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

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

James Atkins · Accepted Answer

#18sqrt3+36sqrt2# The pyramid's surface area = base area + 4 x plane area. Base area #=bcsin A#, where #b=6#, #c=6# and #A=pi/3# Base area #=6*6sin(pi/3)# #=36*sqrt3/2=18sqrt3# Plane area, we have to find the length from peak to the mid point of side. Since the length from mid point of side =3, and the height =3, therefore, the length from peak to the mid point of side #= sqrt(3^2+3^2)# #=sqrt18=3sqrt2# The area of one plane #=1/2*6*3sqrt2=9sqrt2# Total plane area #= 4*9sqrt2=36sqrt2# Total area of pyramid #=18sqrt3+36sqrt2#

Caroline Aucoin · Answer

undefined The surface area $ A $ of a pyramid with a rhombus base can be calculated using the formula:

$$ A = \text{Area of Base} + \text{Area of Each Triangle Face} \times 4 $$

For a rhombus with side length $ s $ and an angle $ \theta $ between adjacent sides, the area is $ A_{\text{rhombus}} = s^2 \sin(\theta) $.

Given that the side length of the rhombus base is $ 6 $ and one angle is $ \frac{\pi}{3} $, the area of the base is:

$$ A_{\text{base}} = 6^2 \sin\left(\frac{\pi}{3}\right) $$

The height of the pyramid is $ 3 $, so the area of each triangular face is:

$$ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 $$

Thus, the total surface area of the pyramid is:

$$ A = A_{\text{base}} + A_{\text{triangle}} \times 4 $$