12 in. )2 โ 2(4 in.ร 4 in. ) + 4(12 in.ร 4 in. )
Example 3 (5 minutes)
Example 3
A right rectangular pyramid has a square base with a side length of ๐๐ inches. The surface area of the pyramid is
๐๐๐ ๐ข๐ง๐. Find the height of the four lateral triangular faces.
Area of base = ๐๐ ๐ข๐ง.ร ๐๐ ๐ข๐ง. โ= ๐๐๐ ๐ข๐ง๐
Area of the four faces = ๐๐๐ ๐ข๐ง๐ โ ๐๐๐ ๐ข๐ง๐ = ๐๐๐ ๐ข๐ง๐
The total area of the four faces is ๐๐๐ ๐ข๐ง๐. Therefore, the area of each triangular face is ๐๐ ๐ข๐ง๐.
Area of lateral side =๐๐
๐๐
๐๐ ๐ข๐ง๐ =๐
๐(๐๐ ๐ข๐ง. )๐
๐๐ ๐ข๐ง๐ = (๐ ๐ข๐ง.)๐
๐ = ๐ ๐ข๐ง.
The height of each lateral triangular face is ๐ inches.
What strategies could you use to help you solve this problem?
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?
We know the total surface area, and we know the length of the sides of the square.
We can use the length of the sides of the square to give us the area of the square base.
How does the area of the base help us determine the height of the lateral triangular face?
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.
Next, we can divide the remaining area by 4 to get the area of just one triangular face.
Finally, we can work backward. We have the area of the triangle, and we know the base is 10 in., so
we can solve for the height.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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Exercises 1โ8 (20 minutes)
Students work in pairs to complete the exercises.
Exercises
Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise.
1.
2. In addition to your calculation, explain how the surface area of the following figure was determined.
The surface area of the prism is found by taking the sum of the
areas of the trapezoidal front and back and the areas of the back of
the four different-sized rectangles that make up the lateral faces.
Area top = ๐๐ ๐๐ฆ ร ๐ ๐๐ฆ
= ๐๐๐ ๐๐ฆ๐
Area bottom = ๐๐ ๐๐ฆ ร ๐ ๐๐ฆ
= ๐๐๐ ๐๐ฆ๐
Area sides = (๐๐ ๐๐ฆ ร ๐ ๐๐ฆ) + (๐๐ ๐๐ฆ ร ๐ ๐๐ฆ)
= ๐๐๐ ๐๐ฆ๐
Area front and back = ๐ (๐
๐(๐๐ ๐๐ฆ + ๐๐ ๐๐ฆ)(๐๐ ๐๐ฆ))
= ๐(๐๐๐ ๐๐ฆ๐)
= ๐๐๐ ๐๐ฆ๐
Surface area = ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐
= ๐, ๐๐๐ ๐๐ฆ๐
Top and bottom = ๐(๐๐ ๐ฆ ร ๐ ๐ฆ) = ๐๐๐ ๐ฆ๐
Extra interior sides = ๐(๐ ๐ฆ ร ๐ ๐ฆ) = ๐๐ ๐ฆ๐
Left and right sides = ๐(๐ ๐ฆ ร ๐๐ ๐ฆ) = ๐๐๐ ๐ฆ๐
Front and back sides = ๐((๐๐ ๐ฆ ร ๐๐ ๐ฆ) โ (๐ ๐ฆ ร ๐ ๐ฆ))
= ๐(๐๐๐ ๐ฆ๐ โ ๐๐ ๐ฆ๐)
= ๐(๐๐๐ ๐ฆ๐)
= ๐๐๐ ๐ฆ๐
Surface area = ๐๐๐ ๐ฆ๐ + ๐๐ ๐ฆ๐ + ๐๐๐ ๐ฆ๐ + ๐๐๐ ๐ฆ๐
= ๐๐๐ ๐ฆ๐
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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Surface area of one of the prisms on the sides:
Area of front and back = ๐(๐ ๐ข๐ง.โร ๐๐ ๐ข๐ง. )
= ๐๐ ๐ข๐ง๐
Area of top and bottom = ๐(๐ ๐ข๐ง.โร ๐๐ ๐ข๐ง. )
= ๐๐ ๐ข๐ง๐
Area of side = ๐๐ ๐ข๐ง.โร ๐๐ ๐ข๐ง. = ๐๐๐ ๐ข๐ง๐
Area of side with hole = ๐๐ ๐ข๐ง.โร ๐๐ ๐ข๐ง. โ ๐ ๐ข๐ง.โร ๐ ๐ข๐ง.
