A regular triangular pyramid has an altitude of 9m and a volume of 46.8cu. meters. Find the length of the edges and the lateral area of the pyramid.
2 answers
please answer this
V = 1/3 base * height
area of the base = 46.8/3 = 15.6
the base is a equilateral triangle.
the area of which is sqrt(3)/4 * edge^2
edge = sqrt(15.6*4/sqrt 3) = 6.002 m
The mid-point of the triangle is 2/3 of the way from the angle to the base.
sqrt(3)/3 * 6 = 2 sqrt 3 ~ 3.46 m
the length of the other 3 edges then using the Pythagorean theorem.
12+81 = edge^2
sqrt 93 ~ 9.64 m
the slant height...
3^2 + slant^2 = 93
slant height = sqrt 84 ~ 9.165 m
lateral area = 3/2 * 6.002 * 9.165 = 82.5 m^2
area of the base = 46.8/3 = 15.6
the base is a equilateral triangle.
the area of which is sqrt(3)/4 * edge^2
edge = sqrt(15.6*4/sqrt 3) = 6.002 m
The mid-point of the triangle is 2/3 of the way from the angle to the base.
sqrt(3)/3 * 6 = 2 sqrt 3 ~ 3.46 m
the length of the other 3 edges then using the Pythagorean theorem.
12+81 = edge^2
sqrt 93 ~ 9.64 m
the slant height...
3^2 + slant^2 = 93
slant height = sqrt 84 ~ 9.165 m
lateral area = 3/2 * 6.002 * 9.165 = 82.5 m^2
2 answers