A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 2 and 8 and the pyramid's height is 2 . If one of the base's corners has an angle of (5pi)/6, what is the pyramid's surface area?

2 Answers
Oct 24, 2017

Total Surface Area # T S A = 34.8328

Explanation:

CH = 2 * sin (5pi/6)= 1
Area of parallelogram base = 8 * b1 = 8 * 1 = 8

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (8/2)^2)= 4.4721
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2*4.4721 = 4.4721

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(2/2)^2 )= 2.2361
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*2.2361= 8.9443

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 4.4721)+ (2* 8.9443)= 26.8328

Total surface area =Area of parallelogram base + Lateral surface area = 8 + 26.8328 = 34.8328

Total Surface Area # T S A = **14.4755**#enter image source hereTotal Surface Area T S A = **14.4755**enter image source here

Oct 24, 2017

The pyramid's surface area is 8+10sqrt(21+4sqrt(3))

Explanation:

I recommend to use coordinates to solve this problem.
enter image source here

The pyramid's base is ABCD and its center is E

The coordinates of A and B are:
A(0;0)
B(8;0)

In order to find the coordinates of C and D, we need to use trigonometry and the pythagorean theorem:

l=2cos(pi/6)=2sqrt(3)/2=sqrt(3)
L=sqrt(2^2-sqrt(3)^2)=sqrt(4-3)=sqrt(1)=1

Therefore the coordinates of C and D are:
C(8+sqrt(3);1)
D(sqrt(3);1)

A parallelogram's diagonals cross mid-length, thus the coordinates of E are:
E(4+sqrt(3)/2;1/2)

If the base is located in the (x;y) plane and the pyramid's peak (called F) is located at 2 above E, then the coordinates of all the pyramid's points are:
A(0;0;0)
B(8;0;0)
C(8+sqrt(3);1;0)
D(sqrt(3);1;0)
F(4+sqrt(3)/2;1/2;2)

The pyramid's base's surface area is AB*L=8*1=8

The ADF and BCF sides' surface areas are:
(AF*AD)/2=(sqrt((4+sqrt(3)/2)^2+(1/2)^2+2^2)*cancel2)/cancel2=sqrt((16+4sqrt(3)+3/4)+1/4+4)=sqrt(21+4sqrt(3))=AF

The ABF and CDF sides' surface areas are:
(AF*AB)/2=AF*4=4sqrt(21+4sqrt(3))

Therefore, the pyramid's total surface area is:
8+2sqrt(21+4sqrt(3))+8sqrt(21+4sqrt(3))=8+10sqrt(21+4sqrt(3))