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 8, its base has sides of length 2 2, and its base has a corner with an angle of (3 pi)/8 3π8. What is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 12, 2017

T S A = 35.9447

Explanation:

AB = BC = CD = DA = a = 2
Height OE = h = 8
OF = a/2 = 2/2 = 1
EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(8^2+1^2) = color(red)(8.0623)EF=√EO2+OF2=√h2+(a2)2=√82+12=8.0623

Area of DCE = (1/2)*a*EF = (1/2)*2*8.0623 = color(red)(8.0623)DCE=(12)⋅a⋅EF=(12)⋅2⋅8.0623=8.0623
Lateral surface area = 4*Delta DCE = 4*8.0623 = color(blue)(32.2492)

/_C = (3pi)/8
Area of base ABCD = a* a * sin /_C = 2^2 * sin (3pi)/8 = 3.6955

T S A = Lateral surface area + Base area
T S A =32.2492 + 3.6955 = color(purple)(35.9447)

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