= ๐๐๐ ๐ข๐ง๐
3.
There are two such rectangular prisms, so the surface area of both is ๐๐๐ ๐ข๐ง๐.
The total surface area of the figure is ๐๐๐ ๐ข๐ง๐ + ๐๐๐ ๐ข๐ง๐ = ๐๐๐ ๐ข๐ง๐.
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.
5. A hexagonal prism has the following base and has a height of ๐ units. Determine the surface area of the prism.
The surface area of the hexagonal prism is ๐๐๐ ๐ฎ๐ง๐ข๐ญ๐ฌ๐.
Area of bases = ๐(๐๐ + ๐ + ๐๐ + ๐๐) = ๐๐๐
Area of ๐ unit sides = ๐(๐ ร ๐)
= ๐๐๐
Area of other sides = (๐ ร ๐) + (๐๐ ร ๐)
= ๐๐๐
Surface area = ๐๐๐ + ๐๐๐ + ๐๐๐
= ๐๐๐
Area of base = ๐ ๐๐ญ.โร ๐ ๐๐ญ.
= ๐๐ ๐๐ญ๐
Area of rectangular sides = ๐(๐ ๐๐ญ.โร ๐ ๐๐ญ. )
= ๐๐๐ ๐๐ญ๐
Area of triangular sides = ๐ (๐
๐(๐ ๐๐ญ. )(๐ ๐๐ญ. ))
= ๐๐๐ ๐๐ญ๐
Surface area = ๐๐ ๐๐ญ๐ + ๐๐๐ ๐๐ญ๐ + ๐๐๐ ๐๐ญ๐
= ๐๐๐ ๐๐ญ๐
๐ ๐ฎ๐ง๐ข๐ญ๐ฌ ๐ ๐ฎ๐ง๐ข๐ญ๐ฌ
Surface area of middle prism:
Area of front and back = ๐(๐ ๐ข๐ง.โร ๐๐ ๐ข๐ง. ) = ๐๐ ๐ข๐ง๐
Area of sides = ๐(๐ ๐ข๐ง.โร ๐๐ ๐ข๐ง. ) = ๐๐ ๐ข๐ง๐
Surface area of middle prism = ๐๐ ๐ข๐ง๐ + ๐๐ ๐ข๐ง๐ = ๐๐๐ ๐ข๐ง๐
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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6. Determine the surface area of each figure.
a.
b. A cube with a square hole with ๐ ๐ฆ side lengths has been cut through the cube.
c. A second square hole with ๐ ๐ฆ side lengths has been cut through the cube.
7. The figure below shows ๐๐ cubes with an edge length of ๐ unit. Determine the surface area.
๐บ๐จ = ๐๐๐
= ๐(๐ ๐ฆ)๐
= ๐(๐๐๐ฆ๐)
= ๐๐๐ ๐ฆ๐
Lateral sides of the hole = ๐(๐ ๐ฆ ร ๐ ๐ฆ) = ๐๐๐ ๐ฆ๐
Surface area of cube with holes = ๐๐๐ ๐ฆ๐ โ ๐(๐ ๐ฆ ร ๐ ๐ฆ) + ๐๐๐ ๐ฆ๐
= ๐๐๐ ๐ฆ๐
๐๐ฎ๐ซ๐๐๐๐ ๐๐ซ๐๐ = ๐๐๐ ๐ฆ๐ โ ๐(๐ ๐ฆ ร ๐ ๐ฆ) + ๐(๐(๐ ๐ฆ ร ๐ ๐ฆ))
= ๐๐๐ ๐ฆ๐
Area top and bottom = ๐๐ units๐
Area sides = ๐๐ units๐
Area front and back = ๐๐ units๐
= ๐๐ units2 + ๐๐ units2 + ๐๐ units2
= ๐๐ units๐
Surface area
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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8. The base rectangle of a right rectangular prism is ๐ ๐๐ญ.ร ๐ ๐๐ญ. The surface area is ๐๐๐ ๐๐ญ๐. Find the height.
