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 has sides of length `5`, and its base has a corner with an angle of `pi/3`. What is the pyramid's surface area?

Answer 1

T S A `= color(purple)(95.98)`

Explanation:

AB = BC = CD = DA = a = 5
Height OE = h = 7
OF = a/2 = 5/2 = 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 = pi/3, /_C/2 = pi/6`
diagonal `AC = d_1` & diagonal `BD = d_2`
#OB = d_2/2 = BCsin (C/2)=5sin(pi/6)= 2.5

#OC = d_1/2 = BC cos (C/2) = 5* cos (pi/6) = 4.33

Area of base ABCD `= (1/2)*d_1*d_2 = (1/2)(2*2.5) (2*4.33) = color (blue)(21.65)`

T S A `= Lateral surface area + Base area`
T S A `=74.33 + 21.65 = color(purple)(95.98)`