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 `5`, its base has sides of length `4`, and its base has a corner with an angle of `pi/3`. What is the pyramid's surface area?

Answer 1

T S A = 56.9376

Explanation:

AB = BC = CD = DA = a = 4
Height OE = h = 5
OF = a/2 = 4/2 = 2
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(5^2+2^2) = color(red)(5.3852)`

Area of `DCE = (1/2)*a*EF = (1/2)*4*5.3852 = color(red)(10.7704)`
Lateral surface area `= 4*Delta DCE = 4*10.7704 = color(blue)(43.0816)`

`/_C = (pi)/3`
Area of base ABCD `= a* a * sin /_C = 4^2 sin (pi/3) = 13.856`

T S A `= Lateral surface area + Base area`
T S A `=43.0816 + 13.856 = color(purple)(56.9376)`