The base of a triangular pyramid is a triangle with corners at $(6, 1)$ $(6 ,1 )$ , $(2, 4)$ $(2 ,4 )$ , and $(4, 3)$ $(4 ,3 )$ . If the pyramid has a height of $2$ $2$ , what is the pyramid's volume?

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The base of a triangular pyramid is a triangle with corners at $(6, 1)$ $(6 ,1 )$ , $(2, 4)$ $(2 ,4 )$ , and $(4, 3)$ $(4 ,3 )$ . If the pyramid has a height of $2$ $2$ , what is the pyramid's volume?

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Answer 1 · 2017-01-25T14:53:30

$\frac{2}{3} u n i t^{3}$ $2/3 unit^3$

Explanation:

let say $x_{1} = 6, x_{2} = 2, x_{3} = 4, y_{1} = 1, y_{2} = 4 and y_{3} = 3$ $x_1=6,x_2=2,x_3=4, y_1=1,y_2=4 and y_3=3$

pyramid volume $= \frac{1}{3}$ $=1/3$ base areaheight

Base area $= \frac{1}{2} | [(x_{1} \cdot y_{2}) + (x_{2} \cdot y_{3}) + (x_{3} \cdot y_{1})] - [(y_{1} \cdot x_{2}) + (y_{2} \cdot x_{3}) + (y_{3} \cdot x_{1})] |$ $=1/2|[(x_1*y_2)+(x_2*y_3)+(x_3*y_1)]-[(y_1*x_2) +(y_2*x_3)+(y_3*x_1)]|$

$= \frac{1}{2} | [(6 \cdot 4) + (2 \cdot 3) + (4 \cdot 1)] - [(1 \cdot 2) + (4 \cdot 4) + (3 \cdot 6)] |$ $=1/2|[(6*4)+(2*3)+(4*1)]-[(1*2)+(4*4)+(3*6)]|$

$= \frac{1}{2} | (24 + 6 + 4) - (2 + 16 + 18) |$ $=1/2|(24+6+4)-(2+16+18)|$

$= \frac{1}{2} | 34 - 36 |$ $=1/2|34-36|$
$= \frac{1}{2} \cdot 2 = 1 u n i t^{2}$ $=1/2*2=1 unit^2$

Therefore the pyramid's volume $= \frac{1}{3} \cdot 1 \cdot 2 = \frac{2}{3} u n i t^{3}$ $=1/3*1*2=2/3 unit^3$

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The base of a triangular pyramid is a triangle with corners at $(6, 1)$ $(6 ,1 )$ , $(2, 4)$ $(2 ,4 )$ , and $(4, 3)$ $(4 ,3 )$ . If the pyramid has a height of $2$ $2$ , what is the pyramid's volume?

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Explanation:

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Science

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Humanities

... and beyond

The base of a triangular pyramid is a triangle with corners at (6 ,1 )(6,1)(6 ,1 ), (2 ,4 )(2,4)(2 ,4 ), and (4 ,3 )(4,3)(4 ,3 ). If the pyramid has a height of 2 22 , what is the pyramid's volume?

1 Answer

Explanation:

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The base of a triangular pyramid is a triangle with corners at $(6, 1)$ $(6 ,1 )$ , $(2, 4)$ $(2 ,4 )$ , and $(4, 3)$ $(4 ,3 )$ . If the pyramid has a height of $2$ $2$ , what is the pyramid's volume?