{"id":4249,"date":"2023-12-19T03:20:51","date_gmt":"2023-12-18T21:50:51","guid":{"rendered":"https:\/\/envisionmathanswerkey.com\/?p=4249"},"modified":"2023-12-20T10:36:30","modified_gmt":"2023-12-20T05:06:30","slug":"envision-math-common-core-grade-8-answer-key-topic-8","status":"publish","type":"post","link":"https:\/\/envisionmathanswerkey.com\/envision-math-common-core-grade-8-answer-key-topic-8\/","title":{"rendered":"enVision Math Common Core Grade 8 Answer Key Topic 8 Solve Problems Involving Surface Area And Volume"},"content":{"rendered":"

Go through the\u00a0enVision Math Common Core Grade 8 Answer Key<\/a>\u00a0Topic 8 Solve Problems Involving Surface Area And Volume<\/strong>\u00a0regularly and improve your accuracy in solving questions.<\/p>\n

enVision Math Common Core 8th Grade Answers Key Topic 8 Solve Problems Involving Surface Area And Volume<\/h2>\n

Topic 8 Essential Question<\/strong>
\nHow can you find volumes and surface areas of three-dimensional figures?
\nAnswer:
\nThe “Surface area” is the sum of the areas of all faces (or surfaces) on a 3D shape.
\nEx:
\nA cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces
\nWe know that,
\nThe volume of a three-dimensional figure = <\/b>Cross-sectional area \u00d7 length<\/p>\n

3-ACT MATH<\/strong><\/p>\n

\"Envision
\nMeasure Up
\nHave you ever heard of the terms griffin beaker, Erlenmeyer flask, or graduated cylinder? Maybe you’ve used them in your science class. Each piece of equipment in a chemistry lab has a specific purpose, so containers come in many shapes. It’s sometimes necessary to pour a solution from one container to another. Think about this during the 3-Act Mathematical Modeling lesson.
\n\"Envision<\/p>\n

Topic 8 \u00ebnVision STEM Project<\/strong><\/p>\n

Did You Know?
\nThe production of packaging is a huge industry employing over five million people with annual sales of more than 400 billion dollars.
\nPackaging materials protect and deliver food and products to consumers.
\n\"Envision<\/p>\n

A plastic bottle takes 450\u20131,000 years to biodegrade.
\n\"Envision
\nSeabirds are dying of starvation with stomachs full of plastic and Styrofoam.
\n\"Envision<\/p>\n

Polystyrene foam lasts forever!
\nEco-friendly packaging materials are being made from mushrooms and bamboo.
\n\"Envision
\nThere is even a drink bottle made from recyclable paper.
\n\"Envision
\nNew technology results in packaging materials that are both affordable and biodegradable.
\nEnvironmentally friendly companies are producing sustainable packaging. In addition to using recyclable materials, they reduce the water, natural resources, and energy needed for production. They minimize waste when designing products.<\/p>\n

Your Task: Wrap it Up!
\n\"Envision
\nEngineers consider several factors when designing product packaging. These factors include cost efficiency and eco-friendly design so that materials are disposable, recyclable, biodegradable, and not wasted. Suppose you are an engineer working for Liquid Assets, an environmentally friendly company that designs, builds, and packages water purifiers. You and your classmates will use your knowledge of volume and surface area to determine an environmentally sound way to package the purifiers.<\/p>\n

Topic 8 GET READY!<\/h3>\n

Review What You Know!<\/strong><\/p>\n

Vocabulary
\nChoose the best term from the box to complete each definition.
\nbase
\ndiameter
\nradius
\nthree-dimensional
\ntwo-dimensional<\/p>\n

Question 1.
\nThe __________ is the distance from the center to the edge of a circle.
\nAnswer:
\nWe know that,
\nThe “Radius” is the distance from the center to the edge of a circle
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is: Radius<\/p>\n

\"Math<\/p>\n

Question 2.
\nA shape that has length, width, and height is ___________.
\nAnswer:
\nWe know that,
\nA shape that has a length, width, and height is known as “Three-dimensional”
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is: Three dimensional<\/p>\n

Question 3.
\nAny side of a cube can be considered a __________.
\nAnswer:
\nWe know that,
\nAny side of a cube can be considered a “Base”
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is: Base<\/p>\n

Question 4.
\nA shape that has length and width, but not height, is ___________.
\nAnswer:
\nWe know that,
\nA shape that has length and width, but not height is known as “Two-dimensional”
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is: Two-dimensional<\/p>\n

Question 5.
\nThe _____________ of a circle is a line segment that passes through its center and has endpoints on the circle.
\nAnswer:
\nWe know that,
\nThe “Diameter” of a circle is a line segment that passed through its center and has endpoints on the circle
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is: Diameter<\/p>\n

Multiplying with Decimals<\/strong><\/p>\n

Find the product.
\nQuestion 6.
\n14 \u2219 3.5 = _______
\nAnswer:
\nThe given expression is:
\n14 \u00d7 3.5
\nSo,
\n14 \u00d7 3.5 = 49.0
\nHence, from the above,
\nWe can conclude that the value for the given expression is: 49<\/p>\n

Question 7.
\n9 \u2219 3.14 = _________
\nAnswer:
\nThe given expression is:
\n9 \u00d7 3.14
\nSo,
\n9 \u00d7 3.14 = 28.26
\nHence, from the above,
\nWe can conclude that the value for the given expression is: 28.26<\/p>\n

Question 8.
\n4.2 \u2219 10.5 = _________
\nAnswer:
\nThe given expression is:
\n4.2 \u00d7 10.5
\nSo,
\n4.2 \u00d7 10.5 = 44.1
\nHence, from the above,
\nWe can conclude that the value for the given expression is: 44.1<\/p>\n

Areas of Circles<\/strong><\/p>\n

Find the area of each circle. Use 3.14 for \u03c0.<\/p>\n

\"Math
\nQuestion 9.
\n\"Envision
\nA = ________
\nAnswer:
\nThe given circle is:
\n\"Envision
\nFrom the given circle,
\nThe radius is: 8 cm
\nNow,
\nWe know that,
\nThe area of the circle = \u03c0r\u00b2
\nSo,
\nThe area of the given circle = 3.14 \u00d7 8\u00b2
\n= 200.96 cm\u00b2
\nHence, from the above,
\nWe can conclude that the area for the given circle is: 200.96 cm\u00b2<\/p>\n

Question 10.
\n\"Envision
\nA = _________
\nAnswer:
\nThe given circle is:
\n\"Envision
\nFrom the given circle,
\nThe diameter is: 12 cm
\nNow,
\nWe know that,
\nRadius = \\(\\frac{Diameter}{2}\\)
\nRadius = \\(\\frac{12}{2}\\)
\nRadius = 6 cm
\nNow,
\nWe know that,
\nThe area of the circle = \u03c0r\u00b2
\nSo,
\nThe area of the given circle = 3.14 \u00d7 6\u00b2
\n= 113.04 cm\u00b2
\nHence, from the above,
\nWe can conclude that the area of the given circle is: 113.04 cm\u00b2<\/p>\n

Use the Pythagorean Theorem<\/strong><\/p>\n

Find the missing side length of the triangle.
\nQuestion 11.
\n\"Envision
\nx = _________
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nNow,
\nFro the given figure,
\nWe can observe that the triangle is a right triangle
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the side lengths
\nSo,
\n13\u00b2 = 12\u00b2+ x\u00b2
\nx\u00b2= 169 – 144
\nx\u00b2 = 25
\nx = \\(\\sqrt{25}\\)
\nx = 5 in.
\nHence, from the above,
\nWe can conclude that the missing side length of the given triangle is: 5 in.<\/p>\n

Question 12.
\n\"Envision
\nx = __________
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nNow,
\nFro the given figure,
\nWe can observe that the triangle is a right triangle
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the side lengths
\nSo,
\n30\u00b2 = 24\u00b2+ x\u00b2
\nx\u00b2= 900 – 576
\nx\u00b2 = 324
\nx = \\(\\sqrt{324}\\)
\nx = 18 in.
\nHence, from the above,
\nWe can conclude that the missing side length of the given triangle is: 18 in.<\/p>\n

Language Development<\/strong><\/p>\n

Complete the word web. Write keywords, ideas, examples, or illustrations that connect to each new vocabulary term.
\n\"\"
\n\"\"<\/p>\n

