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Prisms and Cylinders Key Ideas A prism is a solid with two polygons for bases that are parallel and congruent. A circular cylinder is a solid with two congruent, parallel circles for bases. Both prisms and cylinders may be oblique or right.
(B is the area of the base, h is the height, r is the radius) Definitions The following definitions are used in our discussion of solid figures: A solid is any 3-D figure that is fully enclosed. A face of a solid is any of the surfaces that bound the solid. A polyhedron is any solid whose faces are polygons. An edge is the intersection of two faces in a polyhedron. A lateral face is any face of a solid other than its bases. Volume refers to the amount of space contained within a solid figure. We measure volume in units of length3, such as in.3, cm3, and ft3. These units refer to the amount of space occupied by a cube 1 unit in length on each side. For example, 1 in.3 is the volume of a cube 1 in. × 1 in. × 1 in. A cylinder with a volume of 50 in.3 would be able to contain 50 of these 1 in. × 1 in. × 1 in. cubes. Prisms A prism is a polyhedron with: Two bases that are parallel and congruent polygons Rectangular lateral faces
The prism shown in Figure is a rectangular prism. The edges between the lateral faces are called lateral edges. The height of the prism is the perpendicular distance between the two bases. Prisms are named based on the type of polygons that make up the bases.
Parts of a prism A prism can also be classified as right or oblique. A right prism, such as the one above, has its lateral edges perpendicular to the bases and the bases are aligned one over the other. An oblique prism has lateral edges that are not perpendicular to the bases as shown in the Figure above. Note that the height in an oblique prism is still the perpendicular distance between the bases.
Oblique prism Volume of a Prism Volume (V) = B · h where B is the area of the base and h is the height. Since the base can be any polygon, the volume formula is in terms of the area of the base B. When calculating the volume of a prism, the first step is to identify the bases and determine the correct area formula. Example Find the volume of the right triangular prism in the accompanying figure. Solution: The bases of the prism are the two triangles. So use the formula for the area of a triangle to find B. The base of the triangle is the 6 cm dimension, and the height of the triangle is 8 cm. Now apply the volume formula for the cylinder. The height of the prism is the 12 cm distance between the two triangles. Math Fact When working with prisms be careful not to confuse the height of the polygonal base with the height of the prism.
For example, a prism with a 10 cm height may have a triangular base whose height is 4 cm. The 4 cm height of the triangle is used in the area formula to calculate the area of the triangular base. The 10 cm height of the prism is used in V = Bh to calculate the volume of the prism. Circular Cylinders A circular cylinder is a solid figure with two parallel and congruent circular bases and a curved lateral surface. The height is the perpendicular distance between the bases. As with the prisms, cylinders can be right or oblique.
Figure below shows each type of cylinder.
Cylinder Volume of a Circular Cylinder Volume (V) = πr2h where r is the radius of the base and h is the height. Example 1 A right circular cylinder has a diameter of 10 cm and a height of 5 cm. Find the volume. Express your answers in terms of π. Solution: We can work backward to find the height or the radius given the volume of a cylinder. Example 2 A right circular cylinder has a volume of 200 in.3 If the height of the cylinder is 6 in., what is the radius of the cylinder? Round your answer to the nearest hundredth. Solution: Set up the formula for volume. Substitute the known values. Then solve for the radius.
Cones, Pyramids, and Spheres Key Ideas A circular cone is a solid with one circular base and an apex.
A pyramid is a solid with one polygonal base and an apex.
A sphere is the solid that is the set of all points a fixed distance from a center point.
Circular Cones A circular cone is a solid with one circular base and a curved lateral surface that comes to a point at the apex. The height of the cone is the length of the segment from the apex perpendicular to the base. In a right circular cone, the height will intersect at the center of the circular base. Right circular cone Volume of a Circular Cone where r is the radius of the base and h is the height. These formulas are applied in a manner similar to those for the cylinder. The Pythagorean theorem can be used to relate slant height, height, and radius. So if you know any two, you can find the third. Example Jenna is planting a garden and has a pile of topsoil delivered to her house. A dump truck delivers the topsoil and dumps it in a pile that is cone shaped. The radius of the pile is 18 ft and the height is 12 ft. What is the volume of the pile? Round to the nearest tenth. Solution: Pyramids A pyramid is a solid with one polygonal base and triangular lateral faces that meet at an apex. The height of the pyramid is the length of the segment from the apex perpendicular to the base. In a right pyramid, the height intersects at the center of the base.
