The base of a triangular pyramid is a triangle with corners at $(3, 9)$ $(3 ,9 )$ , $(1, 6)$ $(1 ,6 )$ , and $(2, 8)$ $(2 ,8 )$ . If the pyramid has a height of $4$ $4$ , what is the pyramid's volume?

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Answer 1 · 2016-03-18T12:48:19

Volume of pyramid is $0.661$ $0.661$ units.

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

$a = \sqrt{{(1 - 3)}^{2} + {(6 - 9)}^{2}} = \sqrt{4 + 9} = \sqrt{13} = 3.606$ $a=sqrt((1-3)^2+(6-9)^2)=sqrt(4+9)=sqrt13=3.606$

$b = \sqrt{{(2 - 3)}^{2} + {(8 - 9)}^{2}} = \sqrt{1 + 1} = \sqrt{2} = 1.414$ $b=sqrt((2-3)^2+(8-9)^2)=sqrt(1+1)=sqrt2=1.414$

$c = \sqrt{{(2 - 1)}^{2} + {(8 - 6)}^{2}} = \sqrt{1 + 4} = \sqrt{5} = 2.236$ $c=sqrt((2-1)^2+(8-6)^2)=sqrt(1+4)=sqrt5=2.236$

Now for using Heron's formula, $s = \frac{1}{2} (3.606 + 1.414 + 2.236) = \frac{7.256}{2} = 3.628$ $s=1/2(3.606+1.414+2.236)=7.256/2=3.628$

Hence, area of base triangle is

$\sqrt{3.628 \times (3.628 - 3.606) \times (3.628 - 1.414) \times (3.628 - 2.236)}$ $sqrt(3.628xx(3.628-3.606)xx(3.628-1.414)xx(3.628-2.236)$

= $\sqrt{3.628 \times 0.022 \times 2.214 \times 1.392} = 0.496$ $sqrt(3.628xx0.022xx2.214xx1.392)=0.496$

Hence, Volume of pyramid is $\frac{1}{3} \times 0.496 \times 4 = 0.661$ $1/3xx0.496xx4=0.661$

The base of a triangular pyramid is a triangle with corners at (3,9)(3 ,9 ), (1,6)(1 ,6 ), and (2,8)(2 ,8 ). If the pyramid has a height of 44 , what is the pyramid's volume?