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 9, its base has sides of length 2 2, and its base has a corner with an angle of pi/4 π4. What is the pyramid's surface area?

1 Answer
sankarankalyanam
Oct 24, 2017

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 = 9.0554EF=√EO2+OF2=√h2+(a2)2=√92+(12)=√82=9.0554

Area of DCE = (1/2)*a*EF = (1/2)*2*9.0554 = 9.0554DCE=(12)⋅a⋅EF=(12)⋅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#
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