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

Answer 1

T S A = 19.0133

Explanation:

AB = BC = CD = DA = a = 2
Height OE = h = 3
OF = a/2 = 1/2 = 1
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(3^2+1^2) = color(red)(3.1623)`

Area of `DCE = (1/2)*a*EF = (1/2)*2*3.1623 = color(red)(3.1623)`
Lateral surface area `= 4*Delta DCE = 4*3.1623 = color(blue)(12.6492`)#

`/_C = pi/4, /_C/2 = pi/8`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=3sin(pi/8)= 1.1481

#OC = d_1/2 = BC cos (C/2) = 3* cos (pi/8) = 2.7716

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*1.1481) (2*2.7716) = color (blue)(6.3641)`

T S A `= Lateral surface area + Base area`
T S A `=12.6492 + 6.3641 = color(purple)(19.0133)`