Surface Area - combined straight edge objects

Surface Area - combined straight edge objects exercise

• Find the surface area of the square pyramid.

  Hints

  The height of one of the triangles of the pyramid is the shortest distance from the apex of the pyramid to the base of the triangle.

  The base of one of the triangles of the pyramid is the side of the triangle that is opposite the apex of the pyramid.

  Pay attention to the units. An area is given by $ft^2$, or $in^2$, or ...

  Solution

  To determine the whole area that has to be covered by sticky fly paper, we first have to decide the shape of the roof: it's a quadratic pyramid; i.e. a pyramid with a rectangular base.

  Four triangles of equal dimensions make up the sides of the pyramid. So we have to determine the area of one of the triangles and then multiply it by four to get the area of all of the sides together.

  Do you still know the formula for the area of a triangle? It's $A_{\text{triangle}}=\frac12(\text{base})(\text{height})$.

  What is the base and the height? Just take a look at the picture beside. The base is $45~ft$ and the height is $25.5~ft$.

  We plug these values into the formula above to get $A_{\text{triangle}}=\frac12(45)(25.5)=573.75~ft^2$, the area of one of the triangles.

  We then have to multiply this value by $4$ to get the area of the whole roof: $A=4\times 573.75=2295~ft^2$.

• Calculate the surface area of the staircase.

  Hints

  Just use the formula $A_{\text{rectangle}}=(\text{length})(\text{width})$.

  There are some rectangles of type $A$ on the other side of the staircase.

  You have to add all calculated areas.

  Solution

  How much of the super slippery soap does Calvin need?

  Just take a look at the staircase; it's composed by different rectangles of type $A$, $B$, and $C$. For each type, we use the formula $A_{\text{rectangular}}=(\text{length})(\text{width})$.

  Let's start with $A$:

  • There are $6$ such rectangles visible to us, but on the other side of the staircase, there another $6$. Together we have $12$ rectangles of type $A$.
  • Each of those rectangles has the area $12\times 8=96~in^2$.
  • We multiply this area by $12$ to get $12\times 96=1,152~in^2$.

  Next, we consider $B$:

  • There are $3$ such rectangles.
  • The area of one of these rectangles is $40\times 8=320~in^2$.
  • This value has to multiplied by $3$ to get $3\times 320~in^2=960~in^2$.

  Just one more type of rectangle, type $C$:

  • There are $3$ such rectangles.
  • The area of one of them is $40\times 12=480~in^2$.
  • Multiplying this by $3$ leads to $3\times 480~in^2=1,440~in^2$.

  Last but not least we have to add all these areas to get the staircase area to be covered in slippery soap: $1,152~+~960~+~1,440~=~3,552~in^2$.

• Determine the surface area of the triangular prism.

  Hints

  Use the following formulas: $A_{\text{triangle}}=\frac12(\text{base})(\text{height})$ for a triangle, and $A_{\text{rectangle}}=(\text{length})(\text{width})$ for a rectangle.

  An isosceles triangle has two legs with the same length. The rectangles with identical area have this length as the width and the height of the prism as the length.

  Solution

  To wrap the present Calvin has for his family, he needs to make sure he has enough gift wrap. So first he has to determine the surface area of the triangular prism.

  The prism is composed of $2$ isosceles triangles and $3$ rectangles.

  Let's start with the triangles, which both are of equal size.

  We take the formula,

  $A_{\text{triangle}}=\frac12(\text{base})(\text{height})$,

  and plug the given values for the base, $6~in$, and for the height, $4~in$, in to get $A_{\text{triangle}}=\frac12(6)(4)=12~in^2$ for each triangle.

  We still have to double the area to get $24~in^2$.

  Next, we handle the rectangles: two of them have the same area. The two with the same area have identical width, $5~in$, and length, $15~in$.

  We plug those values into the formula, $A_{\text{rectangular}}=(\text{length})(\text{width})$, to get $A_{\text{rectangular}}=(15)(5)=75~in^2$. We double it to $150~in^2$.

  There is still one rectangle left with width $6~in$ and the length $15~in$.

