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 4 4, its base's sides have lengths of 3 3, and its base has a corner with an angle of (5 pi)/6 5π6. What is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 25, 2017

T S A = 30.132

Explanation:

AB = BC = CD = DA = a = 3
Height OE = h = 4
OF = a/2 = 3/2 = 1.5
EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(4^2+1.5^2) = color(red)(4.272)EF=√EO2+OF2=√h2+(a2)2=√42+1.52=4.272

Area of DCE = (1/2)*a*EF = (1/2)*3*4.272 = color(red)(6.408)DCE=(12)⋅a⋅EF=(12)⋅3⋅4.272=6.408
Lateral surface area = 4*Delta DCE = 4*6.408 = color(blue)(25.632)

/_C = pi - (5pi)/6 = (pi)/6
Area of base ABCD = a* a * sin /_C = 3^2 sin (pi/6) = 4.5

T S A = Lateral surface area + Base area
T S A =25.632 + 4.5 = color(purple)(30.132)

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