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

Answer 1

T S A = 59.5175

Explanation:

AB = BC = CD = DA = a = 4
Height OE = h = 7
OF = a/2 = 1/2 = 2
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(6^2+2^2) = color(red)(6.3246)`

Area of `DCE = (1/2)*a*EF = (1/2)*4*6.3246 = color(red)(12.6492)`
Lateral surface area `= 4*Delta DCE = 4*12.6492 = color(blue)(50.5968)`

`/_C = (pi)/4, /_C/2 = (pi)/8`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=4sin((pi)/8)= 1.5307

#OC = d_1/2 = BC cos (C/2) = 4* cos ((pi)/8) = 3.6955

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*1.5307) (2*3.6955) = color (blue)(8.9207)`

T S A `= Lateral surface area + Base area`
T S A `=50.5958 + 8.9207 = color(purple)(59.5175)`