Surface Area of Prisms, Cubes, and Boxes

Surface Area of Prisms, Cubes, and Boxes exercise

• Determine the surface area of the given rectangular prisms and cubes.

  Hints

  The area of face $ A $ is $ 5 \times 8 $, or $40 ~\text{units squared}.$ There are $2$ of these, so that's $ 80 ~\text{units squared}.$

  The area of face $ B $ is $ 6 \times 8 $, or $48 ~\text{units squared}.$ There are two of those, so that's $ 96 ~\text{units squared}.$

  The area of face $ C $ is $ 5 \times 6 $, or $ 30 ~\text{units squared}.$ There are two of those, so that's $ 60 ~\text{units squared}.$

  The total surface area is $80+96+60$, or $236 ~\text{units squared}.$

  The cube shown has side length $ 3 $. The area of face $ D $ is $3 \times 3 $, that is, $ 3^2 $, or $9$.

  There are six such faces, so the total surface area is $ 6 \times 9$, or $ 54 ~\text{units squared}$.

  The total surface area of the rectangular prism shown is the sum of the areas of each labelled face ($A$, $B$, $C$), DOUBLED.

  The total surface area is given by the following equation:

  $SA = 2lh + 2wh +2lw$

  The total surface area of a cube is $6$ times the area of one face. The area of face $D$ is $s \times s$, or $s^2$. There are six such faces, so the total surface area is $6s^2$.

  Solution

  The surface area of the rectangular prism shown is given by the formula:

  $ SA = 2lh + 2wh +2lw $.

  For the first problem, $l=3$, $w=4$, and $h=5$. Therefore:

  $ SA = 2lh + 2wh +2lw $. $ SA = 2(3\times5) +2(4\times5)+2(3\times4) $. $ SA =2(15) + 2(20) +2(12) $ $ SA =30 +40 +24 $ $ SA = 94 ~\text{units squared}$

  The surface of the cube shown is given by the equation:

  $SA =6s^2$

  For the second problem, $s=2$, therefore:

  $SA = 6(2^2)$ $SA = 6(4)$ $SA =24 ~\text{units squared}$

• Explain how to calculate the surface area of the triangular prisms.

  Hints

  A rectangular prism is composed of two triangular bases (orange in the diagram), and three rectangular faces (blue in the diagram).

  The area of a triangle is $\frac{1}{2}bh$. The height is always perpendicular to the base. If the area of one triangle is $\frac{1}{2}bh$, the area of two such triangles is $2\cdot\frac{1}{2}bh$, or $bh$.

  A triangular prism has three lateral faces that are rectangles. The height of the rectangles is the same for each. The areas of the rectangles in the diagram are $ah$, $bh$, and $ch$.

  Solution

  The triangular base of the first prism has a base length of $6$ and a height of $4$. So the area of the triangle is $\frac12(6)(4)=12$. There are two such triangles and the area of the two triangles combined is $24$. There are two $5\times10$ rectangular faces. Each has an area of $50$. The remaining rectangular $6\times10$ face has an area of $60$. The total surface area is $24 + 100 + 60 =184 ~\text{units}^2$.

  The second prism has a triangular base whose area is $\frac{1}{2}bh$. There are two such triangles, so their combined area is $2\cdot\frac{1}{2}bh$, or $bh$. There are two rectangular faces with area $wl$, and one with area $bl$. The total surface area is $bh+2wl+bl$.

• Solve for the surface area of the 3-D objects.

  Hints

  The surface area of a rectangular prism with dimensions $lwh$ is $2lw+2lh+2wh$.

  The surface area of the triangular prism shown is twice the area of the base, plus the area of the three lateral faces (the rectangles).

  $SA=2\cdot \frac{1}{2}bh + al + bl + cl$

  This simplifies to:

  $SA=bh + al + bl + cl$

  Solution

  The smallest figure is rectangular prism with dimensions $3\times10\times1$.

  $SA=2\cdot(3)(10)+2\cdot(3)(1)+2\cdot(10)(1)$

  $SA=2\cdot30+2\cdot3+2\cdot10$

  $SA=60+6+20$

  $SA=86 ~\text{u}^2$

  The next largest figure is the triangular prism with a triangular base with sides lengths of $5$, $12$, and $13$.