Let ๐ be the height in feet.
Area of one base: ๐ ๐๐ญ.ร ๐ ๐๐ญ. = ๐๐ ๐๐ญ๐
Area of two bases: ๐(๐๐ ๐๐ญ๐) = ๐๐ ๐๐ญ๐
Numeric area of four lateral faces: ๐๐๐ ๐๐ญ๐ โ ๐๐ ๐๐ญ๐ = ๐๐๐ ๐๐ญ๐
Algebraic area of four lateral faces: ๐(๐๐ + ๐๐) ๐๐ญ๐
Solve for ๐.
๐(๐๐ + ๐๐) = ๐๐๐
๐๐๐ = ๐๐๐
๐ = ๐๐
The height is ๐๐ feet.
Closing (2 minutes)
Write down three tips that you would give a friend who is trying to calculate surface area.
Exit Ticket (5 minutes)
Lesson Summary
To calculate the surface area of a composite figure, determine the surface area of each prism separately,
and add them together. From the sum, subtract the area of the sections that were covered by another
prism.
To calculate the surface area with a missing section, find the total surface area of the whole figure. From
the total surface area, subtract the area of the missing parts. Then, add the area of the lateral faces of
the cutout prism.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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Name Date
Lesson 24: Surface Area
Exit Ticket
Determine the surface area of the right rectangular prism after the two square holes have been cut. Explain how you
determined the surface area.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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Exit Ticket Sample Solutions
Determine the surface area of the right rectangular prism after the two square holes have been cut. Explain how you
determined the surface area.
Area of top and
bottom
= ๐(๐๐ ๐๐ฆ ร ๐ ๐๐ฆ)
= ๐๐๐ ๐๐ฆ๐
Area of sides = ๐(๐ ๐๐ฆ ร ๐ ๐๐ฆ)
= ๐๐ ๐๐ฆ๐
Area of front and back = ๐(๐๐ ๐๐ฆ ร ๐ ๐๐ฆ) โ ๐(๐ ๐๐ฆ ร ๐ ๐๐ฆ)
= ๐๐๐ ๐๐ฆ๐
Area inside = ๐(๐ ๐๐ฆ ร ๐ ๐๐ฆ)
= ๐๐๐ ๐๐ฆ๐
Surface area = ๐๐๐ ๐๐ฆ๐ + ๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐
= ๐๐๐ ๐๐ฆ๐
Take the sum of the areas of the four lateral faces and the two bases of the main rectangular prism, and subtract the
areas of the four square cuts from the area of the front and back 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 = ๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ
= ๐๐๐ ๐๐ฆ๐
Area of bottom = ๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ
= ๐๐๐ ๐๐ฆ๐
Area of left and right sides = ๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ + ๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ
= ๐๐๐ ๐๐ฆ๐
Area of front and back sides = ๐ ((๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ) +๐
๐(๐ ๐๐ฆ ร ๐๐ ๐๐ฆ))
= ๐(๐๐๐ ๐๐ฆ๐ + ๐๐ ๐๐ฆ๐)
= ๐(๐๐๐ ๐๐ฆ๐)
= ๐๐๐ ๐๐ฆ๐
Surface area = ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐
= ๐, ๐๐๐ ๐๐ฆ๐
Split the area of the two trapezoidal bases into triangles and rectangles, take the sum of the areas, and then add the
areas of the four different-sized rectangles that make up the lateral faces.
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
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32 m
2.
Area of front and back = ๐(๐๐. ๐ ๐ข๐ง.ร ๐. ๐ ๐ข๐ง. ) = ๐๐๐. ๐๐ ๐ข๐ง๐
Area of sides = ๐(๐. ๐ ๐ข๐ง.ร ๐๐ ๐ข๐ง. ) = ๐๐๐. ๐ ๐ข๐ง๐
Area of top and bottom = ๐((๐๐. ๐ ๐ข๐ง.ร ๐๐ ๐ข๐ง. ) โ (๐. ๐ ๐ข๐ง.ร ๐. ๐ ๐ข๐ง. ))
= ๐(๐๐๐. ๐ ๐ข๐ง๐ โ ๐๐. ๐๐ ๐ข๐ง๐)
= ๐(๐๐๐. ๐๐ ๐ข๐ง๐)
= ๐๐๐. ๐๐ ๐ข๐ง๐
Surface area = ๐๐๐. ๐๐ ๐ข๐ง๐ + ๐๐๐. ๐ ๐ข๐ง๐ + ๐๐๐. ๐๐ ๐ข๐ง๐
= ๐, ๐๐๐. ๐๐ ๐ข๐ง๐
3.
