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 $8$ , 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 · 2018-07-04T10:16:48

$152.75\ \text{unit}^2$

Area of rombus base with each side $8$ & an interior angle ${2\pi}/3$

$=8\cdot 8\sin({2\pi}/3)=55.426$

The rhombus base of pyramid has its semi-diagonals $8\cos(\pi/6)$ & $8\sin(\pi/6)$ i.e. $4\sqrt3$ & $4$

Now, the sides of triangular lateral face of pyramid as given as

$\sqrt{5^2+(4\sqrt3)^2}=\sqrt{73}=8.544$ &

$\sqrt{5^2+(4)^2}=\sqrt{41}=6.403$

Each of 4 identical lateral triangular faces of pyramid has the sides $8, 8.544$ & $6.403$

semi-perimeter of triangle, $s={8+8.544+6.403}/2=11.4735$

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

$=\sqrt{11.4735(11.4735-8)(11.4735-8.544)(11.4735-6.403)}$

$=24.331$

Hence, the total surface area of pyramid

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

$=4(24.331)+55.426$

$=152.75\ \text{unit}^2$

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