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 $8$ , its base has sides of length $4$ , and its base has a corner with an angle of $(3 pi)/8$ . What is the pyramid's surface area?

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Answer 1 · 2017-12-06T12:41:21

T S A = 80.7519

AB = BC = CD = DA = a = 4
Height OE = h = 8
OF = a/2 = 1/2 = 2
$EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(8^2+2^2) = color(red)(8.2462)$

Area of $DCE = (1/2)*a*EF = (1/2)*4*8.2462 = color(red)(16.4924)$
Lateral surface area $= 4*Delta DCE = 4*16.4924 = color(blue)(65.9696)$

$/_C = (3pi)/8, /_C/2 = (3pi)/16$
diagonal $AC = d_1$ & diagonal $BD = d_2$
#OB = d_2/2 = BCsin (C/2)=4sin((3pi)/16)= 2.2223

#OC = d_1/2 = BC cos (C/2) = 4 cos ((3pi)/16) = 3.3259*

Area of base ABCD $= (1/2)*d_1*d_2 = (1/2)(2*2.2223) (2*3.3259) = color (blue)(14.782)$

T S A $= Lateral surface area + Base area$
T S A $=65.9696 + 14.7823 = color(purple)(80.7519)$

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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 4 4 , and its base has a corner with an angle of (3 pi)/8 (3 pi)/8 . What is the pyramid's surface area?