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

1 Answer
Dec 6, 2017

T S A = 39.6354

Explanation:

AB = BC = CD = DA = a = 3
Height OE = h = 5
OF = a/2 = 1/2 = 1.5
# EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(5^2+1.5^2) = color(red)(5.2201)#

Area of #DCE = (1/2)*a*EF = (1/2)*3*5.2201 = color(red)(7.8302)#
Lateral surface area #= 4*Delta DCE = 4*7.8302 = color(blue)(31.3206#)#

#/_C = (3pi)/8, /_C/2 = (3pi)/16#
diagonal #AC = d_1# & diagonal #BD = d_2#
#OB = d_2/2 = BCsin (C/2)=3sin((3pi)/16)= 1.6667

#OC = d_1/2 = BC cos (C/2) = 3* cos ((3pi)/16) = 2.4944

Area of base ABCD #= (1/2)*d_1*d_2 = (1/2)(2*1.6667) (2*2.4944) = color (blue)(8.3148)#

T S A #= Lateral surface area + Base area#
T S A # =31.3206 + 8.3148 = color(purple)(39.6354)#

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