A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #5 # and the pyramid's height is #4 #. If one of the base's corners has an angle of #pi/3#, what is the pyramid's surface area?

Question

A pyramid has a parallelogram shaped base and a peak directly above its center.  Its base's sides have lengths of #7  # and #5  # and the pyramid's height is #4 #.  If one of the base's corners has an angle of #pi/3#, what is the pyramid's surface area?

Ella Armstrong · Accepted Answer

#A=40sin(pi/3)+7sqrt(16+25/4sin^2(pi/3))+5sqrt(28.25)~=93.05#

Let's first try and visualize the pyramid. It would look something like this from an angled top-down perspective:
To be able to solve for the area of this parallelogram, we need to solve for it's height, #h#. Using a bit of trigonometry, we can express the following equation:
#sin(pi/3)=h/5#
If we solve for h, we get that it is equal to:
#h=5sin(pi/3)~=4.33#

The area of a parallelogram is equal to #base*height#, so if we plug in #7# and #5sin(pi/3)# we get:
#A_(parall\e\l\ogram)=7*5sin(pi/3)=40sin(pi/3)~=34.64#

Now that we have the base finished, let's have a look at the triangles in the upper part of the pyramid. We will start with the ones marked in blue:
The bottom left corner is the center of the base parallelogram in the pyramid, so the side with length #4# is the height of the pyramid, the side with length #5/2sin(pi/3)#, is half the width we calculated from before (since the distance from the edge to the center point in the parallelogram is half the width of it), and the hypotenuse with length #h#, is the same #h# in the blue triangle.

Using the pythagorean theorem, #a^2+b^2=c^2#, we can express #h# in the following way:
#4^2+(5/2sin(pi/3))^2=h^2#
Solving for #h#, we get:
#h=sqrt(4^2+(5/2sin(pi/3))^2)=sqrt(16+25/4sin^2(pi/3))~=4.55#

Now we can calculate the area of one of the blue triangles. It will be equal to the height of the blue triangle multiplied by #7#, all over #2#:
#A_(blue)=7/2sqrt(16+25/4sin^2(pi/3))~=15.92#
Let's then multiply this by two to get the area of both of the blue triangles:
#2A_(blue)=7sqrt(16+25/4sin^2(pi/3))~=31.84#

Now, we can go through a similar process to solve the red triangles. This time we will consider a triangle like this:

Sadie Adams · Answer

undefined To find the surface area of the pyramid, we need to calculate the area of the base and the area of the four triangular faces.

1. Area of the base (parallelogram):
   Area = base length * height of the parallelogram
   Area = 7 * 5 = 35 square units

2. To find the area of one of the triangular faces, we need to find the length of the slant height (l) of the triangular face. We can use the Pythagorean theorem to find this.

Let's call one of the base's corners with an angle of π/3 as A, and the midpoint of the base as B. The slant height (l) is the distance from the peak of the pyramid to the midpoint of one of the sides of the base. We can find this using trigonometry.

sin(π/3) = opposite/hypotenuse
   sin(π/3) = 4/l
   l = 4/sin(π/3)

Using the value of sin(π/3) = √3/2:
   l = 4/(√3/2) = (4 * 2)/√3 = (8√3)/3

3. Now, we can calculate the area of one triangular face:
   Area of one triangular face = (1/2) * base * height
   Area = (1/2) * 7 * ((8√3)/3)
   Area = (28√3)/3

4. Since there are four triangular faces, the total area of the triangular faces = 4 * Area of one triangular face = 4 * (28√3)/3 = (112√3)/3 square units

5. Now, we can find the total surface area of the pyramid by adding the area of the base and the area of the four triangular faces:
   Total surface area = Area of the base + Total area of the triangular faces
   Total surface area = 35 + (112√3)/3 square units

So, the surface area of the pyramid is 35 + (112√3)/3 square units.