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 `4`, its base's sides have lengths of `1`, and its base has a corner with an angle of `(5 pi)/6`. What is the pyramid's surface area?

Answer 1

T S A = 8.5624

Explanation:

AB = BC = CD = DA = a = 1
Height OE = h = 4
OF = a/2 = 1/2 = 0.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(4^2+0.5^2) = color(red)(4.0311)`

Area of `DCE = (1/2)*a*EF = (1/2)*1*4.0311 = color(red)(2.0156)`
Lateral surface area `= 4*Delta DCE = 4*2.0156 = color(blue)(8.0624)`

`/_C =(pi) - ((5pi)/6) = (pi)/6`
Area of base ABCD `= a* a * sin /_C = 1^2 sin (pi/6) = 0.5`

T S A `= Lateral surface area + Base area`
T S A `=8.0624 + 0.5 = color(purple)(8.5624)`