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 7 7, its base has sides of length 6 6, and its base has a corner with an angle of (3 pi)/4 3π4. What is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 12, 2017

T S A = 116.8416

Explanation:

AB = BC = CD = DA = a = 6
Height OE = h = 7
OF = a/2 = 6/2 = 3
EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(7^2+3^2) = color(red)(7.6158)EF=√EO2+OF2=√h2+(a2)2=√72+32=7.6158

Area of DCE = (1/2)*a*EF = (1/2)*6*7.6158 = color(red)(22.8474)DCE=(12)⋅a⋅EF=(12)⋅6⋅7.6158=22.8474
Lateral surface area = 4*Delta DCE = 4*22.8474 = color(blue)(91.3896)

/_C = pi - ((3pi)/4) = (pi)/4
Area of base ABCD = a* a * sin /_C = 6^2 sin (pi/4) = 25.452

T S A = Lateral surface area + Base area
T S A =91.3896 + 25.452 = color(purple)(116.8416)

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