Can anyone help me out?? Don't know how to start

Question

Show that a pyramid of height h whose base is an equilateral triangle of side s has volume √3/12 hs²

Answer 1

V = 1/3Bh, where B is area of base. B = s²√3/4 (area of equilateral triangle with side s)

So, V = 1/3s²h√3/4 = s²h√3/12.

Answer 2

Here are some suggestions.

Do you know how to do volume by slicing? ∫ A(x) dx where A(x) is a formula for the area of a slice.

If you picture the pyramid lying horizontally along the x-axis with the base at x = h, then typical slices along the x-axis will be equilateral triangles that are similar to the base. You might set up the ratio: h/s = x/v , where v is the length of a side of a typical slice. Solve for v to get v= xs/h, remembering that s and h are constants.

Look up the formula for the area of an equilateral triangle depending on the length of a side. Substitute xs/h into that formula, then integrate from 0 to h, with respect to x where A(x) is the formula you get when you do the substitution.

Answer 3

formula for volume of a pyramid = Bh/3 where B= area of the base, h= height of the pyramid

for an equilateral triangular base, B = (s^2/4)sqr3

V= (1/3)hs^2(sq3)/4 = hs^2/4sqr3

Answer 4

Here you need to employ the method by slicing. Draw a triangular pyramid with base an equilateral

triangle ABC and name the apex O.

In separate drawing of the equilateral triangle ABC draw the height AK of the triangle which also happens

to be the median. Now look at the right triangle AKB.

AB = s, KB = s/2.

Then by the Pythagorean Theorem AK = (s√3)/2

Hence the area of the triangle is (s²√3)/4.

Now if the problem was just a basic Geometry's problem the calculation of the volume of the pyramid would be

is easy to be found since V = 1/3 (area of base) ( height) = (1/3) (s²√3)/4 H =(1/12)(s²√3)H.

But we need to use Calculus to reach the same result.

Back to the pyramid OABC .

We introduce a rectangular coordinate system in which the y-axis passes through the apex of the pyramid

and is perpendicular to the base , and the x-axis passes through the center of the base say G, perpendicular to a side of the equilateral triangle and contains AK.

At any y in the interval [0,H] on the y-axis , the cross section perpendicular to the y-axis is an equilateral triangle .

If we let a denote the a denote the length of this triangle , then by similar triangles

s/a =H/(H-y) ⇒a = (H-y)(s/h)

Then the area Ω(y) of the cross section at y . Then

Ω(y)=(a²√3)/4 = [(H-y)²s²√3]/(4H²)

and the volume of the pyramid is

V= ∫^H₀ Ω(y) dy =

∫^H₀ [(H-y)²s²√3]/(4H²)dy =

s²√3]/(4H²)∫^H₀ [(H-y)²dy =

s²√3]/(4H²) (H³/3) =

(1/12)(s²√3)H