Topic 8 Pick A Project<\/h3>\n

PROJECT 8A
\nWhat makes a concert rock?
\nPROJECT: DESIGN PROPS OR STAGE STRUCTURES
\n\"Envision<\/p>\n

PROJECT 8B
\nWhat is the most interesting museum you have visited?
\nPROJECT: MAKE A MODEL OF A MUSEUM
\n\"Envision<\/p>\n

PROJECT 8C
\nWhere around the United States can you find quarries?
\nPROJECT: POUR AND MEASURE SAND
\n\"Envision<\/p>\n

PROJECT 8D
\nIf you were cast in a play, would it be a comedy or a drama? Why?
\nPROJECT: WRITE A SKIT
\n\"Envision<\/p>\n

Lesson 8.1 Find Surface Area of Three-Dimensional Figures<\/h3>\n

Explore It!<\/strong>
\nAndrea is designing the packaging for a tube-shaped container.
\n\"Envision
\nI can… find the surface areas of cylinders, cones, and spheres.<\/p>\n

A. Model with Math What two-dimensional shape represents the top and bottom of the container? What two-dimensional shape represents the tube? Draw a net of the tube-shaped container.
\nAnswer:
\nIt is given that Andrea is designing the packaging for a tube-shaped container.
\nNow,
\nThe given arrangement for tube shaped container is:
\n\"Envision
\nNow,
\nFrom the given arrangement,
\nWe can observe that
\nThe two-dimensional shape that represents the top and bottom of the container is: Circle
\nThe two-dimensional shape that represents the tube is: Rectangle
\nHence,
\nThe representation of the tube-shaped container is:
\n\"\"<\/p>\n

B. Look for Relationships The circular top and bottom fit perfectly on the ends of the container. How are the measures of the circles and the rectangle related?
\nAnswer:
\nThe representation of the tube-shaped container is:
\n\"\"
\nNow,
\nFrom the given figure,
\nWe can observe that the tube-shaped container is made up of 2 circles and 1 rectangle
\nSo,
\nThe total surface area of the tube-shaped container is the sum of the areas of 2 circles and 1 rectangle
\nNow,
\nWe know that,
\nThe area of a circle = \u03c0r\u00b2
\nThe area of a rectangle = Height\u00d7 Base
\nWhere,
\nThe base of the rectangle is a circle
\nSo,
\nThe circumference of the circle = 2\u03c0r
\nHence,
\nThe surface area of the tube-shaped container = 2\u03c0r\u00b2 + 2\u03c0rh<\/p>\n

Focus on math practices
\nModel with Math
\nHow can you check whether the net that you drew accurately represents the tube-shaped container?
\nAnswer:
\nThe representation of the tube-shaped container is:
\n\"\"
\nNow,
\nWhen the top and bottom of the container correctly fit the tube,
\nThat is the representation that the net you drew accurately represents the tube-shaped container<\/p>\n

Essential Question<\/strong>
\nHow are the areas of polygons used to find the surface area formulas for three-dimensional figures?
\nAnswer:
\nWe know that,
\nA three-dimensional figure is a combination of some two-dimensional figures
\nEx:
\nWe can make a “Cuboid” from the combination of “Rectangles”
\nWe can make a “Sphere” from the combination of ‘Circles”
\nSo,
\nThe total surface area of a three-dimensional figure can be given as the sum of all the areas of the two-dimensional figures that are used to make the three-dimensional figure<\/p>\n

Try it
\n<\/strong><\/p>\n

What is the surface area of a cylinder with a height of 9.5 inches and a radius of 2.5 inches?
\nThe surface area of the cylinder is __________ square inches.
\nAnswer:
\nIt is given that
\nThe height of a cylinder is: 9.5 inches
\nThe radius of a cylinder is: 2.5 inches
\nNow,
\nWe know that,
\nThe surface area of the cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\n= 2 \u00d7 3.14 \u00d7 (2.5)\u00b2 + 2 \u00d7 3.14 \u00d7 9.5 \u00d7 2.5
\n= 39.25 + 149.15
\n= 188.4 square inches
\nHence, from the above,
\nWe can conclude that the surface area of the given cylinder is: 188.4 square inches<\/p>\n

Convince Me!<\/strong> How can you find the surface area of a cylinder if you only know its height and the circumference of its base?
\nS.A. = 2(\u03c0r2<\/sup>) + (2\u03c0r)h
\n= 2\u03c0(________2<\/sup>) + 2\u03c0(_______)(________)
\n= _______\u03c0 + _______\u03c0
\n= _______\u03c0
\nAnswer:
\nIt is given that
\nWe know only the height of the cylinder and the circumference of its base
\nNow,
\nWe know that,
\nThe surface area of a cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2\u03c0 (r\u00b2) + 2\u03c0 (r) (h)
\nS.A = 2\u03c0 + \u03c0
\nS.A = 3\u03c0
\nHence, from the above,
\nWe can conclude that the surface area of a cylinder with only its height and the circumference of its base is: 3\u03c0<\/p>\n

\"Math<\/p>\n

Try It!<\/strong><\/p>\n

a. What is the surface area of a cone with a radius of 7 feet and a slant height of 9 feet? Use \\(\\frac{22}{7}\\) for \u03c0.
\nAnswer:
\nIt is given that
\nThe radius of the cone is (r): 7 feet
\nThe slant height of the cone is (l): 9 feet
\nNow,
\nWe know that,
\nThe surface area of the cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = \\(\\frac{22}{7}\\) \u00d7 7\u00b2 + \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 9
\n= 154 + 198
\n= 352 square feet
\nHence, from the above,
\nWe can conclude that the surface area of the given cone is: 352 square feet<\/p>\n

b. What is the surface area of a sphere with a diameter of 2.7 inches? Use 3.14 for \u03c0.
\nAnswer:
\nIt is given that,
\nThe diameter of a sphere is: 2.7 inches
\nNow,
\nWe know that,
\nRadius = \\(\\frac{Diameter}{2}\\)
\nSo,
\nRadius of the sphere = \\(\\frac{2.7}{2}\\)
\nSo,
\nThe radius of the sphere (r) = 1.35 inches
\nNow,
\nWe know that,
\nThe surface area of the sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\nS.A = 4 \u00d7 3.14 \u00d7 (1.35)\u00b2
\n= 22.89 square inches
\nHence, from the above,
\nWe can conclude that the surface area of the given sphere is: 22.89 square inches<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

Formulas for finding the area of polygons can be used to find the surface areas of cylinders, cones, and spheres.<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How are the areas of polygons used to find the surface area formulas for three-dimensional figures?
\nAnswer:
\nWe know that,
\nA three-dimensional figure is a combination of some two-dimensional figures
\nEx:
\nWe can make a “Cuboid” from the combination of “Rectangles”
\nWe can make a “Sphere” from the combination of ‘Circles”
\nSo,
\nThe total surface area of a three-dimensional figure can be given as the sum of all the areas of the two-dimensional figures that are used to make the three-dimensional figure<\/p>\n

Question 2.
\nReasoning<\/strong> Why is the length of the base of the rectangle the same as the circumference of the circles in the net of a cylinder?
\nAnswer:
\nThe representation of the cylinder is:
\n\"\"
\nNow,
\nIf you look at the net, the curved surface of the cylinder is rectangular in shape. The length of the rectangle is the same as the circumference of the circle. Since the length of the rectangle wraps around the circle rim, it is the same length as the circumference of the circle.<\/p>\n

Question 3.
\nConstruct Arguments<\/strong> Aaron says that all cones with a base circumference of 8 inches will have the same surface area. Is Aaron correct? Explain.
\nAnswer:
\nIt is given that
\nAaron says that all cones with a base circumference of 8 inches will have the same surface area
\nNow,
\nWe know that,
\nThe surface area of a cone = \u03c0r\u00b2 + \u03c0rl
\nNow,
\nFrom the above formula,
\nWe can conclude that the surface area of the cone does not depend only on the circumference of the base but also we need the side length of the cone part as well
\nSo,
\nAll cones with a base circumference of 8 inches will not have the same surface area.
\nHence, from the above,
\nWe can conclude that Aaron is not correct<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nWhat is the surface area of the cylinder? Use 3.14 for \u03c0, and round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nThe diameter of the cylinder is: 2 mm
\nThe height of the cylinder is: 10 mm
\nSo,
\nRadius of the cylinder = \\(\\frac{Diameter of the cylinder}{2}\\)
\n= \\(\\frac{2}{2}\\)
\n= 1 mm
\nNow,
\nWe know that,
\nThe surface area of the cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2 \u00d7 3.14 \u00d7 1\u00b2 + 2 \u00d7 3.14 \u00d7 1 \u00d7 10
\n= 6.28 + 62.8
\n= 69.08 mm\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cylinder is: 69.08 mm\u00b2<\/p>\n