Figure shows right pyramids with bases in the shape of a triangle and a rectangle. Triangular and square-based pyramids with height h Volume of a Pyramid where B is the area of the base and h is the height. Example Find the volume of a 5 ft high right pyramid with a square base whose side length is 4 ft. Solution: First find the area of the base, which is a square. Now apply the volume formula Math Fact If the base of a pyramid is a regular polygon, then the lateral faces are all isosceles triangles. Spheres A sphere, as shown, is the solid figure comprised of the set of all points a fixed distance (the radius) from a center point. The only dimension needed to define the volume and area of a sphere is its radius. Sphere with radius r Area and Volume of a Sphere Example Find the volume of a sphere whose diameter is 14 inches. Round to the nearest hundredth. Find the radius of a sphere whose volume equals 432.5 cm3. Round to the nearest tenth. Solution: Working backward, write the volume equation and fill in the known quantities.
Cross Sections and Solids of Revolution Key Ideas The intersection of a plane and a solid is an area called a cross section. Cross sections parallel to the base of a prism and cylinder are congruent to the base. Cross sections parallel to the base of a cone or pyramid are similar to the base. Cross sections of spheres are similar circles. Spheres are generated by rotating circles. Cones, cylinders, and spheres are solids of revolution. This means they can be generated by rotating a particular cross section around a line. Cones are generated by rotating triangles. Cylinders are generated by rotating rectangles. Spheres are generated by rotating circles. Cross Sections A cross section of a solid is the 2-dimensional figure created when a plane intercepts a solid. Cross sections are often made parallel or perpendicular to the base of a figure. Figure shows a cross section of a cone. The base lies in plane P, and plane Q is parallel to plane P. Plane Q intercepts a cross section in the shape of a circle. Cross sections of a cone
Prisms and cylinders, which have two parallel bases, have cross sections that are congruent to the bases when the intercepting plane is parallel to the base. The cylinder in Figure below has circular cross sections all congruent to the bases. Cones and pyramids have only one base. Their cross sections are all similar to the base when the plane is parallel to the base. The cross sections of the pyramid in the Figure are all rectangles similar to the base. As the cross sections approach the apex, their size approaches zero until the cross section is just a point. The cross sections of a sphere are all circles, which get smaller as the distance from the center increases. Cross sections of a pyramid, sphere, and cylinder Example 1 A right circular cylinder has a radius of 4 and a height of 10. What is the area of a cross section taken parallel to the bases? Write your answer in terms of π. Solution: The cross section is congruent to the base. So it is a circle with radius of 4. The area is πr2, or 16π.
Cross sections formed by planes perpendicular to a base may have specific shapes as well. Consider the following cross sections of a cone, pyramid, cylinder, and prism. Example 2 Plane Q intersects a prism with a square base perpendicular to the base and passes through the apex. What shape represents the intersection of the plane and the pyramid? Solution: The cross section perpendicular to the base is a triangle. Since the base is a regular polygon, the triangle must be isosceles.
Finally, cross sections can be taken at an angle that is neither parallel nor perpendicular to the base of a solid. Some examples of these types of cross sections are shown in Figure.
The circle, ellipse, parabola, and hyperbola are known as the conic sections because they represent the possible cross sections formed by a plane intersecting a cone. Some cross sections of solids Example Planes A and B both intersect a cylinder. If the cross section formed by plane A is a rectangle and the cross section formed by plane B is an ellipse, which of the following could be true? plane A is parallel to the base of the cylinder, and plane B forms a 45° angle with the base plane A is perpendicular to the base of the cylinder, and plane B is parallel to the base plane A is perpendicular to the base of the cylinder, and plane B forms a 45° angle with the base plane A forms a 45° angle with the base of the cylinder, and plane B is perpendicular to the base Solution:The correct choice is (3). The two cross sections are shown in the accompanying figures. Solids of Revolution
Certain solids can be generated by rotating a planar figure around a line. We call these solids of revolution.
Some examples of solids of revolution and the figures that generate them are shown in the following table.
The dimensions of the planar figure that is being rotated will determine the dimensions of the resulting solid.
For example, when a cone is generated by rotating a right triangle, the lengths of the two legs will be equal to the radius and height of the cone. Some planar figures can be rotated about more than one side.
The right triangle in the table could have been rotated about leg to generate a short, wide cone with height AC and radius AB.
