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 $8$ , its base has sides of length $9$ , and its base has a corner with an angle of $(2 pi)/3$ . What is the pyramid's surface area?

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Answer 1 · 2018-07-14T00:10:44

$230.328\ \text{unit}^2$

Area of rhombus base with each side $9$ & an interior angle ${2\pi}/3$

$=9\cdot 9\sin({2\pi}/3)=70.148$

The rhombus shaped base of pyramid has its semi-diagonals

$9\cos({\pi}/6)=4.5$ &

$9\sin({\pi}/6)=7.794$

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

$\sqrt{8^2+(4.5)^2}=9.179$ &

$\sqrt{8^2+(7.794)^2}=11.169$

There are four identical triangular lateral faces of pyramid each has sides $9, 9.179$ & $11.169$

Area of each of four identical triangular lateral faces with sides $9, 9.179$ & $11.169$
semi-perimeter of triangle, $s={9+9.179+11.169}/2=14.674$

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

$=\sqrt{14.674(14.674-9)(14.674-9.179)(14.674-11.169)}$

$=40.045$

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

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

$=4(40.045)+70.148$

$=230.328\ \text{unit}^2$

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

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 8 8 , its base has sides of length 9 9 , and its base has a corner with an angle of (2 pi)/3 (2 pi)/3 . What is the pyramid's surface area?