The base of a triangular pyramid is a triangle with corners at $(6 ,7 )$ , $(3 ,1 )$ , and $(5 ,9 )$ . If the pyramid has a height of $8$ , what is the pyramid's volume?

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Answer 1 · 2018-01-31T01:53:16

$color(green)(V = 16.5781)$ cubic units

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Triangle Area $A = (1/2) * a * h$

$BC = a = sqrt((5-3)^2 + (9-1)^2) = 8.2462$

Slope of BC $m_a = (9-1) / (5-3) = 4$

Equation of BC

$(y - 1) / (9 - 1) = (x - 3) / (5 - 3)$

$2y - 2 = 8x - 24$

$y - 4x = -11$ Eqn (1)

Slope of altitude through A is $m__h = -1/4$

Equation of altitude h is

$y - 7 = -(1/4) (x - 6)$

$4y + x = 34$ Eqn 2

Solving equations (1) & (2), we get the coordinates of D, the base of altitude h

#D (78/11, 128/17)

Height of altitude $AD = h = sqrt((6-(78/17))^2 + (7 - (128/17))^2) = 1.5078$

Area of Triangle $color(green)A = (1/2) b h = (1/2) * 8.2462 * 1.5078 = color(green)(6.2168)$

Volume of Pyramid V = (1/3) * Base Area * Height

$color(green)(V = (1/3) 6.2168 * 8 = 16.5781)$

The base of a triangular pyramid is a triangle with corners at (6 ,7 )(6 ,7 ), (3 ,1 )(3 ,1 ), and (5 ,9 )(5 ,9 ). If the pyramid has a height of 8 8 , what is the pyramid's volume?