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 has sides of length `1`, and its base has a corner with an angle of `pi/4`. What is the pyramid's surface area?

Answer 1

T S A is 20.8067

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)(5.0249)`
Lateral surface area `= 4*Delta DCE = 4*5.0249 = color(blue)(20.0996)`

`/_C = pi - (3pi)/4 = (pi)/4`
Area of base ABCD `= a* a * sin /_C = 1^2 sin (pi/4) = 0.7071`

T S A `= Lateral surface area + Base area`
T S A `=20.0996 + 0.7071 = color(purple)(20.8067)`