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 `2`, its base has sides of length `1`, and its base has a corner with an angle of `(3 pi)/8`. What is the pyramid's surface area?

Answer 1

T S A = 5.0472

Explanation:

AB = BC = CD = DA = a = 1
Height OE = h = 2
OF = a/2 = 1/2 = 0.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(2^2+0.5^2) = color(red)(2.0616)`

Area of `DCE = (1/2)*a*EF = (1/2)*1*2.0616 = color(red)(1.0308)`
Lateral surface area `= 4*Delta DCE = 4*1.0308 = color(blue)(4.1232`)#

`/_C = (3pi)/8, /_C/2 = (3pi)/16`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=1sin((3pi)/16)= 0.5556

#OC = d_1/2 = BC cos (C/2) = 1 cos ((3pi)/16) = 0.8315*

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*0.5556) (2*0.8315) = color (blue)(0.924)`

T S A `= Lateral surface area + Base area`
T S A `=4.1232 + 0.924 = color(purple)(5.0472)`