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 `4`, its base's sides have lengths of `7`, 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 = `color(purple)(98.9239)`

Explanation:

AB = BC = CD = DA = a = 7
Height OE = h = 4
OF = a/2 = 7/2 = 3.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(4^2+(3.5)^2) = color(red)5.3151`

Area of `DCE = (1/2)*a*EF = (1/2)*7*5.3151 = color(red)(18.6029)`
Lateral surface area `= 4*Delta DCE = 4*18.6029 = color(blue)(74.4116)`

`/_C =5 pi/6, /_C/2 = 5pi/12`
diagonal `AC = d_1` & diagonal `BD = d_2`
`OB = d_2/2 = BC*sin (C/2)=7*sin(5pi/12) = **6.765**`

#OC = d_1/2 = BC cos (C/2) = 7*cos (5pi/12) = 1.8117

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*6.765)(2*1.8117) = color (blue)(24.5123)`

Total Surface Area `= Lateral surface area + Base area`
T S A `=74.4116 + 24.5123 = color(purple)(98.9239)`