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?

2 Answers

#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

Jul 14, 2018

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

Explanation:

https://math.tutorvista.com/geometry/surface-area-of-a-pyramid.html

#"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"#