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 `7`, its base's sides have lengths of `1`, and its base has a corner with an angle of `(7 pi)/8`. What is the pyramid's surface area?

Answer 1

T S A = 14.4183

Explanation:

AB = BC = CD = DA = a = 1
Height OE = h = 7
OF = a/2 = 1/2 = 0.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(7^2+0.5^2) = color(red)(7.0178)`

Area of `DCE = (1/2)*a*EF = (1/2)*1*7.0178 = color(red)(3.5089)`
Lateral surface area `= 4*Delta DCE = 4*3.5089 = color(blue)(14.0356)`

`/_C = (7pi)/8, /_C/2 = (7pi)/16`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=1sin((7pi)/16)= 0.9808

#OC = d_1/2 = BC cos (C/2) = 1* cos ((7pi)/16) = 0.1951

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*0.1951) (2*0.9808) = color (blue)(0.3827)`

T S A `= Lateral surface area + Base area`
T S A `=14.0356 + 0.3827 = color(purple)(14.4183)`