Park Forest Math Team
Meet #5
Self-study Packet
Problem Categories for this Meet (in addition to topics of earlier meets):
1.
2.
3.
4.
5.
Mystery: Problem solving
Geometry: Solid Geometry (Volume and Surface Area)
Number Theory: Set Theory and Venn Diagrams
Arithmetic: Combinatorics and Probability
Algebra: Solving Quadratics with Rational Solutions, including word problems
Important Information you need to know about GEOMETRY:
Solid Geometry (Volume and Surface Area)
Know these formulas!
SHAPE
SURFACE
AREA
VOLUME
Rect.
prism
2(LW + LH +
WH)
LWH
Any prism
sum of areas
of all surfaces
H(Area of Base)
Cylinder
2 R2 + 2 RH
R2H
Pyramid
sum of areas
of all surfaces
1/3 H(Base area)
Cone
R2 + RS
1/3 R2H
Sphere
4 R2
4/3 R3
Surface Diagonal: any diagonal (NOT an edge) that connects two
vertices of a solid while lying on the surface of that solid.
Space Diagonal: an imaginary line that connects any two vertices of a
solid and passes through the interior of a solid (does not lie on the
surface).
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Category 2 ± Geometry
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Answers
1. _______________
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2. _______________
3. _______________
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Answers
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You may use a calculator today.
Category 2
Geometry
Meet #5, March 2009
1.
What is the total surface area of a solid hemisphere with radius 12 cm? Use
3.14 as an estimation for ! and express your answer as a decimal.
2.
The radius of cone A is 6 times as long as the radius of cone B. The height of
cone A is one-ninth the height of cone B. If the volume of cone A is 200 cm3,
how many cm are in the volume of cone B?
3.
In the diagram below, an 8 inch by 8 inch by 8 inch cube has a cylinder with
radius 2 in drilled through the center of one face all the way through the cube
and out the opposite face. What is the surface area of the resulting figure? Use
3.14 as an estimation for ! and express your answer as a decimal.
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 2
Geometry
Meet #5, March 2009
1. The surface area of a sphere is found by the formula
!" # $%& ' , so a hemisphere would have a surface area of (%& ' .
However, there is also a circular base to a hemisphere which has
an area of %& ' ) for a total surface area of *%& ' . Since the radius
is 12 cm, the surface area is !" # *%+,(-' # *%+,$$- #
$*(% . ,*/01$2
Answers
1. 1356.48
2. 50
3. 459.36
5
2. Volume of a cone is found by the formula 34#4 %& ' 71 If &4is the radius of cone
6
8
B, then 0& is the radius of cone A. If 7 is the height of cone B, then is the height of
9
cone A. We can then write a formula for the volume of cone A as 3" #
5
8
5
4 %+0&-' : ;, while the volume of cone B is 3< # 4 %& ' 7. Since we know the
6
9
6
volume of cone A is 200, we can substitute 200 for VA and simplify to get:
5
8
5
8
?
5
3" # 4 %+0&-' : ; =(>> # 4 %*0& ' : ; = (>> # %& ' 7 = /> # 4 %& ' 7 # 3<
6
9
6
9
6
6
3. Describing the surface area in words, there are 6 squares, although 2 of them
have circular holes cut out of two of them. There is also “tube” through the middle of
the cube which is a cylinder without bases. So the surface area of the whole thing is:
0+2-' @ (%+(-' A (%+(-+2- #
*2$ @ 2% A *(% #
*2$ A ($% .
*2$ A ($+*1,$- #
*2$ A B/1*0 #
$/C1*0
Category 2
Geometry
Meet #5, March 2007
You may use a calculator.
1. Each edge of the tetrahedron shown at right
has been trisected. Suppose a tetrahedron with
one-third the edge length is cut from each
vertex of the tetrahedron, using these points to
guide the cutting. How many edges will there
be on the resulting solid?
2. Two teams, the Stackers and the Packers, were each given 1729 unit cubes and
told to arrange all the cubes to make just two cubes. It turns out that the two teams
found different solutions to this problem. How many square units are in the
surface area of the largest of the four cubes that were made?
3. The figure below depicts a cone inside a hemisphere. If the diameters of both
the cone and the hemisphere are 6 centimeters and the height of the cone is 3
centimeters, how many cubic centimeters are in the hemisphere but not in the
cone? Use 3.14 for ! and express your answer to the nearest hundredth of a cubic
centimeter.
