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 `3`, its base has sides of length `9`, and its base has a corner with an angle of `pi/4`. What is the pyramid's surface area?

Answer 1

Total Surface Area `= 200.2119`

Explanation:

AB = BC = CD = DA = a = 9
Height OE = h = 3
OF = a/2 = 9/2
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(3^2+(9/2)^2)=5.4083`

Area of `DCE = (1/2)*a*EF = (1/2)*9*5.4083 = 24.3375`
Lateral surface area `= 4*Delta DCE = 4*24.3375 = 94.3799`

`/_C = pi/4, /_C/2 = pi/8`
diagonal `AC = d_1` & diagonal `BD = d_2`
`OB = d_2/2 = BC*sin (C/2)=9*sin(pi/8) = 6.364`
`OC = d_1/2 = BC cos (C/2) = 9* cos (pi/8) = 8.3149`

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*6.364)(2*8.3149) = 105.832`

Total Surface Area #= Lateral surface area + Base area. T S A # 94.3799 + 105.832 = 200.2119#