The base of a triangular pyramid is a triangle with corners at (6 ,7 ), (3 ,1 ), and (4 ,2 ). If the pyramid has a height of 8 , what is the pyramid's volume?

1 Answer
Martin C.
Mar 4, 2018

V=4

Explanation:

V=1/6*a*h_a*H
H is the hight into the 3-dimension.
p_1=(6,7)
p_2=(3,1)
p_3=(4,2)
H=8
To find the distanc a between p_1 and p_2 we use pythagoras:
a^2+b^2=c^2
Notice that the c from pythagoras equals our a.
((x_1-x_2)^2+(y_1-y_2)^2)^(1/2)=a
((6-3)^2+(7-1)^2)^(1/2)=a
(9+36)^(1/2)=a
sqrt(45)=a
((x_1-x_3)^2+(y_1-y_3)^2)^(1/2)=b
sqrt(29)=b
((x_2-x_3)^2+(y_2-y_3)^2)^(1/2)=c
sqrt(2)=c

c^2=a^2+b^2-2abcos(gamma)
2=45+29-2sqrt(45*29)cos(gamma)
2-45-29=-2sqrt(1305)cos(gamma)
cos^-1(36/sqrt(1305))=gamma
4,76°~~gamma
sin(gamma)=h_a/b|*b
sin(gamma)*b=h_a
0.45~~h_a

V=1/6*sqrt(45)*0.45*8~~4

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