Answers
1. _______________
2. _______________
3. _______________
www.imlem.org
Solutions to Category 2
Geometry
Meet #5, March 2007
Answers
1. 18
2. 864
3. 28.26
1. In place of each of the
four vertices, we will get
three new edges. There will
be 3 × 4 = 12 new edges, in
addition to the original 6
edges for a total of 18 edges.
2. Suppose the Stackers make a 10 by 10 by 10 cube, which
uses 1000 unit cubes, and a 9 by 9 by 9 cube, which uses 729
unit cubes. Then the Packers must have made a 12 by 12 by 12
cube, which uses 1728 unit cubes, and a 1 by 1 by 1 cube,
which uses 1 cube. The surface area of the largest cube is
6 ×12 2 = 6 ×144 = 864 square units.
3. The volume of a sphere is given by the formula
4
VSphere = πr 3 . The volume of a hemisphere would be half that
3
amount. The radius of our hemisphere is 3 cm, so the volume
4
is × π × 3 3 ÷ 2 = 18π . The volume of a cone is given by the
3
1
formula VCone = πr 2 h . The radius and the height of our cone
3
1
are both 3 cm, so the volume is × π × 3 2 × 3 = 9π .
3
The difference between the two volumes is 18! – 9! = 9! "
9 × 3.14 = 28.26 cubic centimeters.
www.imlem.org
Category 2
Geometry
Meet #5, March/April 2005
You may use a calculator
1. The figure at right is a “net” for a cube.
The net is folded up to form a cube, and the
cube is rolled. If C is on top, what letter is on
bottom?
A
B
C
D
E
F
2. A solid cone with radius 2.25 cm and height 4.5
cm is placed inside a cylinder that also has radius
2.25 cm and height 4.5 cm. How many cubic
centimeters of space inside the cylinder are not
occupied by the cone? Use 3.14 for ! and give
your answer to the nearest tenth of a cubic
centimeter.
3. How many square inches are in the surface area of a tetrahedron with side
length 1.5 inches? Reminder: A tetrahedron is a solid with four congruent faces,
each of which is an equilateral triangle. Hint: You can use the Pythagorean
Theorem to find the height of an equilateral triangle. Give your answer to the
nearest tenth of a square inch.
Answers
1. _______________
2. _______________
3. _______________
www.Imlem.org
Solutions to Category 2 Average team got 18.04 points, or 1.5 questions correct
Geometry
Meet #5, March/April 2005 Average number of correct answers: 1.50 out of 3
1. Letters A, B, and D are clearly adjacent to letter C, so
none of these will be on the bottom of the cube. Letter E
will also be adjacent to C, when the net is folded into a
cube. Letter F will be on the opposite side of C and
therefore on the bottom when C is on top.
Answers
1. F
2. 47.7
3. 3.9
2. The volume of a cone is one third of the volume of
the cylinder, so the space not occupied by the cone is two
thirds of the cylinder. This volume is given by the
calculations below:
2
2
V = ⋅ π ⋅ r 2 ⋅ h = ⋅ 3.14 ⋅ 2.25 2 ⋅ 4.5 = 47.68875
3
3
To the nearest tenth of a cubic centimeter, the space
inside the cylinder not occupied by the cone is 47.7 cubic
centimeters.
3. We need to find the area of one equilateral triangle
and then multiply by 4. To find the area of an equilateral
triangle, we need to have a height. The picture at left
shows how an equilateral triangle can be cut into two 3060-90 triangles. The height, h, can be calculated using
the Pythagorean Theorem as follows:
1.5 in.
h
h = 1.5 2 − 0.75 2 = 2.25 − 0.5625 = 1.6875 ≈ 1.299
The area of one of the triangles is thus
0.75 in.
1
Aone triangle = ⋅1.5 ⋅1.299 ≈ 0.974 .
2
The surface area of the tetrahedron is four times this
amount, which is 4 ⋅ 0.974 = 3.896 or 3.9 square inches
to the nearest tenth of a square inch.
www.Imlem.org
Category 2
Geometry
Meet #5, April 2003
You may use a calculator
today!
1. How many surface diagonals can be drawn on a pentagonal prism such as the
one depicted below?
2. The surface area of a cube is 73.5 square inches. How many cubic inches are in
its volume? Express your answer as a decimal to the nearest thousandth.
