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

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Answer 1 · 2018-05-17T06:46:57

Volume of triangular base pyramid $V = 466.47$ $color(violet)(V = 466.47$ cub. units

$A (2, 4), B (3, 2), C (5, 5)$ $A (2,4), B (3,2), C (5,5)$

Using distance formula,

$¯ ¯¯¯¯ ¯ A B = c = \sqrt{{(3 - 2)}^{2} + {(2 - 4)}^{2}} = \sqrt{5} = 2.235$ $bar (AB) = c = sqrt((3-2)^2 + (2-4)^2) = sqrt5 = 2.235$

$¯ ¯¯¯¯ ¯ B C = a = \sqrt{{(5 - 2)}^{2} + {(5 - 4)}^{2}} = \sqrt{10} = 3.162$ $bar (BC) = a = sqrt((5-2)^2 + (5-4)^2) = sqrt 10 = 3.162$

$¯ ¯¯¯¯ ¯ A C = b = \sqrt{{(3 - 5)}^{2} + {(2 - 5)}^{2}} = \sqrt{13} = 3.606$ $bar (AC) = b = sqrt((3-5)^2 + (2-5)^2) = sqrt 13= 3.606$

Area of base triangle $A_{b} = \sqrt{s (s - a) (s - b) (s - c)}$ $A_b= sqrt(s (s-a) (s-b) (s-c))$

Semi perimeter $s \frac{a - + b + c}{2} = \frac{2.235 + 3.162 + 3.606}{2} = 18$ $s (a -+ b+ c) / 2 = (2.235 + 3.162 + 3.606)/2 = 18$

$A_{b} = \sqrt{18 \cdot 15.765 \cdot 16.838 \cdot 16.394} = 279.88$ $A_b = sqrt(18* 15.765 * 16.838 * 16.394) = 279.88$

Formula for volume of pyramid $V = (\frac{1}{3}) \cdot A_{b} \cdot h$ $V = (1/3) * A_b * h$

$V = (\frac{1}{3}) \cdot 279.88 \cdot 5 = 466.47$ $color(violet)(V = (1/3) * 279.88 * 5 = 466.47$ cub. units.

The base of a triangular pyramid is a triangle with corners at (2,4)(2 ,4 ), (3,2)(3 ,2 ), and (5,5)(5 ,5 ). If the pyramid has a height of 55 , what is the pyramid's volume?