Question 5.
\nWhat is the surface area of the cone to the nearest tenth? Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nThe slant height of the cone is (l): 4 ft
\nThe radius of the cone is: 3 ft
\nNow,
\nWe know that,
\nThe surface area of the cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 3\u00b2 + 3.14 \u00d7 3 \u00d7 4
\n= 28.26 + 37.68
\n= 65.94 ft\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cone is: 65.94 ft\u00b2<\/p>\n

Question 6.
\nWhat is the surface area of the sphere in terms of \u03c0?
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nThe diameter of the sphere is: 2 cm
\nNow,
\nWe know that,
\nThe radius of the sphere (r) = \\(\\frac{Diameter of the sphere}{2}\\)
\n= \\(\\frac{2}{2}\\)
\n= 1 cm
\nNow,
\nWe know that,
\nThe surface area of the sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\nS.A = 4 \u00d7 3.14 \u00d7 1\u00b2
\n= 12.56 cm\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given sphere is: 12.56 cm\u00b2<\/p>\n

\"Math<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Leveled Practice In 7-8, find the surface area.
\nQuestion 7.
\nWhat is the surface area of the cylinder? Use 3.14 for \u03c0, and round to the nearest tenth.
\n\"Envision
\nS.A. = 2(\u03c0r2<\/sup>) + (2\u03c0r)h
\n= 2\u03c0(________2<\/sup>) + 2 \u03c0(_______)(________)
\n= 2 \u03c0(________) + 2 \u03c0(_________)
\n= _______\u03c0 + _______\u03c0
\n= _______\u03c0
\n\u2248 ________ cm2<\/sup>
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nThe radius of the cylinder is: 3 cm
\nThe height of the cylinder is: 5 cm
\nNow,
\nWe know that,
\nThe surface area of the cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2 \u00d7 3.14 \u00d7 3\u00b2 + 2 \u00d7 3.14 \u00d7 3 \u00d7 5
\n= 56.52 + 94.2
\n= 150.72 cm\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cylinder is: 150.72 cm\u00b2<\/p>\n

Question 8.
\nWhat is the surface area of the cone? Use 3.14 for \u03c0.
\n\"Envision
\nS.A. = \u03c0r2<\/sup> + \u03c0lr
\n= \u03c0(________2<\/sup>) + \u03c0(_______)(________)
\n= ________ \u03c0 + _________\u03c0
\n= _______\u03c0
\n\u2248 ________ cm2<\/sup>
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nThe radius of the cone (r) is: 7 cm
\nThe slant height of the cone (l) is: 13 cm
\nNow,
\nWe know that,
\nThe surface area of the cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 7\u00b2 + 3.14 \u00d7 7 \u00d7 13
\n= 153.86 + 285.74
\n= 439.6 cm\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cone is: 439.6 cm\u00b2<\/p>\n

Question 9.
\nConstruct Arguments Sasha incorrectly claimed that the surface area of the cylinder is about 76.9 square inches. Explain her likely error and find the correct surface area of the cylinder.
\n\"Envision
\nAnswer
\nThe given figure is:
\n:\"Envision
\nNow,
\nFrom the given figure,
\nThe diameter of the cylinder is: 7 in.
\nThe height of the cylinder is: 19 in.
\nNow,
\nWe know that,
\nRadius = \\(\\frac{Diameter}{2}\\)
\nSo,
\nRadius of the circle = \\(\\frac{7}{2}\\)
\n= 3.5 in.
\nNow,
\nWe know that,
\nThe surface area of the cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2 \u00d7 3.14 \u00d7 (3.5)\u00b2 + 2 \u00d7 3.14 \u00d7 3.5 \u00d7 19
\n= 76.93 + 417.62
\n= 494.55 in.\u00b2
\nHence, from the above,
\nWe can conclude that
\nThe correct surface area of the given cylinder is: 494.55 in.\u00b2
\nThe mistake made by Sasha is:
\nShe adds only the area of the top and bottom but not the area of the rectangle and the area of the top and bottom of the cylinder<\/p>\n

Question 10.
\nA theme park has a ride that is located in half a sphere. The ride goes around the widest part of the sphere, which has a circumference of 514.96 yards. What is the surface area of the sphere? Estimate to the nearest hundredth using 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nIt is given that
\nA theme park has a ride that is located in half a sphere. The ride goes around the widest part of the sphere, which has a circumference of 514.96 yards
\nNow,
\nThe given sphere is:
\n\"Envision
\nNow,
\nWe know that,
\nCircumference = 2\u03c0r
\nSo,
\nCircumference of the theme park = 514.96 yd
\n2\u03c0r = 514.96
\nr = \\(\\frac{514.96}{2 \u00d7 3.14}\\)
\n= 2.38 yd
\nNow,
\nWe know that,
\nThe surface area of the sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\nS.A = 4 \u00d7 3.14 \u00d7 (2.38)\u00b2
\n= 71.14 yd\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given sphere is: 71.14 yd\u00b2<\/p>\n

Question 11.
\nFind the amount of wrapping paper you need to wrap a gift in the cylindrical box shown. You need to cover the top, the bottom, and all the way around the box. Use 3.14 for \u03c0, and round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nThe radius of the cylinder is 9 in.
\nThe height of the cylinder is: 8 in
\nNow,
\nWe know that,
\nThe surface area of the cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2 \u00d7 3.14 \u00d7 9\u00b2 + 2 \u00d7 3.14 \u00d7 9 \u00d7 8
\n= 508.68 + 452.16
\n= 960.84 in.\u00b2
\nHence, from the above,
\nWe can conclude that the amount of wrapping paper you need to wrap a gift in the cylindrical box is: 960.84 in.\u00b2<\/p>\n

Question 12.
\nDonna paints ornaments for a school play. Each ornament is made up of two identical cones, as shown. How many bottles of paint does she need to paint 70 ornaments?
\n\"Envision
\nAnswer:
\nIt is given that
\nDonna paints ornaments for a school play. Each ornament is made up of two identical cones, as shown
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe radius of a cone is: 4.1 cm
\nThe slant height of a cone is: 8.9 cm
\nNow,
\nWe know that,
\nThe surface area of a cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 (4.1)\u00b2 + 3.14 \u00d7 4.1 \u00d7 8.9
\n= 52.78 + 114.57
\n= 167.35 cm\u00b2
\nSo,
\nThe surface area of the second cone (S.A) = 167.35 cm\u00b2
\nSo,
\nThe S.A of 70 ornaments = 70 \u00d7 (167.35 \u00d7 2)
\n= 23,429 cm\u00b2
\nSo,
\nThe number of bottles of paint she needed to paint 70 ornaments = \\(\\frac{23,429}{2,000}\\)
\n= 11.7
\n\u2245 12 bottles
\nHence, from the above,
\nWe can conclude that she need 12 bottles of paint to paint 70 ornaments<\/p>\n

Question 13.
\nHigher-Order Thinking<\/strong>
\na. What is the surface area of the cone? Use 3.14 for \u03c0, and round to the nearest whole number.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nThe diameter of the cone is: 6 cm
\nThe slant height of the cone is: 12 cm
\nNow,
\nWe know that,
\nRadius = \\(\\frac{Diameter}{2}\\)
\nRadius = \\(\\frac{6}{2}\\)
\nRadius = 3 cm
\nNow,
\nWe know that,
\nThe surface area of the cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 3\u00b2 + 3.14 \u00d7 3 \u00d7 12
\n= 28.26 + 113.04
\n= 141.3 cm\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cone is: 141.3 cm\u00b2<\/p>\n

b. Reasoning Suppose the diameter and the slant height of the cone are cut in half. How does this affect the surface area of the cone? Explain.
\nAnswer:
\nIt is given that the diameter and the slant height of the cone are cut in half
\nSo,
\nS.A = \u03c0(\\(\\frac{d}{2}\\))\u00b2 + \u03c0 (\\(\\frac{d}{2}\\)) (\\(\\frac{l}{2}\\))
\n= 3.14 \u00d7 \\(\\frac{6\u00b2}{4}\\) + 3.14 \u00d7 3 \u00d7 6
\n= 28.26 + 56.52
\n= 84.78 cm\u00b2
\nNow,
\nFrom part (a),
\nThe S.A of the cone is: 141.3 cm\u00b2
\nSo,
\nThe ratio of S.A of the cones obtained from part (a) and part (b) respectively = \\(\\frac{141.3}{84.78}\\)
\n= 1.666
\nHence, from the above,
\nWe can conclude that the S.A of the cone we obtained in part (b) is 1.666 times of the S.A of the cone we obtained in part (b)<\/p>\n

