Please review
Measurements first.
Surface area usually refers to three dimensional
objects, however one sometimes talks about the surface area of a flat surface
such as a table top, a wall, a field, that is, just one side of the object.
However the term "surface area" refers to the measure
of the "amount of surface" enclosing some object, or the amount of surface
within some object. At the end I provide ways to visualize these shapes.
Consider these shapes:
Each face has an area, and each face is in a pair
that have equal areas. The front and back of this box have an area equal
to L·H, so the area of the front and the back equals 2LH. Likewise
the top and bottom forms a pair each with an area of LW; they contribute
2LW, and finally the left and right ends contribute an area of 2WH. We
have no other surfaces on the exterior of this box, therefore the total surface
area is 2LH + 2LW + 2WH = 2(LY + LW + WH).
If we have a cube then all dimensions are
equal, that is, L=W=H. In this case
let s equal this dimension and The surface
are of a cube would be
6( s·s ) = 6s2.
(6 sides to the square each with area s2.)
The surface area of a tennis ball is about
(4)(3.14)(1.5in)2 = 28 square
inches. The surface area of a #10
soccer ball is about (4)(3.14)(4in)2
= 201 square inches. So, the soccer
ball is about 7 times larger than the
tennis ball as far as surface area is
concerned.
These numbers man seem large, but recall
that we are dealing with a curved surface,
not a flat on, when you place a square
inch on the surface the corners are pulled
in; imagine placing a stamp on a
tennis ball, the stamp gets deformed
because of the curvature of the ball.
A right circular cylinder is a cylinder
that has circular ends and the wall is
perpendicular to both of these ends
Now the surface area of this cylinder is
comprised of the areas of the ends
(circles) and the area of the wall.
The wall is a rectangle with the width
equal to h and the length equal to
the circumference of either end. The
are of a circle is πr2
and its circumference is 2πr; se we
get 2(πr2) + (2πr)h
as the surface area. The 2, the π,
and the r can be factored out to arrive at
the equation above.
A right circular cone is a cone that has
a circular base and whose altitude (height)
from circle center to the apex of the cone
is perpendicular to the base. In the
figure above the radius r is perpendicular
to the height h. For example if the
radius equals 10 cm, and the height is 25
cm, then the surface area is
(25cm)2 + (10 cm)2 = 625cm2 +
100cm2 = 725cm2; taking the square root we
get s = 26.9 cm
so the surface area would be
2(3.14)(10cm)(26.9cm) + (3.14)(10cm)2 = 1690.9 cm2
+ 314 cm2 = 2005 cm2
Here are ways to visualize the surface
areas above and are great ways to provide
a hands on experience for your student.
First the box. Find a
cardboard box and label the sides.
Cut it then unfold it until it looks like
the unfolded image below. Or, start
with the unfolded image, have your student
create that image with the dimensions you
provide, find the total area, then fold it
back into place to create the box.
Below is a right circular cylinder that is
being rolled along a sheet of paper whose
width is the height of the cylinder.
The numbers are provided to show the
positions along the circumference of an
end to keep track of when to stop rolling.
As you can see the rectangular sheet of
paper has a length equal to the
circumference of the one end of the
cylinder.
Finally, shown is a cone rolled along a
sheet of paper with its point remaining in
one place. One complete revolution
of the cone traces out a sector of a
larger circle on the paper. The area
of this sector is the area of the curved
surface of the cone. To carry out
this exercise you must make sure the point
of the cone does not change position as
the cone rotates around the paper.
This is another FREE ALGEBRA PRINTABLE presented to you from the
Algebra section of
K12math.com