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 `3`, its base has sides of length `5`, and its base has a corner with an angle of `pi/4`. What is the pyramid's surface area?

Answer 1

`color(green)(T S A)` `color(green)(A_T = A_L + A_R = 74.4063)`

Explanation:

Total Surface Area of the pyramid is sum of the areas of the rhombus base `(A_R)` and the four side triangles `(A_L= 4 * A_t)`
`A_T = A_L + A_R = 4A_t + A_R`

`A_R = a*asin theta = 5 * 5 sin (pi/4) = 25sqrt2 = 35.3553`#

A_L = 4 A_t = 4 * (1/2) (a * l)# where l is the slant height of the triangle.

But `l = sqrt((a/2)^2 + h^2) = sqrt((5/2)^2 + 3^2) = 3.9051`

`:. A_L = 4 * (1/2) * 5 * 3.9051 = 39.051`

`color(green)(T S A)` `color(green)(A_T = A_L + A_R = 39.051 + 35.3553 = 74.4063)`