Assessment Practice<\/strong>
\nQuestion 14.
\nWhat is the surface area of the sphere? Use 3.14 for \u03c0, and round to the nearest tenth.
\n\"Envision
\nA 254.5 cm2<\/sup>
\nB. 56.55 cm2<\/sup>
\nC. 1,017.4 cm2<\/sup>
\nD. 4,071.5 cm2<\/sup>
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nThe radius of the sphere is: 9 cm
\nNow,
\nWe know that,
\nThe surface area of the sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\nS.A = 4 \u00d7 3.14 \u00d7 9\u00b2
\n= 1,017.36 cm\u00b2
\n\u2248 1,017.4 cm\u00b2
\nHence, from the above,
\nWe can conclude that the S.A of the given sphere is: 1,017.4 cm\u00b2<\/p>\n

Question 15.
\nWhat is the approximate surface area of the cone, in square inches? Use 3.14 for \u03c0, and round to the nearest whole number.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nThe slant height of the cone (l) is 40 in.
\nThe diameter of the cone (d) is 40 in.
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{40}{2}\\)
\nr = 20 in.
\nNow,
\nWe know that,
\nThe surface area of the cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 20\u00b2 + 3.14 \u00d7 20 \u00d7 40
\n= 1,256 + 2,512
\n= 3,768 in.\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cone is: 3,768 in\u00b2<\/p>\n

Lesson 8.2 Find Volume of Cylinders<\/h3>\n

Explain It!<\/strong>
\nJenna and Ricardo are buying a new fish tank for the growing population of zebrafish in their science lab. Jenna says the tanks hold the same amount of water because they have the same dimensions. Ricardo says that he can fill the bottom of the rectangular tank with more cubes, so it can hold more water.
\n\"Envision
\nI can… use what I know about finding volumes of rectangular prisms to find the volume of a cylinder.<\/p>\n

A. Look for Relationships How are the shapes of the two fish tanks alike? How are they different?
\nAnswer:
\nWhen we observe the fish tanks of Jenna and Ricardo,
\nWe can observe that
\na. The heights of the two fish tanks are the same
\nb. The number of cubes that can be filled at the bottom is different
\nc. The number of cubes filled in Jenna’s fish tank is less than that of the number of cubes filled in Ricardo’s fish tank
\nd. The amount of water that can hold in Jenna’s fish tank is less than the amount of water that can hold in Ricardo’s fish tank<\/p>\n

B. Critique Arguments Who do you think is correct, Ricardo or Jenna? Explain.
\nAnswer:
\nIt is given that
\nJenna and Ricardo are buying a new fish tank for the growing population of zebrafish in their science lab. Jenna says the tanks hold the same amount of water because they have the same dimensions. Ricardo says that he can fill the bottom of the rectangular tank with more cubes, so it can hold more water.
\nNow,
\nTo find which fish tank holds more water, find the volume of the two fish tanks
\nNow,
\nFor Jenna’s fish tank:
\nThe fish tank is in the form of a cylinder
\nNow,
\nWe know that,
\nThe volume of the cylinder (V) = \u03c0r\u00b2h
\n= \\(\\frac{\u03c0d\u00b2h}{4}\\)
\nSo,
\nV = \\(\\frac{3.14 \u00d7 24 \u00d7 24 \u00d7 48}{4}\\)
\n= \\(\\frac{86,814.72}{4}\\)
\n= 21,073.68 in\u00b3
\nFor Ricardo’s fish tank:
\nThe fish tank is in the form of a rectangular prism
\nNow,
\nWe know that,
\nThe volume of a rectangular prism (V) = Length \u00d7 Width \u00d7 Height
\nSo,
\nV = 24 \u00d7 24 \u00d7 48
\n= 27,648 in\u00b3
\nHence, from the above,
\nWe can conclude that Ricardo is correct on the basis of volumes of their fish tanks<\/p>\n

Focus on math practices
\nUse Structure How can you use what you know about areas of two-dimensional figures and volumes of prisms to compare the volumes of the fish tanks?
\nAnswer:
\nWe know that,
\nVolume = Area \u00d7 Length (or) Height (or) Depth
\nWhere,
\n“Area” is the area of two-dimensional figures like rectangles, circles, etc.<\/p>\n

Essential Question<\/strong>
\nHow is the volume of a cylinder related to the volume of a rectangular prism?
\nAnswer:
\nRectangular prisms and cylinders\u00a0are somewhat\u00a0similar\u00a0because they both have two bases and a height.
\nThe formula for the volume of a rectangular solid is
\nV=Bh
\ncan also be used to find the volume of a cylinder
\nWhere,
\n“B” in the rectangular prism is the area of the rectangle
\n“B” in the cylinder is the area of the circle<\/p>\n

Try It!<\/strong><\/p>\n

The area of the base of the cylinder is 78.5 in.2<\/sup>. What is the volume of the cylinder?
\n\"Envision
\nV = Bh
\n= _______ \u2219 _______
\n= _______
\nThe volume of the cylinder is ________ cubic inches.
\nAnswer:
\nIt is given that
\nThe area of the base of the cylinder is 78.5 in.2<\/sup>
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = Bh
\nWhere,
\n“B” is defined as the area of the circle
\nSo,
\nV = 78.5 \u00d7 11
\n= 863.5 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 863.5 cubic inches<\/p>\n

Convince Me!
\nWhy can you use the formula V = Bh to find the volume of a cylinder?
\nAnswer:
\nThe representation of the cylinder is:
\n\"\"
\nNow,
\nFrom the given figure,
\nWe know that,
\nThe two circles that are in the top and bottom positions are congruent
\nSo,
\nThe area for both the circles is also the same
\nNow,
\nWe know that,
\nVolume = Area \u00d7 Height
\nNow,
\nWe know that,
\nArea of the circle = \u03c0r\u00b2
\nHence,
\nThe volume of the cylinder (V) = \u03c0r\u00b2h<\/p>\n

Try It!<\/strong><\/p>\n

Lin is building a cylindrical planter with a base diameter of 15 inches. She has 5,000 cubic inches of soil to fill her planter. What is the height of the largest planter Lin can build? Use 3.14 for \u03c0, and round to the nearest inch.
\nAnswer:
\nIt is given that
\nLin is building a cylindrical planter with a base diameter of 15 inches. She has 5,000 cubic inches of soil to fill her planter.
\nSo,
\nFrom the given information,
\nThe volume of the cylindrical planter = 5,000 cubic inches
\nThe diameter of the cylindrical planter = 15 inches
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\n= \\(\\frac{\u03c0d\u00b2h}{4}\\)
\nSo,
\n5,000 = \\(\\frac{3.14 \u00d7 15 \u00d7 15 \u00d7 h}{4}\\)
\n5,000 = 176.625h
\nh = \\(\\frac{5,000}{176.625}\\)
\nh = 28.3 inches
\nHence, from the above,
\nWe can conclude that the height of the largest planter Lin can build is: 28.3 inches<\/p>\n

KEY CONCEPT<\/strong>
\nThe formula for the volume of a cylinder is the same as the formula for the volume of a prism. The formula for the volume of a cylinder is V= Bh, where B is the area of the circular base and h is the height of the cylinder.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How is the volume of a cylinder related to the volume of a rectangular prism?
\nAnswer:
\nRectangular prisms and cylinders\u00a0are somewhat\u00a0similar\u00a0because they both have two bases and a height.
\nThe formula for the volume of a rectangular solid is
\nV=Bh
\ncan also be used to find the volume of a cylinder
\nWhere,
\n“B” in the rectangular prism is the area of the rectangle
\n“B” in the cylinder is the area of the circle<\/p>\n