  So we get $A_{\text{rectangular}}=(15)(6)=90~in^2$.

  Last, we add all those areas together to get the area of the triangular prism: $24~+~150~+~90~=~265~in^2$.

• Identify the formulas for surface areas of the different shapes.

  Hints

  First, decide which two dimensional shapes each three dimensional shape is composed of.

  There are only two three-dimensional shapes which are composed of only one type of two-dimensional shape.

  Here you see a trapezoid. Two sides, $b_1$ and $b_2$, are parallel. The distance of those sides is the height $h_T$ of the trapezoid.

  The area formula is given by $A_{\text{trapezoid}}=\frac12(b_2+b_2)(h_T)$.

  Solution

  Let's start with the tetrahedron. It's only composed of triangles with area given by the formula,

  $A_{\text{triangle}}=\frac12bh$.

  Because the pyramid consists of four such triangles, we get that the area of the tetrahedron is $2\frac12bd=2bh$.

  Next we have a rectangular prism, which is composed of rectangles with area equal to either $lw$, $lh$, or $wh$.

  There are two rectangles with area $lw$, two with area $lh$, and two with area $wh$.

  Adding these areas together, we get that the surface area of the rectangular prism is $2(lw+lh+wh)$.

  The next shape is a prism with a trapezoid as the base. The area of the trapezoids is given by the formula,

  $A_{\text{trapezoid}}=\frac12(b_2+b_2)(h_T)$,

  where

  • $b_1$ and $b_2$ are the lengths of the parallel sides, and
  • $h_T$ is the height of the trapezoid, or distance between the parallel sides.

  For the rectangles, we have one with an area of $b_1l$, one with an area of $b_2l$, and two with an area of $sl$.

  Adding the areas of the four rectangles and two trapezoids, we get that the surface area is $(b_2+b_2)(h_T)+(b_1+b_2+2s)l=(h_T+l)b_1+(h_T+l)b_2+2sl$.

  The last shape is a pyramid with a square as the base. For this we need the formula for the area of a triangle as well as the formula for the area of a square with side length $a$:

  $A_{\text{square}}=a^2$.

  Together with the area formula for a triangle with height $h$, we can see that the area of the pyramid is $2ah+a^2$.

• Recall the area formulas for a triangle and rectangle and find the areas of the ones displayed.

  Hints

  You get the values for each shape in the header. Just plug those values into the formula.

  If you "cut" a rectangle along the diagonal into two, you get two triangles. Those triangles have the same area.

  Solution

  For the calculations of different areas we need a few formulas:

  The formula for the area of a triangle is $A_{\text{triangle}}=\frac12(\text{base})(\text{height})$.

  So for the triangle given, the area is $A_{\text{triangle}}=\frac12(45)(25.5)=573.75$.

  The formula for the area of a rectangle is $A_{\text{rectangular}}=(\text{length})(\text{width})$.

  So for the rectangle given, the area is $A_{\text{rectangular}}=(40)(12)=480$.

• Figure out the surface area of the sculpture.

  Hints

  The surface area of one of the sides of the bigger tetrahedron is given by subtracting the area of one of the smaller tetrahedron's sides.

  Calculate the total surface area of the bigger tetrahedron, then subtract the area of one of the sides of the smaller one from this total. Then add the area of three of the sides of the smaller tetrahedron to get the surface area of the sculpture.

  Solution

  Let's start with the bigger tetrahedron: the $4$ triangles all have a base of $20~in$ and a height of $17.3~in$. Using the formula for the area of a triangle,

  $A_{\text{triangle}}=\frac12(\text{base})(\text{height})$

  We get that the total area of the bigger tetrahedron is $4\times\frac12(20)(17.3)=692~in^2$.

  The area of one of the sides of the smaller tetrahedron is $\frac12(5)(4.3)=10.75~in^2$.

  Since only three of the sides of the smaller tetrahedron are included in the surface area of the sculpture, and that part of the surface area of one of the sides of the bigger tetrahedron is not included in the surface area of the sculpture, we have that the total surface area of the sculpture is $(692-10.75)+(3\cdot 10.75)=713.5~in^2$.