Park Forest Math Team. Meet #5. Self-study Packet. Problem Categories for this Meet (in addition to topics of earlier meets):

Transcription

1 Park Forest Math Team Meet #5 Self-study Packet Problem Categories for this Meet (in addition to topics of earlier meets): 1. Mystery: Problem solving 2. : Solid (Volume and Surface Area) 3. Number Theory: Set Theory and Venn Diagrams 4. Arithmetic: Combinatorics and Probability 5. Algebra: Solving Quadratics with Rational Solutions, including word problems

2 Important Information you need to know about GEOMETRY: Solid (Volume and Surface Area) Know these formulas! SHAPE Rect. prism SURFACE AREA 2(LW + LH + WH) VOLUME LWH Any prism sum of areas of all surfaces H(Area of Base) Cylinder 2 R RH R 2 H Pyramid sum of areas of all surfaces 1/3 H(Base area) Cone R 2 + RS 1/3 R 2 H Sphere 4 R 2 4/3 R 3 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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5 You may use a calculator today. Category 2 Meet #5, March 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 cm 3, 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

6 Solutions to Category 2 Meet #5, March 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$2 2. Volume of a cone is found by the formula 34#4 5 6 %&' 71 If &4is the radius of cone B, then 0& is the radius of cone A. If 7 is the height of cone B, then 8 is the height of 9 cone A. We can then write a formula for the volume of cone A as 3 " # %+0&-' : 8 ;, while the volume of cone B is 3 9 < #4 5 6 %&' 7. Since we know the volume of cone A is 200, we can substitute 200 for V A and simplify to get: 3 " #4 5 6 %+0&-' : 8 ; =(>> # %*0&' : 8 ; 9 =(>>#? 6 %&' 7=/>#4 5 6 %&' 7#3 < 3. Describing the surface area in words, there are 6 squares, although 2 of them have circular holes cut out of two of them. There is also tube through the middle of the cube which is a cylinder without bases. So the surface area of the whole thing is: 0+2- ' A(%+(-+2- # 2% A *(% # *2$ A ($%. *2$ A ($+*1,$- # *2$ A B/1*0 # $/C1*0

7 Category 2 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

8 Solutions to Category 2 Meet #5, March 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 = = 864 square units. 3. The volume of a sphere is given by the formula V Sphere = 4 3 πr 3. The volume of a hemisphere would be half that amount. The radius of our hemisphere is 3 cm, so the volume is 4 3 π 33 2 = 18π. The volume of a cone is given by the formula V Cone = 1 3 πr 2 h. The radius and the height of our cone are both 3 cm, so the volume is 1 3 π 32 3 = 9π. The difference between the two volumes is 18! 9! = 9! " = cubic centimeters.

9 Category 2 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

10 Solutions to Category 2 Average team got points, or 1.5 questions correct Meet #5, March/April 2005 Average number of correct answers: 1.50 out of 3 1. F 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. 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: V = 2 3 π r 2 h = = To the nearest tenth of a cubic centimeter, the space inside the cylinder not occupied by the cone is 47.7 cubic centimeters. 1.5 in in. h 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 triangles. The height, h, can be calculated using the Pythagorean Theorem as follows: h = = = The area of one of the triangles is thus A one triangle = The surface area of the tetrahedron is four times this amount, which is = or 3.9 square inches to the nearest tenth of a square inch.

11 Category 2 Meet #5, April 2003 You may use a calculator 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 for pi and, sinc volume, round your answer down to the previous whole number of ounces

12 Solutions to Category 2 Meet #5, April 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! " "! #$ #$ 20 surface diagonals. 2. All six sides of the cube have the same area, so the area of one side would be = square inches. The side length of the cube is the square root of 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! "! #. We will need half our diameter for the radius of the cylinder, or = inches. Using this value for r h, we calculate the volume of the cylinder as follows:! %&#'#( #&%#!"! )&*)" '! &(#*) cubic inches. To convert this volume to ounces, we multiply by 128 and divide by 231. This gives us: '! &(#*) #!*!%#!% &(#"( ounces. Rounding this number down to the previous whole number of ounces, we can say that the cylinder holds about 23 ounces of water.

13 Category 2 Meet #5, April How many 8-inch by 8-inch by 16- inch 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.) 24 ft. You may use a calculator today! 32 ft. 8 ft. 2. The label from each of the two cans shown here is a 4-inch by 12-inch rectangle which 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. 4 in. A 12 in. B 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. 50 feet

14 Solutions to Category 2 Meet #5, April : 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 or 24 blocks that are 16 inches long on each layer. Similarly, the sides that are 24 feet long will require 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 = 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 12 6 =. Similarly, we can find the 2π radius of can B, which is 4 π 2π =. Using the volume formula V = Areabase height = πr 2 h, we find: Can A: π( ) Can B: π( ) The ratio 2 π V = 4 = π 4 = π 2 π π V = 12 = π 12 = π π π 2 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 2 radius of 25, so = = 3925 square feet.