Question 2.
\nUse Structure What two measurements do you need to know to find the volume of a cylinder?
\nAnswer:
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nNow,
\nFrom the given formula,
\nWe can observe that
\n\u03c0 is a constant
\nHence, from the above,
\nWe can conclude that the two measurements that needed to be known are:
\na. Radius of the cylinder
\nb. The height of the cylinder<\/p>\n

Question 3.
\nReasoning<\/strong> Cylinder A has a greater radius than Cylinder B. Does Cylinder A necessarily have a greater volume than Cylinder B? Explain.
\nAnswer:
\nIt is given that
\nCylinder A has a greater radius than Cylinder B
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nNow,
\nFro the above,
\nWe can observe that
\nVolume (V) \u221d Radius\u00b2
\nSo,
\nWhen we increase the value of the radius, the value of the volume will automatically increase
\nHence, from the above,
\nWe can conclude that cylinder A has a greater volume than Cylinder B<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nWhat is the volume of the cylinder? Express your answer in terms of \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = Bh
\nWhere,
\nB = \u03c0r\u00b2
\nSo,
\nV = 4\u03c0 \u00d7 10
\nV = 40\u03c0 mm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder in terms of \u03c0 is: 40\u03c0 mm\u00b3<\/p>\n

Question 5.
\nWhat is the approximate height of the cylinder? Use 3.14 for \u03c0, and if necessary, round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe volume of a cylinder (V) = 314 ft\u00b3
\nThe radius of a cylinder = 10 ft
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n314 = 3.14 \u00d7 10 \u00d7 10 \u00d7 h
\nh = \\(\\frac{314}{3.14 \u00d7 10 \u00d7 10}\\)
\nh = 1 ft
\nHence, from the above,
\nWe can conclude that the height of the given cylinder is: 1 ft<\/p>\n

Question 6.
\nWhat is the volume of the cylinder? Use 3.14 for \u03c0, and if necessary, round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of a cylinder = 4 cm
\nThe circumference of a circle = 22.4 cm
\nNow,
\nWe know that,
\nThe circumference of a circle = 2\u03c0r
\nSo,
\n2\u03c0r = 22.4
\nr = \\(\\frac{22.4}{2 \u00d7 3.14}\\)
\nr = 3.56 cm
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = 3.14 \u00d7 3.56 \u00d7 3.56 \u00d7 4
\n= 159.2 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 159.2 cm\u00b3<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Question 7.
\nLeveled Practice What is the volume of a cylinder with a radius of 5 centimeters and height of 2.5 centimeters? Use 3.14 for \u03c0.
\nV = \u03c0 ________ 2<\/sup> \u2219 ________
\n= \u03c0 ______ \u2219 ________
\n= _______ \u03c0
\nThe volume of the cylinder is about _______ cubic centimeters.
\nAnswer:
\nIt is given that
\nThe radius of a cylinder (r) = 5 cm
\nThe height of a cylinder (h) = 2.5 cm
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = 3.14 \u00d7 5 \u00d7 5 \u00d7 2.5
\n= 196.25 cm
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 196.25 cm\u00b3 (or) 62.5\u03c0 cm\u00b3<\/p>\n

Question 8.
\nFind the volume of each cylinder in terms of. Which cylinder has the greater volume?
\nCylinder A: Area of Base = 61 ft2<\/sup>, height = 10 ft
\nCylinder B: Circumference = 6\u03c0 ft, height = 6 ft
\nAnswer:
\nThe given data is:
\nCylinder A: Area of Base = 61 ft2<\/sup>, height = 10 ft
\nCylinder B: Circumference = 6\u03c0 ft, height = 6 ft
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = Bh
\nWhere,
\nB is the area of the circle
\nNow,
\nFor Cylinder A:
\nV = 61 \u00d7 10
\n= 610 ft\u00b3
\nFor Cylinder B:
\nWe know that,
\nCircumference = 2\u03c0r
\nSo,
\n2\u03c0r = 6\u03c0
\nr = \\(\\frac{6\u03c0}{2\u03c0}\\)
\n= 3 ft
\nSo,
\nV = 3.14 \u00d7 3 \u00d7 3 \u00d7 6
\n= 169.56 ft\u00b3
\nHence, from the above,
\nWe can conclude that Cylinder A has the greatest volume when we compare the volumes of Cylinder A and Cylinder B<\/p>\n

Question 9.
\nThe volume of a cylinder is 2251 cubic inches, and the height of the cylinder is 1 inch. What is the radius of the cylinder?
\nAnswer:
\nIt is given that
\nThe volume of a cylinder is 2251 cubic inches, and the height of the cylinder is 1 inch
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n2251 = 3.14 \u00d7 r\u00b2 \u00d7 1
\nr\u00b2 = \\(\\frac{2251}{3.14}\\)
\nr\u00b2 = 716.87 in\u00b2
\nr = 26.77 in
\nHence, from the above,
\nWe can conclude that the radius of the given cylinder is: 26.77 in.<\/p>\n

Question 10.
\nA company is designing a new cylindrical water bottle. The volume of the bottle is 103 cubic centimeters. What is the radius of the water bottle? Estimate using 3.14 for \u03c0, and round to the nearest hundredth.
\n\"Envision
\nAnswer:
\nIt is given that
\nA company is designing a new cylindrical water bottle. The volume of the bottle is 103 cubic centimeters
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n103 = 3.14 \u00d7 r\u00b2 \u00d7 8.1
\nr\u00b2 = \\(\\frac{103}{3.14 \u00d7 8.1}\\)
\nr\u00b2 = 4.05cm\u00b2
\nr = 2.01 cm
\nHence, from the above,
\nWe can conclude that the radius of the given cylinder is: 2.01 cm<\/p>\n

Question 11.
\nUse the figure at the right.
\n\"Envision
\na. Find the volume of the cylinder in terms of \u03c0.
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe radius of the cylinder (r) = 4 in.
\nThe height of the cylinder (h) = 3 in
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = \u03c0 \u00d7 4 \u00d7 4 \u00d7 3
\nV = 48\u03c0 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder in terms of \u03c0 is: 48\u03c0 in.\u00b3<\/p>\n

b. Is the volume of a cylinder, which has the same radius but twice the height, greater or less than the original cylinder? Explain.
\nAnswer:
\nFrom part (a),
\nThe radius of the cylinder (r) = 4 in.
\nThe height of the cylinder (h) = 3 in
\nSo,
\nNow,
\nFor part (b),
\nThe radius of the cylinder (r) = 4 in.
\nThe height of the cylinder (h) = 6 in.
\nSo,
\nV = \u03c0 \u00d7 4 \u00d7 4 \u00d7 6
\n= 96\u03c0 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the cylinder we obtained in part (b) is greater than the volume of the cylinder we obtained in part (a)<\/p>\n

Question 12.
\nReasoning<\/strong> A rectangular piece of cardboard with dimensions 6 inches by 8 inches is used to make the curved side of a cylinder-shaped container. Using this cardboard, what is the greatest volume the cylinder can hold? Explain.
\nAnswer:
\nIt is given that
\nA rectangular piece of cardboard with dimensions 6 inches by 8 inches is used to make the curved side of a cylinder-shaped container.
\nNow,
\nLet the height of the cylinder be: 6 inches (or) 8 inches
\nNow,
\nWe know that,
\nThe volume of a rectangular prism (V) = Length \u00d7 Width \u00d7 Height
\nSo,
\nFor h = 6 inches:
\nV = 6 \u00d7 8 \u00d7 6
\n= 288 cubic inches
\nFor h = 8 inches:
\nV = 6 \u00d7 8 \u00d7 8
\n= 384 cubic inches
\nHence, from the above,
\nWe can conclude that the greatest volume the given cylinder can hold is: 384 cubic inches<\/p>\n

