A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 1 1 and 4 4 and the pyramid's height is 2 2. If one of the base's corners has an angle of (5pi)/125π12, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 12, 2017

T S A =/14.9704

Explanation:

CH = 1 * sin ((5pi)/12) = 0.9659CH=1⋅sin(5π12)=0.9659
Area of parallelogram base = a * b1 = 4*0.9659 = color(red)(3.876 )=a⋅b1=4⋅0.9659=3.876

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (4/2)^2)= 2.8284EF=h1=√h2+(a2)2=√22+(42)2=2.8284
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 2.828= color(red)(1.414)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(1/2)^2 )= 2.0616
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*4*2.0616 = color(red)( 4.1232)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 1.414)+ (2* 4.1232) = color(red)(11.0744)

Total surface area =Area of parallelogram base + Lateral surface area = 3.876 + 11.0744 = 14.9707

Total Surface Area # T S A = **14.9704**#enter image source here

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