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

Answer 1

TVS A = 43.4721

Explanation:

AB = BC = CD = DA = a = 3
Height OE = h = 6
OF = a/2 = 3/2 = 1.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(6^2+1.5^2) = color(red)(6.1847)`

Area of `DCE = (1/2)*a*EF = (1/2)*3*6.1847 = color(red)(9.2771)`
Lateral surface area `= 4*Delta DCE = 4*9.2771 = color(blue)(37.1084)`

`/_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 `=37.1084 + 6.3641 = color(purple)(43.4721)`