Question 13.
\nThe cylinder shown has a volume of 885 cubic inches.
\n\"Envision
\na. What is the radius of the cylinder? Use 3.14 for \u03c0.
\nAnswer:
\nIt is given that
\nThe cylinder shown has a volume of 885 cubic inches.
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of a cylinder (h) = 11.7 in.
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n885 = 3.14 \u00d7 r\u00b2 \u00d7 11.7
\nr\u00b2 = \\(\\frac{885}{3.14 \u00d7 11.7}\\)
\nr\u00b2 = 24.08
\nr = 4.9 in.
\nHence, from the above,
\nWe can conclude that the radius of the given cylinder is: 4.9 in.<\/p>\n

b. Reasoning If the height of the cylinder is changed, but the volume stays the same, then how will the radius change? Explain.
\nAnswer:
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nNow,
\nIt is given that
\nVolume —–> Constant
\nHeight —– > Changed
\nNow,
\nFrom the given formula,
\nIf the volume is constant, then
\nh \u221d \\(\\frac{1}{r\u00b2}\\)
\nHence, from the above relation,
\nWe can conclude that
\nIf the value of height increases, then the value of radius decreases
\nIf the value of height decreases, then the value of radius increases<\/p>\n

Question 14.
\nToy rubber balls are packaged in a cylinder that holds 3 balls. Find the volume of the cylinder. Use 3.14 for \u03c0, and round to the nearest tenth.
\n\"Envision
\nAnswer:
\nIt is given that
\nToy rubber balls are packaged in a cylinder that holds 3 balls
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cylinder (h) = 20.7 cm
\nThe diameter of the cylinder (d) = 6.9 cm
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\n= \\(\\frac{\u03c0d\u00b2h}{4}\\)
\nSo,
\nV = \\(\\frac{3.14 \u00d7 20.7 \u00d7 20.7 \u00d7 6.9}{4}\\)
\n= 2,321 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 2,321 cm\u00b3<\/p>\n

Question 15.
\nHigher-Order Thinking<\/strong>
\nAn insulated collar is made to cover a pipe. Find the volume of the material used to make the collar. Let r = 3 inches, R= 5 inches, and h = 21 inches. Use 3.14 for \u03c0, and round to the nearest hundredth.
\n\"Envision
\nAnswer:
\nIt is given that
\nAn insulated collar is made to cover a pipe.
\nNow,
\nThe given data is:
\nr = 3 inches, R= 5 inches, and h = 21 inches
\nNow,
\nThe radius of insulated collar = R – r
\n= 5 – 3
\n= 2 inches
\nNow,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nThe volume of the material that is used to make the collar (V) = 3.14 \u00d7 2 \u00d7 2 \u00d7 21
\n= 263.76 cubic inches
\nHence, from the above,
\nWe can conclude that the volume of the material that is used to make the collar is: 263.76 cubic inches<\/p>\n

Assessment Practice<\/strong>
\nQuestion 16.
\nThe volume of a cylinder is 1,029\u03c0 cubic centimeters. The height of the cylinder is 21 centimeters. What is the radius, to the nearest centimeter, of the cylinder?
\nAnswer:
\nIt is given that
\nThe volume of a cylinder is 1,0291 cubic centimeters. The height of the cylinder is 21 centimeters.
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n1,029\u03c0 = \u03c0r\u00b2 \u00d7 21
\nr\u00b2 = \\(\\frac{1,029}{21}\\)
\nr\u00b2 = 49
\nr = 7 cm
\nHence, from the above,
\nWe can conclude that the radius of the given cylinder is: 7 cm<\/p>\n

Question 17.
\nThe diameter of a cylinder is 7 yards. The height is 12 yards. What is the volume, in terms of \u03c0 and to the nearest cubic yard, of the cylinder?
\nAnswer:
\nIt is given that
\nThe diameter of a cylinder is 7 yards. The height is 12 yards
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\n= \\(\\frac{\u03c0d\u00b2h}{4}\\)
\nSo,
\nV = \\(\\frac{3.14 \u00d7 7 \u00d7 12 \u00d7 7}{4}\\)
\n= 1,846.32 yards\u00b3 (or) 588\u03c0 yards\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 1,846.32 cubic yards (or) 588\u03c0 cubic yards<\/p>\n

Question 18.
\nA cylinder is shown. What statements about the cylinder are true?
\n\"Envision
\n\u2610 The radius of the cylinder is 2 ft.
\n\u2610 The diameter of the cylinder is 4 yd.
\n\u2610 The height of the cylinder is 8 in.
\n\u2610 The volume of the cylinder is 32 in.2<\/sup>.
\n\u2610 The volume of the cylinder is 32\u03c0 in.3<\/sup>.
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe diameter of the cylinder = 4 in.
\nThe height of the cylinder = 8 in.
\nSo,
\nThe radius of the cylinder = \\(\\frac{Diameter}{2}\\)
\n= 2 in.
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = 3.14 \u00d7 2 \u00d7 2 \u00d7 8
\n= 32\u03c0 in.\u00b3
\n= 100.48 in.\u00b3
\nHence, from the above,
\nWe can conclude that the statements that are true about the given cylinder are:
\n\"\"<\/p>\n

Topic 8 MID-TOPIC CHECKPOINT<\/h3>\n

Question 1.
\nVocabulary Select all the statements that describe surface area and volume. Lessons 8-1 and 8-2
\n\u2610 Surface area is the sum of the areas of all the surfaces of a figure.
\n\u2610 Volume is the distance around a figure.
\n\u2610 The surface area is a three-dimensional measure.
\n\u2610 Volume is the amount of space a figure occupies.
\n\u2610 Volume is a three-dimensional measure.
\nAnswer:
\nThe above statements that describe the surface area and volume are:
\n\"\"<\/p>\n

In 2-4, use the figure at the right. Sallie packed a cone-shaped cup inside of a cylindrical package.
\n\"Envision
\nQuestion 2.
\nThe cone-shaped cup is made out of paper. How much paper was used to make the cup, excluding the opening at the top of the cup? Use 3.14 for \u03c0, and round to the nearest tenth. Lesson 8-1
\nAnswer:
\nIt is given that
\nSallie packed a cone-shaped cup inside of a cylindrical package and the cone-shaped cup is made out of paper
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe slant height of cone (l) is: 33 cm
\nThe diameter of the cone (d) is: 20 cm
\nSo,
\nThe radius of cone (r) = \\(\\frac{Diameter}{2}\\)
\n= 10 cm
\nNow,
\nWe know that,
\nThe surface area of a cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 10 \u00d7 10 + 3.14 \u00d7 10 \u00d7 33
\n= 314 + 1,036.2
\n= 1,350.2 cm\u00b2
\nHence, from the above,
\nWe can conclude that
\nWe have to use 1,350.2 cm\u00b2 of paper was used to make the cup, excluding the opening at the top of the cup<\/p>\n

Question 3.
\nThe cylindrical package is made out of cardboard. In terms of \u03c0, how much cardboard was used to make the package? Lesson 8-1
\nAnswer:
\nIt is given that the cylindrical package is made out of cardboard
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cylinder (h) is: 33 cm
\nThe diameter of the cylinder (d) is: 20 cm
\nSo,
\nRadius = \\(\\frac{Diameter}{2}\\)
\n= \\(\\frac{20}{2}\\)
\n= 10 cm
\nNow,
\nWe know that,
\nThe surface area of a cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2\u03c0 \u00d7 10\u00b2 + 2\u03c0 \u00d7 10 \u00d7 33
\n= 200\u03c0 + 660\u03c0
\n= 860\u03c0 cm\u00b2
\nHence, from the above,
\nWe can conclude that 860\u03c0 cm\u00b2 of cardboard was used to make the package<\/p>\n

Question 4.
\nHow much space does the package occupy in terms of \u03c0? Lesson 8-1
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the above,
\nWe can observe that the given figure is a cylinder
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cylinder (h) is: 33 cm
\nThe diameter of the cylinder (d) is: 20 cm
\nSo,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{20}{2}\\)
\nr = 10 cm
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = \u03c0 \u00d7 10\u00b2 \u00d7 33
\n= 3,300\u03c0 cm\u00b3
\nHence, from the above,
\nWe can conclude that 3,300\u03c0 cm\u00b3 of space does the package occupy in terms of \u03c0<\/p>\n

Question 5.
\nWhat is the surface area of the sphere in terms of \u03c0? Lesson 8-1
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe radius of the sphere (r) is: 3 ft
\nNow,
\nWe know that,
\nThe surface area of a sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\nS.A = 4\u03c0 \u00d7 3\u00b2
\n= 36\u03c0 ft\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given sphere is: 36\u03c0 ft\u00b2<\/p>\n

