# Math Help - Geometry help-surface area

1. ## Geometry help-surface area

"Beginning with the Surface Area of a regular pyramid, show algebraically how to derive the forumula for the Surface Area of a cone. Write a sentence or two explaining your solution.

now i know surface area for pyramid is 1/2 perimiter times slant height +base

and cone is pi times radius times slant height times pi times radius squared

but how do it derive formula?

2. Hello,

I guess that you should prove that both formulas are similar:

The surface area of a regular pyramid is

$A= A_{base}+\frac{1}{2} \cdot perimeter_{base} \cdot slant\ height$

Now take a cone:

$A= A_{base}+\frac{1}{2} \cdot perimeter_{base} \cdot slant\ height$

$A= \pi \cdot r^2+\frac{1}{2} \cdot 2 \cdot \pi \cdot r \cdot slant\ height$

So the formula for the surface of the cone is special form of the formula of a pyramid's surface.

Bye

3. Originally Posted by Geometry_sucks
"Beginning with the Surface Area of a regular pyramid, show algebraically how to derive the forumula for the Surface Area of a cone. Write a sentence or two explaining your solution.

now i know surface area for pyramid is 1/2 perimiter times slant height +base

and cone is pi times radius times slant height times pi times radius squared

but how do it derive formula?
The pyramid, if there is only one slant height, or if the pyramid is a right pyramid---not oblique or leaning pyramid where there are many slant heights of varying lengths.
Surface Area = (area of base) + (areas of slanting triangular faces).
You want to know why the (areas of the slanting triangular faces) is equal to (1/2 perimeter of base times slant height).
You the area of any triangle. It is 1/2 times base times height, the height being perpendicular to the base. So get one of the triangular faces. Make it erect. Now the base remains the the same but the slanting height becomes the "height" of the erect triangle. So its area is (1/2)(base)("height").
Return the erect triangle to its original slanting position. The "height" becomes the slant height again. And the area of this slanting triangle is (1/2)(base)(slant height).
That is just for one of the slanting triangular faces.
If the base of the pyramid has 5 edges, the areas of the slanting triangular faces is
= (1/2)(edge 1)(slant height) +(1/2)(edge 2)(slant height) +(1/2)(edge 3)(slant height) + ....
= (1/2)(slant height)(edge 1 +edge 2 +edge 3 +edge 4 +edge 5)
= (1/2)(slant height)(perimeter of base)
= (1/2)(perimeter of base)(slant height).

That is it.

------------------------
For a right cone, meaning, the apex is vertically above or below the center of the horizontal circular base.

Why is the lateral surface area is
= (1/2)(2pi *r)(slant height)
= (pi)(r)(slant height) ?

Analogous to a right pyramid, that is
= (1/2)(perimeter of circular base)(slant height).

To derive that, without using the analogy to a pyramid, there are many ways. Some of them are
---by Calculus or integration,
---by "average" of lateral area of a right cylinder whose two circular bases have radii of r and zero,
---by " the area generated by a straight line is the product (multiplication) of the length of the line and the path of the centerpoint of the line".

4. Hello ticbol,

you wrote:

For a right cone, meaning, the apex is vertically above or below the center of the horizontal circular base.

Why is the lateral surface area is
= (1/2)(2pi *r)(slant height)
= (pi)(r)(slant height) ?

You suggested very powerful instruments to derive the formula of the lateral surface of a right cone. Here is another go:

At a right cone every point of the base's circle has the same distance to the apex of the cone. Thus those points are situated on a circle with the radius r= slant height. The lateral surface cann't be a complete circle because than the lateral surface would be flat as a dish. Therefore the lateral surface must be a sector. The arcus(?) of this sector fits exactly around the circular base of the cone.
Let r = radius of the base, s = slant height, a = arcus of the sector, p = perimeter of the base. (I've attached a drawing to illustrate)

Than you get:
$p=2\cdot\pi\cdot r = a$

and for the area of the sector you may use a proportion:

$\frac{area_{sector}}{area_{fullcircle}}=\frac{arcu s}{perimeter_{circle}}$
$\frac{area_{sector}}{\pi \cdot s^2}=\frac{2\cdot \pi r}{2\cdot\pi\cdot s}$
$area_{sector}=\frac{(2\cdot \pi \cdot r) \cdot (\pi \cdot s^2)}{2\cdot\pi\cdot s}$
$area_{sector}=\frac{(2\cdot \pi \cdot r) \cdot (\pi \cdot s^2)}{2\cdot\pi\cdot s}$

$area_{sector}=\pi \cdot r \cdot s$

Bye