In a regular triangular prism, the side of the base is 6 dm, the lateral edge is 7 dm. Find the volume of the prism

1) The volume of the prism is: V = S * h, where S is the area of ​​the base of the prism h is the height.

The area of ​​the base of the prism is S = 1/2 * 6 * 6 * sin60˚, since at the base of the regular prism there is a regular triangle, whose sides are equal to 6 dm by the condition, and the angles are equal

180˚ / 3 = 60˚. S = 9√3 dm2. Therefore, V = 9√3 * 7 = 56√3 dm3.

2) The volume of a regular triangular pyramid is V = (S * h) / 3, where S is the area of ​​the base of the pyramid h is the height of the pyramid.

To find the area of ​​the base, first we find the radius of a circle circumscribed around a regular triangle: r = √ (10 ^ 2 – 8 ^ 2) = √ (100 – 64) = √36 = 6 dm.

We use a corollary from the theorem of sines, which says that 2 * r = a / sin 60˚, where a is the side of the triangle. Hence a = 2 * r * sin60˚ = 2 * 6 * √ 3/2 = 6√3 dm. Then the area of ​​the base is S = 1/2 * 6√3 * 6√3 * sin60˚ = 27√3 dm2.

It remains to calculate V = 27√3 * 8/3 = 56√3 dm3.



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