Question 6.
\nThe volume of the cylinder is 400\u03c0 cm3<\/sup>. What is the height of the cylinder? Lesson 8-2
\nA. 5 cm
\nB. 16 cm
\nC. 25 cm
\nD. 80 cm
\nAnswer:
\nIt is given that
\nThe volume of the cylinder is 400\u03c0 cm\u00b3
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe diameter of the cylinder (d) is: 10 cm
\nSo,
\nRadius = \\(\\frac{Diameter}{2}\\)
\n= \\(\\frac{10}{2}\\)
\n= 5 cm
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n400\u03c0 = \u03c0 \u00d7 5\u00b2 \u00d7 h
\nh = \\(\\frac{400}{25}\\)
\nh = 16 cm
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 16 cm<\/p>\n

Topic 8 MID-TOPIC PERFORMANCE TASK<\/h3>\n

Melissa designed a sculpture in which a cylinder-shaped section was removed from a cube.
\n\"Envision<\/p>\n

PART A
\nBefore painting the surface of the sculpture, Melissa wants to sand the surface where the cylinder section was removed. What is the surface area of the section she will sand? Use 3.14 for \u03c0. Explain how you found the surface area.
\nAnswer:
\nIt is given that
\nMelissa designed a sculpture in which a cylinder-shaped section was removed from a cube.
\nNow,
\nThe given figure is:
\n\"EnvisionNow,
\nFrom the given figure,
\nThe diameter of the cylinder (d) is: 5 cm
\nThe height of the cylinder (d) is: 10 cm
\nSo,
\nThe radius of the cylinder (r) = \\(\\frac{Diameter of the cylinder}{2}\\)
\nr = 2.5 cm
\nNow,
\nWe know that,
\nWe know that,
\nThe surface area of a cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2 \u00d7 3.14 \u00d7 2.5 \u00d7 2.5 + 2 \u00d7 3.14 \u00d7 2.5 \u00d7 10
\n= 39.25 + 157
\n= 196.25 cm\u00b2
\nHence, from the above,
\nWe can conclude that 196.25 cm\u00b2\u00a0 is the surface area of the section Melissa will sand<\/p>\n

PART B
\nMelissa has a can of spray paint that covers about 6,500 square centimeters. Can Melissa apply two coats of paint to the entire sculpture? Explain. Use 3.14 for \u03c0.
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the above,
\nWe can observe that
\nThe scripture is the combination of the cube and the cylinder
\nNow,
\nWe know that,
\nThe surface area of a cube (S.A) = 6a\u00b2
\nWhere,
\na is the side of the cube
\nSo,
\nS.A of the cube = 6 \u00d7 10\u00b2
\n= 600 cm\u00b2
\nNow,
\nFrom part (a),
\nThe S.A of the cylinder = 196.25 cm\u00b2
\nSo,
\nThe S.A of the scripture = 600 + 196.25
\n= 796.25 cm\u00b2
\nSo,
\n6,500 > 796.25 \u00d7 2
\nHence, from the above,
\nWe can conclude that Melissa can apply two coats of paint to the entire sculpture<\/p>\n

PART C
\nWhat is the volume of the sculpture? Use 3.14 for \u03c0.
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the above,
\nWe can observe that
\nThe scripture is the combination of the cube and the cylinder
\nNow,
\nThe diameter of the cylinder (d) is: 5 cm
\nThe height of the cylinder (h) is: 10 cm
\nSo,
\nThe radius of the cylinder (r) is: 2.5 cm
\nNow,
\nWe know that,
\nThe volume of the cylinder (V) = \u03c0r\u00b2h
\nThe volume of the cube (V) = a\u00b3
\nWhere,
\na is the side of the cube
\nSo,
\nThe volume of the scripture (V) = \u03c0r\u00b2h + a\u00b3
\nSo,
\nV = 3.14 \u00d7 2.5 \u00d7 2.5 \u00d7 10 + 10\u00b3
\n= 196.25 + 1,000
\n= 1,196.25 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the scripture is: 1,196.25 cm\u00b3<\/p>\n

Lesson 8.3 Find Volume of Cones<\/h3>\n

Solve & Discuss It!<\/strong>
\nA landscape architect uses molds for casting rectangular pyramids and rectangular prisms to make garden statues. He plans to place each finished pyramid on top of a prism. If one batch of concrete mix makes one prism or three pyramids, how does the volume of one pyramid compare to the volume of one prism? Explain.
\n\"Envision
\nI can… find the volume of cones.
\nAnswer:
\nIt is given that
\nA landscape architect uses molds for casting rectangular pyramids and rectangular prisms to make garden statues. He plans to place each finished pyramid on top of a prism and one batch of concrete mix makes one prism or three pyramids
\nNow,
\nWe know that,
\nThe prism is in the shape of the cylinder
\nThe pyramid is in the shape of the cone
\nNow,
\nWe know that,
\nThe volume of a prism (V) = \u03c0r\u00b2h
\nThe volume of a pyramid (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nThe volume of a prism = 3 \u00d7 The volume of a pyramid
\nHence, from the above,
\nWe can conclude that
\nThe volume of a prism is 3 times the volume of a pyramid<\/p>\n

Look for Relationships
\nWhat do you notice about the dimensions of the bases of the pyramid and prism? How are the heights of the two solids related?
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\na. The dimensions of the bases of the prism and the pyramid are the same
\nb. The heights of the pyramid and the prism are the same<\/p>\n

Focus on math practices
\nMake Sense and Persevere If the architect mixes 10 batches of concrete, how many sculptures combining 1 prism and 1 pyramid could he make? Explain.
\nAnswer:
\nIt is given that
\none batch of concrete mix makes one prism or three pyramids
\nSo,
\nThe ratio of the mix of prism and pyramid in 1 batch is: 1 : 3
\nHence,
\nFor 10 batches of the mix of concrete,
\nWe can make sculptures combining 1 prism and 1pyramid are: 10 prisms and 30 pyramids<\/p>\n

Essential Question<\/strong>
\nHow is the volume of a cone related to the volume of a cylinder?
\nAnswer:
\nThe volumes of a cone and a cylinder are related in the same way as the volumes of a pyramid and a prism is related. If the heights of a cone and a cylinder are equal, then the volume of the cylinder is 3 times as much as the volume of a cone<\/p>\n

Try It!<\/strong>
\nFind the volume of the cone. Use 3.14 for \u03c0.
\n\"Envision
\nThe volume of the cone is about ________ cubic inches.
\nV = ____ \u03c0r\u00b2h
\n\u2248 _______(3.14) (______)2<\/sup> (4)
\n= ________ (3.14) (______)(4)
\n= _________
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the above figure,
\nWe can observe that
\nThe radius of the cone (r) is 1.5 in.
\nThe height of the cone (h) is 4 in.
\nNow,
\nWe know that,
\nThe volume of the cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 (1.5)\u00b2 \u00d7 4
\n= 9.42 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 9.42 in.\u00b3<\/p>\n

Convince Me!
\nIf you know the volume of a cone, how can you find the volume of a cylinder that has the same height and radius as the cone?
\nAnswer:
\nWe know that,
\nIf the cylinder and the cone has the same height and radius, then
\nThe volume of cone = \\(\\frac{1}{3}\\) \u00d7 The volume of cylinder<\/p>\n

Try It!<\/strong><\/p>\n

Find the volume of each cone.
\na. Use \\(\\frac{22}{7}\\) for \u03c0. Express the answer as a fraction.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe slant height (l) is: 5 mm
\nThe height of the cone (h) is: 3 mm
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2+ h\u00b2
\nSo,
\n5\u00b2 = r\u00b2 + 3\u00b2
\nr\u00b2 = 25 – 9
\nr\u00b2 = 16
\nr = 4 mm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 4\u00b2 \u00d7 3
\n= \\(\\frac{1,056}{21}\\) mm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone in the fraction form is: \\(\\frac{1,056}{66}\\) mm\u00b3<\/p>\n

b. Express the volume in terms of \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nWe can observe that
\nThe height of the cone (h) is: 21 ft
\nThe circumference of the circle (C) is: 16\u03c0 ft
\nNow,
\nWe know that,
\nThe circumference of the circle (C) = 2\u03c0r
\nSo,
\n2\u03c0r = 16\u03c0
\nr = 8 ft
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \u03c0 \u00d7 8\u00b2 \u00d7 21
\n= 448\u03c0 ft\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone in terms of \u03c0 is: 448\u03c0 ft\u00b3<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

