The base of a triangular pyramid is a triangle with corners at $(5, 1)$ $(5 ,1 )$ , $(2, 3)$ $(2 ,3 )$ , and $(9, 4)$ $(9 ,4 )$ . If the pyramid has a height of $4$ $4$ , what is the pyramid's volume?

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Answer 1 · 2016-03-18T12:58:33

Volume of pyramid is $11.336$ $11.336$ units

First to find area of base, let us find all the sides of base triangle.

$a = \sqrt{{(2 - 5)}^{2} + {(3 - 1)}^{2}} = \sqrt{9 + 4} = \sqrt{13} = 3.606$ $a=sqrt((2-5)^2+(3-1)^2)=sqrt(9+4)=sqrt13=3.606$

$b = \sqrt{{(9 - 2)}^{2} + {(4 - 3)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = 7.071$ $b=sqrt((9-2)^2+(4-3)^2)=sqrt(49+1)=sqrt50=7.071$

$c = \sqrt{{(9 - 5)}^{2} + {(4 - 1)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$ $c=sqrt((9-5)^2+(4-1)^2)=sqrt(16+9)=sqrt25=5$

Now for using Heron's formula, $s = \frac{1}{2} (3.606 + 7.071 + 5) = \frac{15.677}{2} = 7.8385$ $s=1/2(3.606+7.071+5)=15.677/2=7.8385$

Hence, area of base triangle is

$\sqrt{7.8385 \times (7.8385 - 3.606) \times (7.8385 - 7.071) \times (7.8385 - 5)}$ $sqrt(7.8385xx(7.8385-3.606)xx(7.8385-7.071)xx(7.8385-5)$

= $\sqrt{7.8385 \times 4.2325 \times 0.7675 \times 2.8385} = 8.502$ $sqrt(7.8385xx4.2325xx0.7675xx2.8385)=8.502$

Hence, Volume of pyramid is $\frac{1}{3} \times 8.502 \times 4 = 11.336$ $1/3xx8.502xx4=11.336$

The base of a triangular pyramid is a triangle with corners at (5,1)(5 ,1 ), (2,3)(2 ,3 ), and (9,4)(9 ,4 ). If the pyramid has a height of 44 , what is the pyramid's volume?