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 `8`, its base has sides of length `3`, and its base has a corner with an angle of `( pi)/6`. What is the pyramid's surface area?

Answer 1

T S A = 56.185

Explanation:

AB = BC = CD = DA = a = 3
Height OE = h = 8
OF = a/2 = 3/2 = 1.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(8^2+(1.5)^2) = color(red)8.1394`

Area of `DCE = (1/2)*a*EF = (1/2)*3*8.1394 = color(red)(12.2091)`
Lateral surface area `= 4*Delta DCE = 4*12.2091 = color(blue)(48.8364)`

`/_C = pi/6, /_C/2 = pi/12`
diagonal `AC = d_1` & diagonal `BD = d_2`
`OB = d_2/2 = BC*sin (C/2)=3*sin(pi/12) = **0.7765**`

#OC = d_1/2 = BC cos (C/2) = 3*cos (pi/12) = 2.8978

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*0.7765)(2*2.8978) = color (blue)(7.3486)`

Total Surface Area `= Lateral surface area + Base area`
T S A `=48.8364 + 7.3486 = color(purple)(56.185)`