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

Answer 1

`34.82\ \text{unit}^2`

Explanation:

Area of rhombus base with each side `3` & an interior angle `{\pi}/6`

`=3\cdot 3\sin({\pi}/6)=4.5`

The rhombus shaped base of pyramid has its semi-diagonals

`3\cos({\pi}/12)=2.898` &

`3\sin({\pi}/12)=0.776`

Now, two unequal lateral edges of each triangular lateral face of pyramid are given as

`\sqrt{5^2+(2.898)^2}=5.779` &

`\sqrt{5^2+(0.776)^2}=5.06`

There are four identical triangular lateral faces of pyramid each has sides `3, 5.779` & `5.06`

Area of each of four identical triangular lateral faces with sides `3, 5.779` & `5.06`

semi-perimeter of triangle, `s={3+5.779+5.06}/2=6.919`

Now, using heron's formula the area of lateral triangular face of pyramid

`=\sqrt{6.919(6.919-3)(6.919-5.779)(6.919-5.06)}`

`=7.58`

Hence, the total surface area of pyramid (including area of base)

`=4(\text{area of lateral triangular face})+\text{area of rhombus base}`

`=4(7.58)+4.5`

`=34.82\ \text{unit}^2`

Note:
Height of pyramid is taken as vertical height of pyramid

Answer 2

`color(maroon)(T S A = A_R = 31.32 + 4.5 = 35.82 " sq units"`

Explanation:


`"Total Surface Area of Rhombic Pyramid ' = A_R = L S A (A_S) + Area of Base (A_T)"`

`L S A = A_S = 4 * (1/2) * a * S, A_B = a^2 sin theta`

`A_S = 4 * (1/2) * a * sqrt((a/2)^2 + h^2) = 6 * sqrt((3/2)^2 + 5^2) = 31.32`

`A_B = 3^2 sin (pi/6) = 9/2 = 4.5`

`color(maroon)(T S A = A_R = 31.32 + 4.5 = 35.82 " sq units"`