Rectangle ABCD could have been rotated about to generate a cylinder with height AD and radius CD. Example Right triangle JGH has a right angle at G and legs with lengths GH = 7 cm and GJ = 6 cm. What solid figure is generated when △GHJ is rotated 360° about ? What is the volume in terms of π? Solution: The solid figure is a cone. The radius equals GH, and the height equals GJ. Apply the formula for the volume of a cone with the radius = 7 and height = 6. Math Fact The largest cross section of a sphere is called a “great circle.” A great circle is also the circle with the largest circumference that can be traced on the sphere. The center of a great circle is always the center of the sphere. On a flat surface, the shortest distance between two points is a straight line.
The corresponding theorem on the curved surface of a sphere is that the shortest distance between two points is along the great circle that contains those points.
Proving Volume and Area by Dissection, Limits, and Cavalieri’s Principle Key Ideas The formulas for circumference may depend on some parameter, such as the number of sides and area of a circle. The formulas for the volume of a cylinder, cone, pyramid, and sphere can all be derived using dissection, limits, and Cavalieri’s principle and the set of properties. Dissection—Any region or solid can be divided into smaller, nonoverlapping regions or solids such that the sum of the smaller areas or volumes equals the whole area or volume. Limit—A figure’s properties may depend on some parameter, such as the number of sides. A limit in geometry describes the set of properties that are approached as the parameter approaches some new value. We sometimes let the parameter approach infinity. For example, as the number of sides in a polygon approaches infinity, the shape of the polygon approaches that of a circle. Cavalieri’s principle—If two solids are bounded by the same two parallel planes and if any plane parallel to those two planes intercepts regions of equal area, then the two solids have the same volume. Alternatively, if two parallel planes intercept solids of equal volume, the original solids have equal volumes. Dissection The proofs of the area and volume formulas we will look at in this section rely on the mathematical concepts that we will define here—dissection, limits, and Cavalieri’s principle. The term “dissection” describes dividing a figure or solid into multiple parts. It is really just another way of saying “the whole equals the sum of the parts.”
Figure below shows the dissection of a regular hexagon into 6 equilateral triangles.
The area of the hexagon is equal to the sum of the areas of the 6 triangles.
This type of dissection was the basis for deriving the formula for the area of a regular hexagon, perimeter ⋅ apothem.
Figure b below shows the dissection of a right circular cylinder into three smaller cylinders. The volume of the large cylinder is equal to the volume of the three smaller cylinders. (a) Dissection of a hexagon into 6 triangles (b) Dissection of a cylinder into smaller cylinders Limits An algebraic limit is the value that an expression approaches as a variable gets closer and closer to some specified value.
A simple example is the limit of the value of as x approaches infinity.
We can make a table showing values of the expression as a function of x. You can see from the pattern that the limit of approaches 4 as x approaches infinity. Of course, the value of x cannot actually equal infinity since infinity is not a real number.
We can consider limits in the context of geometry.
Figure shows △ABC for three values of the measure of angle A. As m∠A approaches 90°, △ABC approaches a right triangle. In this case, we can actually reach the limit and △ABC will become a right triangle. Figure 12.12 shows a geometric example in which we do not actually reach the limit.
Consider and , which are not parallel. Now translate point D in the direction of vector . The slope of will begin to approach the slope of . The farther we translate D in the direction of , the closer the two slopes will be to each other. The limit of the slope of as the translation distance approaches infinity is the slope of . △ABC as m ∠A approaches 90° The slope of approaches the slope of as D translates away from C in the direction of
The final concept we will use in proving area and volume formulas is Cavalieri’s principle. Cavalieri’s Principle If two solids are contained between two parallel planes and if every parallel plane between these two planes intercepts regions of equal area, then the solids have equal volume. Also, any two parallel planes intercept two solids of equal volume.
Both the triangular prism and the rectangular prism shown in Figure are bounded by planes R and T. “Bounded” means that the solids have at least one point contained in each of planes R and T and that the solids do not pass through to the other sides of the planes. As plane S sweeps upward from R toward T, corresponding cross sections are formed in the triangular prism and the rectangular prism.
If every pair of corresponding cross sections are equal in area (though not necessarily congruent), then the two solids have the same volume.
Cavalieri’s principle applied to a prism with a triangular base and to a prism with a rectangular base
The second part of Cavalieri’s principle is illustrated below.
If the conditions required for Cavalieri’s principle are true, then any pair of parallel planes R and S will intercept regions of equal volume. Equal volumes contained between two parallel planes Math Fact The part of the cone and the part of the pyramid intercepted by planes R and S in Figure are called a truncated cone and truncated pyramid, respectively.