  $A_\text{triangle}=\frac{1}{2}(12)(5)$

  $A_\text{triangle}=(6)(5)$

  $A_\text{triangle}=30$

  $A_\text{two triangles}=2\cdot30$

  $A_\text{two triangles}=60$

  $SA=60+(5)(1)+(12)(1)+(13)(1)$

  $SA=60+5+12+13$

  $SA=90 ~\text{u}^2$

  The third largest figure is the rectangular prism with dimensions $3\times4\times5$.

  $SA=2\cdot(3)(4)+2\cdot(4)(5)+2\cdot(3)(5)$

  $SA=2\cdot12+2\cdot20+2\cdot15$

  $SA=24+40+30$

  $SA=94 ~\text{u}^2$

  The largest figure is the triangular prism with a triangular base with base length $12$ and height $8$.

  $A_\text{triangle}=\frac{1}{2}(12)(8)$

  $A_\text{triangle}=(6)(8)$

  $A_\text{triangle}=48$

  $A_\text{both triangles}=96$

  $SA=96+(12)(1)+10(1)+10(1)$

  $SA=96+12+10+10$

  $SA=128 ~\text{u}^2$

• Apply surface area to real world problems.

  Hints

  If the top and bottom of a rectangular prism have dimensions $l\times w$, with a height of $h$, we can represent such a prism with the diagram shown. The surface area is given by the formula:

  $SA=2lw+2wh+2lh$

  The prism shown has dimensions $l\times w\times h$. Suppose we are only interested in the lateral area, the four shaded rectangles. If we unfold the figure as shown, we can see that they form a large rectangle, the base of which is equal to the perimeter of the base of the original prism. That's $2l+2w$. To find the lateral area, we multiply this perimeter times the height. So the lateral area of a rectangular prism is the perimeter of the base times the height.

  $A_\text{lateral}=(2l+2w)h$

  How much greater is the lateral area of the prism on the right than that of the prism on the left? The perimeter of each base is $10$ units. The prism on the left is 5 units tall, while the prism on the right is $7$ units tall. The prism on the right is $2$ units taller than the one on the left. To find the difference in lateral area, shaded red, we multiply the perimeter of $10$ times a height of $2$. That's $20 ~\text{u}^2$.

  Solution

  For the first problem, we know the prism has a base that is $5\times7$ and a height of $10$.

  $SA=2\cdot(5)(7)+2\cdot(5)(10)+2\cdot(7)(10)$

  $SA=2\cdot35+2\cdot50+2\cdot70$

  $SA=75+100+140=315 ~\text{in}^2$

  For the second problem, the pond is $10\times20$ with a height of $5$. We are interested in the lateral area, that is, the area of the four rectangles that are NOT the top and bottom. We can calculate the area of two of the adjacent rectangles then double those values:

  $A_1=(10)(5)=50$

  $A_2=(20)(5)=100$

  $A_\text{lateral}=2\cdot50+2\cdot100$

  $A_\text{lateral}=100+200=300 ~\text{ft}^2$

  Another way is to calculate the perimeter of the base, then multiply times the height:

  $P=2\cdot10+2\cdot20=60$

  $A_\text{lateral}=60\cdot5=300 ~\text{ft}^2$

  For the third problem, the perimeter of each base is $50$. The height of the first prism is $3$ and the height of the second is $5$, and we are interested in the difference. That's the lateral area of the prism shaded in red, with a a height of $2$.

  $A_\text{lateral}=50\cdot2=100 ~\text{m}^2$

  For the fourth problem, we have a prism that is $10\times12\times3$. We can frost all faces except the bottom. We could find the area of each rectangular face and add them. We can also find the lateral area using the perimeter of the base, then add the area of one base.

  $P=2\cdot10+2\cdot12=44$

  $A_\text{lateral}=44\cdot3=132$

  $A_\text{top}=(12)(10)=120$

  $A_\text{frosting}=252 ~\text{in}^2$

• Find the area of rectangles and triangles.

  Hints

  The area of a rectangle is base times height. That's $A = bh $.

