An octagon based pyramid, perhaps.
A triangle with side a: 8, side b: 8, and side c: 8 cm has an area of 27.71 cm2
A quadrilateral bipyramid is one possible answer. This is two pyramids with the same quadrilateral base - one on each side of the "base".
root square of 73
hypotenuse= 16.24
8 inches.
= 1/2 × Perimeter × [Side Length] + [Base Area] = 0.5*(3*8)*4*sqrt(10)+0.5*8*4tan60o = 158.72 (base area is 0.5*base length*perp ht, per ht is just 0.5*side*tan60o as triangel is equilateral) alex
The base of a cube is a square, so the area of the base of the cube is the area of that square. The area of a square is s2, where s is the length of on side. Side all edges of a cube have the same length, it doesn't matter which edge you use. Example: Find the area of the base of a cube with edges of length 7: A = 72 = 49 Example: Find the area of the base of a cube with volume 8 cubic meters. The volume of a cube is equal to s3, where s is the length of a side. 23=8, so 2 is the length of a side, and the area of the base is 22 = 4.
Such a pyramid cannot exist. If it is a regular pyramid with side length 8, its slant height MUST be less than 8. In fact, it is approx 6.39.
462 base 8 - 177 base 8 = 263 base 8 , (that's 179 in base 10) I used the octal feature on the windows calculator.
503 base 10 = 767 base 8 503 base 8 = 323 base 10
Okay. A square pyramid has 5 faces. So a pyramid with twice that would have 10 faces. You subtract 1 for the base and you get nine. Therefore you have 9 sides. Divide 72 by 9 and you get 8. So one side is 8 ft.
If the sides of the top and base of the pyramidal frustum are 3 and 8 metres units then the radius of the cylinder is 3.2081 metres.