Another term used to describe these shapes is “frustum.” We can use Cavalieri’s principle to infer information about cross sections of solids with equal volume or about the volume of solids with corresponding cross sections of the same area. Example 1 Two stacks of quarters are shown in the accompanying figure. If each stack contains 7 identical quarters, what must be true about the volume of the two stacks? Explain your answer in terms of Cavalieri’s principle. Solution: Each of the quarters has a uniform and congruent cross section, and any plane parallel to the bases of the two stacks must intercept cross sections of equal area. Cavalieri’s principle states that the volumes of the stacks must be equal. Example 2 A right circle cylinder has a radius of 3 in. and a height of 4 in. A prism has bases in the shape of a square with side lengths in. and a height of 4 in. In terms of Cavalieri’s principle, explain why the two solids have the same volume. Solution: Every cross section of a cylinder parallel to its base is congruent, and the base of the given cylinder has an area of A = πr2 = 9π in.2
Every cross section of a prism parallel to its base is congruent, and its base has an area of A = s2 = 9π in.2 Since the solids have the same height, they are contained by the same two parallel planes. Every cross section of the two solids are equal in area. Therefore, Cavalieri’s principle states the two solids have the same volume. Circumference and Area of a Circle Circumference of a Circle The formula for the area of a circle can be derived using a limit argument by considering a polygon inscribed in a circle and letting the number of sides increase towards infinity. Consider the three polygons inscribed in circle P, as shown in Figure. The perimeter of the polygon will depend on both the radius of the circle and the number of sides of the polygon.
We can write the equation where r is the radius of the circle and kN is a number that depends on N. As N gets larger, the perimeter of the polygon approaches the circumference of the circle.
The value of kN also gets larger, but will never exceed a value of 2π. This fact can be shown by using trigonometry to write an expression for kN in terms of N, but it is outside the scope of this course. Proving the area of a circle using inscribed polygons
We can say that in the limit as the number of sides of an inscribed polygon approaches infinity, the figure will approach a circle and its perimeter will approach 2πr. In other words, the circumference of a circle equals 2πr. Area of a Circle The area of a circle can also be derived from a limit argument and inscribed polygons, as shown above.
The formula for the area of a polygon is
perimeter · apothem.
The apothem is the length a shown in the figure. As the number of sides increases towards infinity, the perimeter approaches the circumference of the circle and the apothem approaches the radius of the circle. Volume of a Cylinder The formula for the volume of a cylinder can be derived starting with the formula for a prism. Figure a shows a prism with a triangular base inscribed inside a right circular cylinder of the same height.
The volume of the cylinder is larger than the volume of the prism. Prisms with a triangular base and with a hexagonal base inscribed inside a right circular cylinder
Figure b shows the same cylinder with a 6-sided prism inscribed inside it. The volume of the cylinder is still larger but by not as much. As the number of sides of the prism approaches infinity, the area of the base of the prism approaches the area of the circular base of the cylinder.
Therefore the prism and cylinder must have the volume formula:
The formula for the area of a circle can be substituted for B: Volume of a Pyramid The volume of a pyramid can be derived by the dissection of a cube with side length s as shown in Figure below. Let point P be the center of the cube. The four bottom vertices and point P form a pyramid with a square base as shown in Figure a. Since P is the center of the cube, a congruent pyramid is formed by the top four vertices and point P as shown in Figure b below. Four more congruent pyramids can be made using point P and four vertices at the front, back, left, and right. None of the pyramids overlap. So the volume of the cube must be equal to the volume of the 6 pyramids: Dissection of a cube into 6 pyramids
The height of each pyramid, shown in Figure b, is equal to , or s = 2h. We can separate the s3 term into s2·s and then substitute s = 2h to get the final formula: The area of the base of each pyramid is the same as the area of the base of the cube, or s2. Volume of a Cone The same approach that was used for the volume of a cylinder can be applied to derive the formula for the volume of a cone. Starting with a pyramid whose volume is , let the number of sides in the polygonal base approach infinity.
The shape of the base will approach a circle, but the apex will be unchanged.
So the formula remains the same:
Modeling and Design Key Ideas Physical objects can be modeled using the basic solids—prisms, cylinders, cones, pyramids, and spheres.
Volume and area considerations can be used to design physical objects or to meet a condition.
Dimensions can be found to maximize or minimize area, volume, cost, and so on.
Density can be used to find mass, population, or any other quantity specified on a per-area or per-volume basis.
