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 `4`, 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 = 80.7519

Explanation:

AB = BC = CD = DA = a = 4
Height OE = h = 8
OF = a/2 = 1/2 = 2
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(8^2+2^2) = color(red)(8.2462)`

Area of `DCE = (1/2)*a*EF = (1/2)*4*8.2462 = color(red)(16.4924)`
Lateral surface area `= 4*Delta DCE = 4*16.4924 = color(blue)(65.9696)`

`/_C = (3pi)/8, /_C/2 = (3pi)/16`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=4sin((3pi)/16)= 2.2223

#OC = d_1/2 = BC cos (C/2) = 4* cos ((3pi)/16) = 3.3259

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*2.2223) (2*3.3259) = color (blue)(14.782)`

T S A `= Lateral surface area + Base area`
T S A `=65.9696 + 14.7823 = color(purple)(80.7519)`