Question

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

Answer 1

T S A `= color(purple)(190.8252)`

Explanation:

AB = BC = CD = DA = a = 8
Height OE = h = 3
OF = a/2 = 8/2 = 4
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(3^2+4^2) = color(red)5`

Area of `DCE = (1/2)*a*EF = (1/2)*8*5 = color(red)(20)`
Lateral surface area `= 4*Delta DCE = 4*20 = color(blue)(80)`

`/_C = pi/3, /_C/2 = pi/6`
diagonal `AC = d_1` & diagonal `BD = d_2`
`OB = d_2/2 = BC*sin (C/2)=8*sin(pi/6)= **4**`

#OC = d_1/2 = BC cos (C/2) = 8 cos (pi/6) = 6.9282*

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*8) (2*6.9282) = color (blue)(110.8512)`

Total Surface Area `= Lateral surface area + Base area`
T S A `=80 + 110.8252 = color(purple)(190.8252)`