  The area of a triangle is half the base times the height. That's $A = \frac{1}{2}bh $.

  Even if the triangle is obtuse, and the height of the triangle is drawn outside of the triangle, the area is still half the base times the height. Extending the base with a dotted line helps us visualize the height.

  The units for area are square units, like $\text{ft}^2$, $\text{km}^2$, and $\text{cm}^2$. Cubic units like $\text{ft}^3$ are for volume, and units like $\text{ft}$ are for length.

  Solution

  The area of the first triangle is half the base times height.

  $A = \frac{1}{2}(4)(3) $

  $ A= 6 ~\text{cm}^2 $

  The area of the rectangle is base times height.

  $A = (4)(3) $

  $A = 12 ~\text{cm}^2 $

  The area of the next figure, a triangle, is half base times height.

  $A = \frac{1}{2}(4)(3) $

  $ A = 6 ~\text{cm}^2 $

  The area of the last figure, a triangle, is half base times height.

  $ A = \frac{1}{2}(6)(4) $

  $ A = 12 ~\text{cm}^2 $

• Find the surface area of prisms and cylinders.

  Hints

  To find the surface area, we can always find the area of each face, but with fractions like this, there is a better way. First, let's choose the $\frac14 \times \frac14$ face as the base.

  $A_\text{base}=(\frac14)(\frac14)=\frac18$

  $A_\text{two bases}=\frac18+\frac18=\frac14$

  The lateral area (the area of the remaining rectangles that are not the base) is the perimeter of the base, times the height.

  $P=4\cdot\frac14=1$

  $A_\text{lateral}=(1)(7 \frac12)=7 \frac12$

  $SA=\frac14+7\frac12$

  $SA=0.25+7.5=7.75~\text{units}^2$

  The surface area of a cylinder can be found by finding area of the bases, and then finding the lateral area. To find the lateral area, we will use the method described in the last tip. We find the perimeter of the base, and multiply times the height. The base of a cylinder is a circle, and the perimeter of a circle is called circumference. Here are two useful formulas:

  $A_\text{circle}=\pi r^2$

  $P_\text{circle}=C=2 \pi r$

  The area of the base in this diagram is $\pi r^2$. There are two of them, so that's $2\pi r^2$. The lateral area is the circumference of $2\pi r$ times the height of $h$. That's $2 \pi r h$. Therefore:

  $SA_\text{cylinder}=2 \pi r^2 + 2 \pi r h$.

  Solution

  For the first figure, let's select the $\frac12 \times \frac12$ face as the base.

  $A_\text{base}=(\frac12)(\frac12)=\frac14$

  $A_\text{two bases}=2\cdot\frac14=\frac12$

  Let's find the lateral area by finding the perimeter of the base and multiplying by the height of $9 \frac12$.

  $P=4\cdot\frac14=1$

  $A_\text{lateral}=(2)(9 \frac12)=19$

  $SA=19 +\frac12=19\frac12=19.5~\text{units}^2$

  For the second figure, let's first find the area of the triangular base:

  $A_\text{base}=\frac12(1)(\frac34)$

  $A_\text{two bases}=2\cdot\frac12(1)(\frac34)$

  $A_\text{two bases}=\frac34$

  Now let's find the lateral area by finding the perimeter of the base and multiplying by the height.

  $P=1+\frac34+1\frac14=3$

  $A_\text{lateral}=3\cdot2=6$

  $SA=6+\frac34=6.75~\text{units}^2$

  For the third figure, we find the area of the base, a circle, and double it. Then we find the perimeter of the base, otherwise known as the circumference, and multiply times the height.

  $A_\text{circle}=\pi(5)^2=25\pi$

  $A_\text{two circles}=2\cdot25\pi=50\pi$

  $C=2\pi(5)=10\pi$

  $A_\text{lateral}=10\pi\cdot20=200\pi$

  $SA=200\pi+50\pi=250\pi~\text{units}^2$

  For the last figure, we are given the area and the perimeter of the base.

  $A_\text{base}=16$

  $A_\text{two bases}=32$

  $P=20$

  $A_\text{lateral}=20 \cdot 10 =200$

  $SA=200+32=232~\text{units}^2$