3. A cylinder that used to contain three tennis balls is now
filled with water. The diameter of the cylinder is ! "# inches
and the height is $ $# inches. A gallon is defined as a unit of
liquid capacity equal to 231 cubic inches or 128 ounces.
How many ounces of water does the cylinder hold? Use
3.1416 for pi and, sincHWKHF\OLQGHUFDQ¶WKROGPRUHWKDQLWV
volume, round your answer down to the previous whole
number of ounces.
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 2
Geometry
Meet #5, April 2003
Answers
1. 20
2. 42.875
1. There are two surface diagonals on each of the five
rectangular faces of the pentagonal prism and there are
five surface diagonals on each of the two pentagonal
faces of the pentagonal prism. That makes a total of
! u " " u ! #$ #$ 20 surface diagonals.
3. 23
2. All six sides of the cube have the same area, so the
area of one side would be 73.5 ÷ 6 = 12.25 square
inches. The side length of the cube is the square root of
12.25 or 3.5 inches and the volume of the cube is
%
%&"
!"#$%& cubic inches to the thousandth.
3. The volume
of a cylinder is given by the formula
!
! S" # . We will need half our diameter for the radius
of the cylinder, or 2.625 ÷ 2 = 1.3125 inches. Using this
value for r IRU ʌ DQG IRU h, we calculate
the volume of
the cylinder as follows:
!
%&#'#( u #&%#!"
!
u )&*)"
'! &(#*)
cubic inches.
To convert this volume to ounces, we multiply by 128
and divide by 231. This gives us:
'! &(#*) u
#!*
!%#
| !% &(#"(
ounces.
Rounding this number down to the previous whole
number of ounces, we can say that the cylinder holds
about 23 ounces of water.
Category 2
Geometry
Meet #5, April 2001
1. How many 8-inch by 8-inch by 16inch cement blocks will be required to
construct a foundation that is 32 feet
long, 24 feet wide, and 8 feet tall?
(Each wall is 8 inches thick.)
You may use a
calculator today!
8 ft.
32 ft.
24 ft.
2. The label from each of the two cans shown
here is a 4-inch by 12-inch rectangle which
4 in.
exactly covers the rounded surface of each can,
but not the top and bottom. If the height of can
A is 4 inches and the height of can B is 12
inches, what is the ratio of the volume of can A
to the volume of can B? Express your answer
as a ratio in the form a:b, where the greatest
common factor of a and b is 1.
3. A exterior surface of a hemispherical capital
dome is to be covered with gold leaf. If the diameter
of the dome is 50 feet, how many square feet of gold
leaf will be required to cover the dome? Use 3.14
for Pi and give your answer to the nearest whole
number of square feet.
Answers
1. _____________
2. _____________
3. _____________
A
12 in.
50 feet
B
Solutions to Category 2
Geometry
Meet #5, April 2001
Answers
1. 984
2. 3:1
3. 3925
1. The foundation wall is 8 feet, or 96 inches,
tall all the way around and will require 12 layers
of cement blocks that are 8 inches tall. Each of
the sides that are 32 feet long will require
32 × 12 ÷ 16 or 24 blocks that are 16 inches long
on each layer. Similarly, the sides that are 24
feet long will require 24 × 12 ÷ 16 or 18 blocks
on each layer. Watch out! The half-block in
each corner on each layer is shared by two walls.
If we count our way around one level of the
foundation— 24 + 18 + 24 + 18 = 84 blocks—we
have overcounted by 4 halves or 2 blocks. We
will need only 82 blocks on each of 12 layers of
block, for a grand total of 984 blocks.
2. We know from the dimensions of the label
that can A must have a circumference of 12
inches. From this we can find that the radius of
can A must be 212π = π6 . Similarly, we can find the
radius of can B, which is
4
2π
=
2
π
. Using the
volume formula V = Areabase × height = πr 2 h ,
we find:
Can A: V = π ( π6 ) × 4 = π ×
2
Can B: V = π ( π2 ) × 12 = π ×
2
The ratio
× 4 = 144
π
36
π2
4
π2
× 12 =
48
π
144 48
π
: π can be simplified to 3:1.
3. The formula for the surface area of a sphere is
4πr 2 , so the surface of a hemisphere will be half
of that. Our dome has a diameter of 50 and a
radius of 25, so 2 × 314
. × 252 = 6.28 × 625 =
3925 square feet.