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Study Notes 025, Solid Geometry 3:
By William E. Steinman:
We are still working on the surface area of solid figures.
The lateral area (S) of a right circular cone is equal to ½ the product of the circumference (C) of its base and the slant height (L).
Thus S = ½ CL = πrL.
Oh yeah!
How do we know that?
Let us consider our definition of a cone. A cone is formed by joining with straight-line segments the points on a closed curve to a point outside the plane of the curve.
Now, in a right circular cone the curve of which we speak is a circle. A few study notes ago we discovered we could think of a circle as a polygon with an infinite number of sides. If we apply that same logic here, we find that the straight line segments can be thought of as an infinite number of triangles with bases on the infinite number of sides of the polygon.
From plane geometry, we know the area of a triangle is ½ bh
In this case h is the slant height of the cone, so the equation becomes ½ bL.
The lateral area then, is the sum of the areas of all of these tiny triangles.
You could lay out the equation for a large number of triangles and reduce the factors, but why bother.
Since all of the bases added together are equal to the perimeter of the polygon, they are also equal to the circumference of the circle.
Hence, the lateral area S = ½ CL
The total area of a right circular cone is the sum of the lateral area and the area of the base.
Or A = S + B = πrL + πr2 = πr(L + r)
Now imagine a cone with a lateral height of 3 inches and a base radius of 1.5 inches.
Let’s plug the values into the equation above and do the math.
A = 3.14x1.5(3 + 1.5) = 4.71x4.5 = 21.195 inches
We could put a lot of ice cream in that one.
The area of a sphere is four times the surface area of the great circle of the sphere.
Or A = 4πr2
Sure, how do we know that? We don’t. This formula can be derived only by using calculus. I have looked at two different methods of intuitively showing the truth of this equation. Both of them were so convoluted and condition that they looked like double talk to me. The one that seemed most reasonable has us treating the sphere as two hemispheres attached together. Then we treat one hemisphere as though it were a cone instead of half of a sphere. Then, by taking very thin slices of the cone and doing some very tedious manipulations of various formulas we can derive the above equation.
If you want to look at one of these, try this site. http://mathforum.org/dr.math/
Now, let’s go on.
Solving the equation, once we have it, is simple. We just plug in our radius value and do the math.
For example, given a sphere of radius 3 meters the surface area will be 4πr2 or 4x3.14x32 = 113.04 meters.
Two solid figures are similar if all of their corresponding dimensions are in he same ratio.
The lateral and total surface areas of similar figures are to each other as the square of their corresponding dimensions.
So, if the slant height and base radius of one right circular cone are twice as large as those of another right circular cone the total area of the larger cone will be 22 times the total area of the smaller cone. Okay!
Now how about volumes .In mathematics, volume is defined as the space occupied by any three-dimensional object.
For example the volume of a 1 inch cube is given as the base area times the height.
The base area will be 1inch x 1inch = 1square inch.
The volume will be 1 square inch x 1 inch = 1 cubic inch = 13.
The volume of any prism is equal to the product of its base area times the height.
V= Bh.
This really is intuitive. It is implicit in the definition of volume.
Given a right triangular prism with base dimensions of 3 mm, 4 mm, and 5 mm with a height of 10 mm, let’s compute the volume.
The base area will be ½ x 3 x 4 = 6 square mm.
The volume will be Bh = 6 x 10 = 60 cubic mm.
The Volume of a pyramid is equal to 1/3 of the base area B times altitude h.
V = Bh
This one is not so intuitive. We can get to it by considering a right triangular prism like the one below.
Thing of this as a solid transparent object.
We can divide the prism up into three equal sized pyramids.
The bases are not at the ends of the prism, but on the faces.
ABD is then a triangular base of a pyramid with vertex at C.
AED is he base of a pyramid with vertex at C.
Finally CDF is the base of a pyramid with vertex at .E.
When we examine the faces of the pyramids we will see that they are congruent and the pyramids are, in fact, equal sized.
Since there are three pyramids taking up the entire volume of the prism we can see that the volume of each pyramid is equal to 1/3 of the volume of the prism.
We already know the volume of the prism is Bh where B is the base area and h is the height.
Ipso: The volume of the pyramid is 1/3 Bh
So given a pyramid with base area of 10 sq meters and a height of 5 meters we can compute the volume directly just by plugging in the values.
V = 1/3 10x5 = 1/3 50 = 16.666 cubic inches.
Had enough?
Me too!