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

Answer 1

T S A = 20.5387

Explanation:

AB = BC = CD = DA = a = 5
Height OE = h = 2
OF = a/2 = 1/2 = 2.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(2^2+2.5^2) = color(red)(3.2016)`

Area of `DCE = (1/2)*a*EF = (1/2)*5*3.2016 = color(red)(2.5125)`
Lateral surface area `= 4*Delta DCE = 4*2.5125 = color(blue)(8.04)`

`/_C = pi/6, /_C/2 = pi/12`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=5sin(pi/12)= 1.294

#OC = d_1/2 = BC cos (C/2) = 5* cos (pi/12) = 4.8295

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*1.294) (2*4.8295) = color (blue)(12.4987)`

T S A `= Lateral surface area + Base area`
T S A `=8.04 + 12.4987 = color(purple)(20.5387)`