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 `3`, 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 = 39.6354

Explanation:

AB = BC = CD = DA = a = 3
Height OE = h = 5
OF = a/2 = 1/2 = 1.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(5^2+1.5^2) = color(red)(5.2201)`

Area of `DCE = (1/2)*a*EF = (1/2)*3*5.2201 = color(red)(7.8302)`
Lateral surface area `= 4*Delta DCE = 4*7.8302 = color(blue)(31.3206`)#

`/_C = (3pi)/8, /_C/2 = (3pi)/16`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=3sin((3pi)/16)= 1.6667

#OC = d_1/2 = BC cos (C/2) = 3 cos ((3pi)/16) = 2.4944*

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*1.6667) (2*2.4944) = color (blue)(8.3148)`

T S A `= Lateral surface area + Base area`
T S A `=31.3206 + 8.3148 = color(purple)(39.6354)`