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 `5`, and its base has a corner with an angle of `(5 pi)/8`. What is the pyramid's surface area?

Answer 1

T S A = 97.4271

Explanation:

AB = BC = CD = DA = a = 5
Height OE = h = 7
OF = a/2 = 1/2 = 2.5
`EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(7^2+2.5^2) = color(red)(7.433)`

Area of `DCE = (1/2)*a*EF = (1/2)*5*7.433 = color(red)(18.5825)`
Lateral surface area `= 4*Delta DCE = 4*18.5825 = color(blue)(74.33)`

`/_C = (5pi)/8, /_C/2 = (5pi)/16`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=5sin((5pi)/16)= 4.1573

#OC = d_1/2 = BC cos (C/2) = 5* cos ((5pi)/16) = 2.7779

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*4.1573) (2*2.7779) = color (blue)(23.0971)`

T S A `= Lateral surface area + Base area`
T S A `=74.33 + 23.0971 = color(purple)(97.4271)`