Need help finding the Volume of a pyramid.?

Question

Find the volume, to the nearest tenth, of a pyramid that is 6 ft. tall & whose base is an equilateral triangle w/ sides each 10 ft. long.

Answer 1

the volume of a regular triangular pyramid is:
= (1/3) * area of the base * height of pyramid.
The height is know (6 ft), so the problem is the area of the base. The area of the base is the area of a triangle =
(1/2)bh, One side is 10, but h = sqrt(10^2 -5^2) = sqrt(75)
= sqrt(25*3) = 5sqrt(3).

So the area of the base triangle is (1/2)*10 * 5*sqrt(3)
= 25Sqrt(3)
now the volume become = (1/3) * 25*Sqrt(3) * 6 =
50 * sqrt(3) = 86.60

Answer 2

Volume of a pyramid is Base area * Height / 3. Since the base is an equilateral triangle, its area can be found by this great formula (remember it, you won't regret it!). A = s²√3 / 4, where s is a side of the triangle.

So the area of the triangular base = 10²√3 / 4 = 25√3. Multiply this by the pyramid's height and then divide by 3 --> 150√3 / 3 = 50√3 ft³

Answer 3

Volume of pyramid = (1/3) * area of base * height

The base is an equilateral triangle of side 10 ft.
So, the area is [100*sqrt(3)]/4 = 25 * sqrt (3) sq. ft

Volume of pyramid = (1/3) * [25 * sqrt (3)] * 6
= [50 * sqrt (3)] cubic feet.

Answer 4

Use the basic formula for the volume of a pyramid:

V=(base x height) / 3

And the area of an equilateral triangle:

A = ((s^2)√3)/4

So V = ((((10^2)√3)/4)(6))/3
= 50√3
≈ 86.6 cu. ft.

Answer 5

Mathsmanretired is proper - the respond is (40 4, 117, 240) certainly and is ordinary some time past. Your question is regarding the so referred to as 'Euler brick' - a cuboid with integer edges and face diagonals (stick to the link in supplies under for a Wiki Article on the difficulty). In some supplies I honestly have met additionally the requirement the indoors diagonal to have integer length too (defined as 'proper Cuboid' interior the object - certainly one of nonetheless unsolved in call for issues), yet as quickly as we settle for the definition interior the object, the questions with regard to the cuboid and the splendid tetrahedron are equivalent. it is an occasion while an elementary computing gadget seek (no choose one to be a wizard as my chum Mr. Daftary has meant) does each and all the interest with minimum efforts. A code like here, written in under a minute: for x = one million to 1200 do for y = x to 1200 do for z = y to 1200 do if frac(sqrt(x*x + y*y) = 0 and frac(sqrt(y*y + z*z) = 0 and frac(sqrt(z*z + x*x) = 0 then Output(x, y, z), produced for under 5 seconds on my computing gadget all ideas, under 1200: (40 4, 117, 240), (eighty 5, 132, 720), (88, 234, 480)*, (132, 351, 720)*, (a hundred and forty, 480, 693), (one hundred sixty, 231, 792), (176, 468, 960)*, (220, 585, 1200)*, (240, 252, 275), (480, 504, 550)*, (720, 756, 825)*, (960, 1008, 1100), (1008, 1100, 1155) /non-primitive ideas with asterisks/ Such occasion of computing gadget superiority over a human concepts makes me sense undesirable, yet this is the life! very final feedback: one million) probable greater counsel (approximately analytical tips on a thank you to handle the difficulty etc.) is obtainable googling 'Euler Brick' 2) yet another difficulty (i think of exciting additionally) purely got here to my concepts: evaluate a tetrahedron, whose faces are 4 top triangles (3 of its edges are 2 via 2 perpendicular - like a Greek '?', certainly one of whose legs is twisted at a top attitude, its longest element is a diameter of the circumscribed sphere), in assessment to the splendid tetrahedron in question (it has 3 top and one million acute triangle as faces, the three 2 via 2 perpendicular edges meet in a vertex). Can each physique locate such tetrahedron with all 6 edges having integer length? hint: examine heavily the Wiki article.

Answer 6

volume of the pyramid is defined by V=bh/3 which means base times hieght divided by three equals volume so your answer is 20=volume

Answer 7

V = A_base*h*1/3

A_base = 1/2*b*h

So:

V = 1/6*b*h_base*h

Now we plug it all in:

b = 10
h_base = 5sqrt(5)
h = 6

So we plug it in:

V = 1/6*6*10*5sqrt(5)

V = 10*5sqrt(5)

V = 50sqrt(5)

And you can solve it from there if you like.

Answer 8

http://mathworld.wolfram.com/TriangularPyramid.html