Question

A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is `8`, its base has sides of length `8`, and its base has a corner with an angle of `pi/3`. What is the pyramid's surface area?

Answer 1

T S A = 198.5344

Explanation:

AB = BC = CD = DA = a = 8
Height OE = h = 8
OF = a/2 = 8/2 = 4
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(8^2+4^2) = color(red)(8.9443)`

Area of `DCE = (1/2)*a*EF = (1/2)*8*8.9443 = color(red)(35.7772)`
Lateral surface area `= 4*Delta DCE = 4*35.7772 = color(blue)(143.1088)`

`/_C = pi/3, /_C/2 = pi/6`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=8sin(pi/6)= 4

#OC = d_1/2 = BC cos (C/2) = 8* cos (pi/6) = 6.9282

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*4) (2*6.9282) = color (blue)(55.4256)`

T S A `= Lateral surface area + Base area`
T S A `=143.1088 + 55.4256 = color(purple)(198.5344)`