Posted by **kitOy** on Sunday, February 19, 2012 at 9:12pm.

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.

- MATH -
**mariano**, Friday, June 12, 2015 at 11:32pm
please answer this

- MATH -
**CEA ARCHITECTURE**, Monday, September 5, 2016 at 10:15pm
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

- MATH solid mensuration -
**CEA ARCHITECTURE**, Monday, September 5, 2016 at 10:16pm
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

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