JH WEEKLIES ISSUE #19
2011-2012
Mathematics—Volume and Surface Area of Geometric Solids
The Sphere
In calculating attributes of the sphere, you only need one
measurement: the radius. The formulas that relate the volume and
surface area of any sphere require the radius and an irrational constant
known as “pi,” symbolized by the Greek character π, and which is
approximately equal to 3.14. Many quizbowl questions will ask you to
give your answer in terms of this constant, which means that you need
to multiply out your answer except for π.
Calculating the Volume of a sphere: V = 4/3 π r3
Figure 4
Where “r” is the radius of the sphere, multiply four-thirds with pi
and the radius cubed to find the volume.
Calculating the Surface Area of a sphere:
S.A. = 4 π r2
Note: The surface area can also be calculated very easily from the area of the Great
Circle of a sphere. The “Great Circle” is the largest circle that can be drawn through a
sphere, and divides a sphere into two equal hemispheres. The area of the Great Circle is
equal to π r2, so multiplying that area by 4 will result in the sphere’s surface area.
The Cone
A cone is a simple shape consisting of a circular base that pinches to a
single point, known as the vertex.
Calculating the Volume of a cone:
V = ⅓ π r2 h
Where “r” is the radius of the circular base and “h” is the height
of the cone from the center of the base to the vertex. Multiply
one-third, pi, the radius squared, and the height of the cone for
the volume. If we break this equation down, we find that all you
really are doing is multiplying the area of the base (π r2) by onethird of the height.
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Figure 5
Calculating the Surface Area of a cone:
S.A. = π r (l + r)
S.A. = π r2 + π r l
Where “r” is the radius of the circular base and “l” is the slant height. The “slant height”
is a measurement up the side of the cone from the base to the vertex, and can be found
using the Pythagorean Theorem: l = √(r2 + h2). Add the slant height and the radius, then
multiply that by the radius and pi to find the slant height. The second equation that is
listed for Surface Area illustrates the addition of the base area to the lateral cone area
(lateral area is the surface area of a geometric solid that excludes the base).
The Cube
A cube is one of the simplest solids; it consists of six square sides
connected at right angles. It is a special case of a rectangular prism, in
which the length, width and height are all the same. This greatly
simplifies calculations.
Calculating the Volume of a cube:
V = l3
Where “l” is the length of one of the sides; simply multiply the
length by itself twice (l x l x l).
Figure 6
Calculating the Surface Area of a cube:
S.A. = 6 l2
Where “l” is the length of one of the sides. Multiply the length of the side by itself to
find the area of one face, then multiply that by six to calculate the surface area.
The Cylinder
A cylinder is a circle that has been stretched into three dimensions along a
single axis; it consists of two connected circles of the same radius connected
along all edges.
Calculating the Volume of a cylinder:
V = π r2 h
Where “r” is the radius of the circular base and “h” is the height of
the cylinder. Multiply pi by the radius squared and the height to
compute the volume of the cylinder. Here, you find the area of the
base and multiply it by the height of the total cylinder.
Calculating the Surface Area of a cylinder:
Figure 7
S.A. = 2π r h + 2π r2
Where “r” is the radius of the circular base and “h” is the height of the cylinder. This
formula illustrates that the lateral area of the cylinder is effectively a rectangle with
height h and length 2πr, the circumference of the base (think of the shape of the label of a
soup can that is removed from the can—it’s rectangular!). The base of the cylinder is
2
simply a circle with the formula for its area as π r . Therefore, you are simply adding the
lateral area of the solid to two times the base area.
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The Square Pyramid
A square pyramid is a 3D shape created by taking the corners of a
square base and bringing them together to a single vertex.
Calculating the Volume of a Square Pyramid:
V = ⅓ b2 h
Where “b” is the length of one side of the square base, and
“h” is the height of the pyramid. Find the area of the base
(b2), multiply by the height (h), and then multiply by ⅓.
Note that this formula is very similar to that of the cone—
multiplying the base area by the height and ⅓.
Figure 8
Calculating the Surface Area of a Square Pyramid:
S.A. = 2sb + b2
Where “s” is the slant height and “b” is the length of one side of the square base. To find
the slant height, which is the length from the central vertex to the center of one side of the
square base, use the Pythagorean theorem (s = √(h2 + ¼b2)). For the surface area, find
the area of each of the triangular sides (4 x (½sb) = 2sb) and add the area of the square
base (b2).
The Triangular Prism
A Triangular prism is a triangle brought directly into three
dimensions.
Calculating the Volume of a Triangular Prism:
V=½bhl
Where “b” is the length of the triangular face, “h” is the height
(or altitude) of the triangular face, and “l” is the length of the
prism. Compute the area of the triangular base (½bh) and Figure 9
multiply by the prism length (l). Similar to any other type of
prism or a cylinder, you find the area of the base and multiply it by the height.
Calculating the Surface Area of a Triangular Prism:
S.A. = bh+l(s1 + s2 + s3)
Where “b” is the length of the triangular face, “h” is the height (or altitude) of the
triangular face, “s1” “s2” and “s3” are the lengths of the sides of the triangular face, and
“l” is the length of the prism. Compute the area of the triangular faces (or bases)
(2 x ½bh = bh) and add the areas of the rectangular prism faces (similar to the lateral
area) (s1 x l + s2 x l + s3 x l = l(s1 + s2 + s3)).
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Example/Exercises I:
Example:
Compute the volume of the following geometric solids.
Find the volume of a sphere, in terms of pi, that has a radius of 3
centimeters.
Vsphere = 4/3 π r3 = 4/3 π (33) = 4/3 π (9) = 12π cm3
Exercises:
1.
2.
3.
4.
5.
6.
7.
Example/Exercises II:
Example:
A cylinder with height 3 meters and radius 2 meters.
A cube with side length 2 inches.
A cone with base radius 2 centimeters and height 7.5 centimeters.
A sphere with radius 6 meters.
A cylinder with height 1 foot and radius 4 feet.
A triangular prism with base length 2 inches, height 4 inches, and
prism length 6 inches.
A square pyramid with base length 3 feet and height 2 feet.
Compute the surface area of the following geometric solids.
Find the surface area of a sphere with radius 2 feet.
S.A.sphere = 4 π r2 = 4 π (22) = 4 π (4) = 16π ft2
Exercises:
1.
2.
3.
4.
5.
6.
7.
A cone with base radius 3 meters and height 4 meters.
A sphere with radius 7 inches.
A cube with side length 5 feet.
A cylinder with base radius 2 inches and height 4 inches
A cone with base radius 5 inches and height 12 inches.
A square pyramid with base length 6 centimeters and height of 4
centimeters.
A triangular prism with an equilateral triangle base of side length 2
meters (therefore height of √3 by special triangles) and prism
length of 4 meters
Answers at back of packet
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Answers to Exercises
Exercises I Answers:
1.
12π meters3 (≈37.7 meters3)
2.
8 inches3
3.
10π centimeters3 (≈31.4 centimeters3)
4.
288π centimeters3 (≈904.8 centimeters3)
5.
16π feet3 (≈50.3 feet3)
6.
24 inches3
7.
6 feet3
Exercises II Answers:
1.
24π meters2 (≈75.4 meters2)
2.
196π inches2 (≈615.8 inches2)
3.
150 feet2
4.
24π inches2 (≈75.4 inches2)
5.
90π inches2 (≈282.7 inches2)
6.
96 centimeters2
7.
24 + 2√3 meters2
Questions Galore 319 S. Naperville Road Wheaton, IL 60187
Phone: (630) 580-5735 E-Mail: [email protected] Fax: (630) 580-5765