Numerical and graphical methods may be needed to find solutions to problems that do not result in equations that are easily solved. Modeling with Volume Many physical objects can be modeled using the basic solids discussed in this chapter. Some examples include: Cylinders—trees, cans, barrels, people Spheres—balls, balloons, planets Prisms—bricks, boxes, aquariums, swimming pools, rooms, books
The dimensions needed for a solid figure can be found in the text of the problem.
An added layer to these problems is that volumes or areas are used to calculate some other quantity.
Some common relationships to look for include: Mass = density × volume Total cost = cost per unit volume × volume Energy contained in a material = heat content × volume
For example, if a can has a volume of 100 cm3 and it is filled with a liquid whose density is 1.2 grams/cm3, the mass of the liquid inside the can can be found:
In some problems, a volume of material may be transferred from one container to another or transformed into another shape. In these cases, the total volume remains unchanged and the volumes before and after must be equal. Example 1 A cylindrical piece of metal has a radius of 15 in. and a height of 2 in. It goes through a hot-press machine that reduces the height of the cylinder to What is the new radius of the cylinder, assuming no material is lost? Round to the nearest thousandth. Solution: Strategy—Find the volume of the cylinder before pressing, and set it equal to the expression for the volume after pressing. Use this equation to solve for the radius after pressing. Example 2 An engineer for a construction company needs to calculate the total volume of a home that is under construction so the proper-sized heating and air conditioning equipment can be installed. The house is modeled as a rectangular prism with a triangular prism for the roof as shown in the accomanying figure. The roof is symmetric with EF = ED. The dimensions are AB = 30 ft, BC = 50 ft, and BD = 25 ft, and the measure of ∠EDF = 40°. What is the total volume of the house? Round to the nearest cubic foot. Solution: Calculate the volume of each section.
Rectangular prism:
Triangular prism: The first step is to find the area of the triangular base, △FED. FD has a length equal to AB, or 30 ft. The height of the isosceles triangle divides the base into two 15 ft lengths. The tangent ratio can be used to find the height of △FED.
The volume of a triangular prism is equal to Bh, where B is the area of the triangular base and h is the length of the prism, which is equal to BC. Add the two volumes to get the total volume of the house. Example 3 A monument consists of a pyramid with a square base on top of a prism with a square base. A central cylindrical shaft in the center houses a spiral staircase. Point H indicates the center of the circular base of the shaft. The dimensions of the monument are AB = 24 ft, BE = 84 ft, GE = 30 ft, and m∠GEH = 60°. The central shaft has a diameter of 6 ft. The monument is solid stone except for the shaft. What was the weight of the stone used to construct the monument, to the nearest ton? The stone has a density of 120 lb/ft3. Solution: The volume of the monument is the sum of the volumes of the prism and pyramid minus the volume of the shaft. Find each volume separately.
Prism:
Pyramid: First calculate the height of the pyramid, GH, using △GHE and the sine ratio. The base of the pyramid is a square with side length 24 ft. Cylinder: The cylindrical shaft has a diameter of 6 ft, a radius of 3 ft, and a height of 84 ft.
Calculate the weight of the stone
Finally, convert the weight to tons using the conversion factor found on the Common Core reference sheet.
The stone has a weight of 3,060 tons. Density and Area Density is usually defined as the amount of some quantity per unit volume. However, it can also be calculated on a per unit area basis.
Some example of densities per unit area that appear in real-world applications include the following:
Population density—The number of people living in a region divided by the area of the region is the population density. High population density is typical of an urban area, while low population density may be found in a rural area. Power density—Solar power panels produce power that is proportional to their area. The power produced per unit area is an important measure of the quality and efficiency of a solar panel. Computer applications—Storage devices such as hard drives may be described in terms of number of gigabytes of data per unit area. Example The city of Deerfield sits along the bank of the Fox Run River. A map of the city is shown. The downtown region, indicated by region A, is mile wide and 1 mile long. The population density of Deerfield, except for the city center, is 800 people per square mile. The population density of the downtown region is 4,000 people per square mile. What is the population of Deerfield?
Solution: We can model the town as a semicircle and the downtown region as a rectangle.
To calculate the population, we first find the area of each region.
Let the area of the downtown region be represented by AreaA and the area of the remainder of the town be represented by AreaB.
The area of the remainder of the town is a semicircle minus the area of the downtown region.
The population of each region is the product of the area and the population density.
The population of Deerfield is 6,626 people.
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