Area of front and back = ๐ (๐
๐(๐๐ ๐ฆ + ๐๐ ๐ฆ)๐๐ ๐ฆ)
= ๐๐๐ ๐ฆ๐
Area of top = ๐๐ ๐ฆ ร ๐๐ ๐ฆ
= ๐๐๐ ๐ฆ๐
Area of left and right sides = ๐(๐๐ ๐ฆ ร ๐๐ ๐ฆ)
= ๐(๐๐๐ ๐ฆ๐)
= ๐, ๐๐๐ ๐ฆ๐
Area of bottom = ๐๐ ๐ฆ ร ๐๐ ๐ฆ
= ๐, ๐๐๐ ๐ฆ๐
Surface area = ๐๐๐ ๐ฆ๐ + ๐, ๐๐๐ ๐ฆ๐ + ๐, ๐๐๐ ๐ฆ๐ + ๐๐๐ ๐ฆ๐
= ๐, ๐๐๐ ๐ฆ๐
4. Determine the surface area after two square holes with a side length of ๐ ๐ฆ are cut through the solid figure
composed of two rectangular prisms.
Surface area of top prism before the hole is drilled:
Area of top = ๐ ๐ฆ ร ๐ ๐ฆ
= ๐๐ ๐ฆ๐
Area of front and back = ๐(๐ ๐ฆ ร ๐ ๐ฆ)
= ๐๐ ๐ฆ๐
Area of sides = ๐(๐ ๐ฆ ร ๐ ๐ฆ)
= ๐๐ ๐ฆ๐
Surface area of bottom prism before the hole is drilled:
Area of top = ๐๐ ๐ฆ ร ๐๐ ๐ฆ โ ๐๐ ๐ฆ๐
= ๐๐ ๐ฆ๐
Area of bottom = ๐๐ ๐ฆ ร ๐๐ ๐ฆ
= ๐๐๐ ๐ฆ๐ Surface area of interiors:
Area of front and back = ๐(๐๐ ๐ฆ ร ๐ ๐ฆ)
= ๐๐ ๐ฆ๐
Area of
interiors
= ๐(๐ ๐ฆ ร ๐ ๐ฆ) + ๐(๐ ๐ฆ ร ๐ ๐ฆ)
= ๐๐ ๐ฆ๐
NYS COMMON CORE MATHEMATICS CURRICULUM 7โข6 Lesson 24
Lesson 24: Surface Area
277
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Area of sides = ๐(๐๐ ๐ฆ ร ๐ ๐ฆ)
= ๐๐ ๐ฆ๐
Surface
Area
= ๐๐๐ ๐ฆ๐ + ๐๐๐ ๐ฆ๐ + ๐๐ ๐ฆ๐ โ ๐๐ ๐ฆ๐
= ๐๐๐ ๐ฆ๐
5. The base of a right prism is shown below. Determine the surface area if the height of the prism is ๐๐ ๐๐ฆ. 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 of rectangles of different sizes.
Area of sides = (๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ) + (๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ) + (๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ) + (๐ ๐๐ฆ ร ๐๐ ๐๐ฆ)
= ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐ + ๐๐ ๐๐ฆ๐
= ๐๐๐ ๐๐ฆ๐
Area of bases = ๐ (๐
๐(๐ ๐๐ฆ ร ๐๐ ๐๐ฆ) +
๐
๐(๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ))
= (๐ ๐๐ฆ ร ๐๐ ๐๐ฆ) + (๐๐ ๐๐ฆ ร ๐๐ ๐๐ฆ)
= ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐
= ๐๐๐ ๐๐ฆ๐
Surface area = ๐๐๐ ๐๐ฆ๐ + ๐๐๐ ๐๐ฆ๐
= ๐, ๐๐๐ ๐๐ฆ๐