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 `5`, its base's sides have lengths of `1`, and its base has a corner with an angle of `(5 pi)/6`. What is the pyramid's surface area?

Answer 1

T S A = 10.5499

Explanation:

AB = BC = CD = DA = a = 1
Height OE = h = 5
OF = a/2 = 1/2 = 0.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(5^2+0.5^2) = color(red)(5.0249)`

Area of `DCE = (1/2)*a*EF = (1/2)*1*5.0249 = color(red)(2.5125)`
Lateral surface area `= 4*Delta DCE = 4*2.5125 = color(blue)(10.05`)#

`/_C = (5pi)/6, /_C/2 = (5pi)/12`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=1sin((5pi)/12)= 0.9659

#OC = d_1/2 = BC cos (C/2) = 1* cos ((5pi)/12) = 0.2588

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*0.2588) (2*0.9659) = color (blue)(0.4999)`

T S A `= Lateral surface area + Base area`
T S A `=10.05 + 0.4999 = color(purple)(10.5499)`