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 `9`, its base has sides of length `2`, and its base has a corner with an angle of `pi/4`. What is the pyramid's surface area?

Answer 1

Total surface area T S A Z= 39.95#

Explanation:

AB = BC = CD = DA = a = 2
Height OE = h = 9
OF = a/2 = 2/2 = 1
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(9^2+(1^2)=sqrt82 = 9.0554`

Area of `DCE = (1/2)*a*EF = (1/2)*2*9.0554 = 9.0554`
Lateral surface area `= 4*Delta DCE = 4*9.0554 = 36.2215`

`/_C = pi/4, /_C/2 = pi/8`
diagonal `AC = d_1` & diagonal `BD = d_2`
`OB = d_2/2 = BC*sin (C/2)=2*sin(pi/8) = 0.7654`
`OC = d_1/2 = BC cos (C/2) = 2* cos (pi/8) = 1.8478`

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*0.7654)(2*1.8478)= 2.8286`

Total Surface Area #= Lateral surface area + Base area. T S A # 36.22115 + 2.8286 =39.05#