The base of a triangular pyramid is a triangle with corners at (8 ,5 )(8,5), (2 ,7 )(2,7), and (3 ,1 )(3,1). If the pyramid has a height of 8 8, what is the pyramid's volume?

1 Answer
Mar 7, 2016

V_(Pr) = BA xx HVPr=BA×H
A_Delta = 1/2 (6.08*6.40) sin60.8 ~~ 17
V_(Py) = 1/3V_(Pr)
V_(Py) = 1/3 (17*8) = 45.3 cubic units
you can also use determinant to get the area of the triangle:
A_Delta = 1/2 det|vec(v_1)xxvec(v_2)| = 1/2det|(-1, 6),(5, 4)| = 17
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Explanation:

This is a straightforward implementation of the Pyramid volume formula: V_(Py) = 1/3V_(Pr) where
V_(Py) = Volume of Pyramid and V_(Pr) = Volume of Prism
with same base and same height.
Now the volume of any prism is straightforward accumulation of the base area over the dimension of the height of the prism, H.
V_(Pr) = BA xx H where
BA = Base Area of prism and H = Height of prism.
Now find the vector that enclose the triangle and use the cross product between them to find the Area, A_Delta = 1/2 A_square .
The vectors are:
vec(v_1) = (2i + 7j) - (3i+ j) = -i +6j; |vec(v_1)|=6.08
vec(v_2) = (8i + 5j) - (3i+ j) = -5i +4j; |vec(v_2)|=6.40
vec(v_3) = (2i + 7j) - (8i+ 5j) = -6i +2j; |vec(v_3)|=6.32

Now the cross product of any of this three vectors will yield the area of a parallelogram, A_square. Let's pick vecv_1 and vecv_2
The angle between them is alpha = 60.8^0

A_Delta = 1/2A_(square) =1/2 |vec(v_1)| xx |vec(v_2)| sin60.8
Theta now since vecv_1 and vecv_2 the area is now simply:
A_Delta = 1/2 (6.08*6.40) sin60.8 ~~ 17
Now we know height, H = 8
Thus the V_(Py) = 1/3 (17*8) = 45.3 cubic units

You can use vectors vecv_2 and vecv_3 and get the same
area ~~17
The angle between them is beta= 57.09^0
A_Delta = 1/2 (6.32*6.40) sin57.09 ~~ 17 as before

The other way is simply determine the cross product using the determinant. Again let's consider vec(v_1) and vec(v_2)
Then A_Delta = 1/2 det|vec(v_1)xxvec(v_2)| = 1/2det|(-1, 6),(5, 4)| = 17