The volume of a cone is \\(\\frac{1}{3}\\) the volume of a cylinder with the same base and height. The formula for the volume of a cone is V = \\(\\frac{1}{3}\\)Bh, where B is the area of the base and his the height of the cone.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How is the volume of a cone related to the volume of a cylinder?
\nAnswer:
\nThe volumes of a cone and a cylinder are related in the same way as the volumes of a pyramid and a prism is related. If the heights of a cone and a cylinder are equal, then the volume of the cylinder is 3 times as much as the volume of a cone<\/p>\n

Question 2.
\nUse Structure What dimensions do you need to find the volume of a cone?
\nAnswer:
\nTo find the volume of a cone,
\nThe dimensions we need are:
\na. The radius of the cone
\nb. The height of the cone<\/p>\n

Question 3.
\nLook for Relationships If you know a cone’s radius and slant height, what must you do before you can find its volume?
\nAnswer:
\nIf you know a cone’s radius (r) and slant height(l), then
\nThe volume of a cone (V) is given as:
\nV = \\(\\frac{1}{3}\\)\u03c0r\u00b2 \u00d7 \\(\\sqrt{l\u00b2 – r\u00b2}\\)<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nWanda found a cone-shaped seashell on the beach. The shell has a height of 63 millimeters and a base radius of 8 millimeters. What is the volume of the seashell? Estimate 63 mm using \\(\\frac{22}{7}\\) for \u03c0.
\n\"Envision
\nAnswer:
\nIt is given that
\nWanda found a cone-shaped seashell on the beach. The shell has a height of 63 millimeters and a base radius of 8 millimeters
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 8\u00b2 \u00d7 63
\n= 4,224 mm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the seashell is: 4,224 mm\u00b3<\/p>\n

Question 5.
\nWhat is the volume of the cone? Estimate using 3.14 for \u03c0, and round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cone (h) is: 40mm
\nThe slant height of the cone (l) is: 41 mm
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2 + h\u00b2
\nSo,
\n41\u00b2 = r\u00b2+ 40\u00b2
\nr\u00b2 = 81
\nr = 9 mm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 9\u00b2 \u00d7 40
\n= 3,391.2 mm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 3,391.2 mm\u00b3<\/p>\n

Question 6.
\nWhat is the volume of the cone in terms of \u03c0 if the circumference of the base is 1.4\u03c0 feet?
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cone (h) is: 2.7 ft
\nNow,
\nWe know that,
\nCircumference (C) = 2\u03c0r
\nSo,
\n2\u03c0r = 1.4\u03c0
\nr = 0.7 ft
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \u03c0 \u00d7 (0.7)\u00b2 \u00d7 2.7
\n= 0.441\u03c0 ft\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 0.441\u03c0 ft\u00b3<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Multimedia Leveled Practice In 7 and 8, find the volumes of the cones.
\nQuestion 7.
\nWhat is the volume of the cone? Write your answer in terms of \u03c0.
\n\"Envision
\nAnswer:
\nFrom the given figure,
\nWe can observe that
\nThe radius of the cone (r) is: 3 cm
\nThe height of the cone (h) is: 4 cm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \u03c0 \u00d7 3\u00b2 \u00d7 4
\n= 12\u03c0 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 12\u03c0 cm\u00b3<\/p>\n

Question 8.
\nWhat is the volume of the cone to the nearest hundredth? Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nFrom the given figure,
\nWe can observe that
\nThe radius of the cone (r) is: 16 units
\nThe height of the cone (h) is: 36 units
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 16\u00b2 \u00d7 36
\n= 9,646.08 units\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 9,646.08 units\u00b3<\/p>\n

Question 9.
\nIf a cone-shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet, what is the volume of the hole? Use \\(\\frac{22}{7}\\) for \u03c0.
\nAnswer:
\nIt is given that
\nA cone-shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet
\nNow,
\nWe know that,
\nCircumference (C) = 2\u03c0r
\nSo,
\n2\u03c0r = 44
\nr = \\(\\frac{44}{2\u03c0}\\)
\nr = 7 feet
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 7\u00b2 \u00d7 3
\n= 154 feet\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone-shaped hole is: 154 feet\u00b3<\/p>\n

Question 10.
\nThe volume of the cone is 462 cubic yards. What is the radius of the cone? Use \\(\\frac{22}{7}\\) for \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cone (h) is: 9 yd
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\n462 = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 r\u00b2 \u00d7 9
\nr\u00b2 = \\(\\frac{462}{9.428}\\)
\nr = 7 yd
\nHence, from the above,
\nWe can conclude that the radius of the given cone is: 7 yd<\/p>\n

Question 11.
\nA city engineer determines that 5,500 cubic meters of sand will be needed to combat erosion at the city’s beach. Does the city have enough sand to combat erosion? Use \\(\\frac{22}{7}\\) for \u03c0. Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\nA city engineer determines that 5,500 cubic meters of sand will be needed to combat erosion at the city’s beach.
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe slant height of the cone(l) is: 37 m
\nThe height of the cone (h) is: 35 m
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2+ h\u00b2
\n37\u00b2 = r\u00b2 + 35\u00b2
\nr\u00b2 = 37\u00b2 – 35\u00b2
\nr = 12 m
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 12\u00b2 \u00d7 35
\n= 5,280 m\u00b3
\nSo,
\n5,500 > 5,280
\nHence, from the above,
\nWe can conclude that the city has enough sand to combat erosion<\/p>\n

Question 12.
\nA water tank is shaped like the cone shown.
\n\"Envision
\na. How much water can the tank hold? Use 3.14 for \u03c0, and round to the nearest tenth.
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe slant height of the cone (l) is: 61 ft
\nThe height of the cone (h) is: 60 ft
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2+ h\u00b2
\n61\u00b2 = r\u00b2 + 60\u00b2
\nr\u00b2 = 61\u00b2 – 60\u00b2
\nr = 11 ft
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 11\u00b2 \u00d7 60
\n= 7,598.8 ft\u00b3
\nHence, from the above,
\nWe can conclude that the amount of water that the tank can hold is: 7,598.8 ft\u00b3<\/p>\n

b. If water is drained from the tank to fill smaller tanks that each hold 500 cubic feet of water, how many smaller tanks can be filled?
\nAnswer:
\nIt is given that the water is drained from the tank to fill smaller tanks that each holds 500 cubic feet of water
\nNow,
\nFrom part (a),
\nWe know that,
\nThe amount of water that the tank can hold is: 7,598.8 ft\u00b3
\nNow,
\nThe number of smaller tanks that can be filled = \\(\\frac{The amount of water that the tank can hold}{The amount of water that the smaller tank can hold}\\)
\n= \\(\\frac{7,598.8}{500}\\)
\n\u2245 15
\nHence, from the above,
\nWe can conclude that the number of smaller tanks that can be filled is: 15<\/p>\n

Question 13.
\nAn ice cream cone is filled exactly level with the top of a cone. The cone has a 9-centimeter depth and a base with a circumference of 91 centimeters. How much ice cream is in the cone in terms of \u03c0?
\nAnswer:
\nIt is given that
\nAn ice cream cone is filled exactly level with the top of a cone. The cone has a 9-centimeter depth and a base with a circumference of 91 centimeters
\nNow,
\nWe know that,
\nCircumference (C) = 2\u03c0r
\nSo,
\n2\u03c0r = 91
\nr = \\(\\frac{91}{2\u03c0}\\)
\nr = 14.49 cm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \u03c0 \u00d7 (14.49)\u00b2 \u00d7 9
\n= 629.88\u03c0 cm\u00b3
\nHence, from the above,
\nWe can conclude that the amount of ice cream that is in the cone in terms of \u03c0 is: 629.88\u03c0 cm\u00b3<\/p>\n

Question 14.
\nIn the scale model of a park, small green cones represent trees. What is the volume of one green cone? Use \\(\\frac{22}{7}\\) for \u03c0.
\n\"Envision
\nAnswer:
\nThe given green cone is:
\n\"Envision
\nNow,
\nFrom the given green cone,
\nWe can observe that
\nThe slant height of the cone (l) is: 65 mm
\nThe height of the cone (h) is: 63 mm
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2+ h\u00b2
\n65\u00b2 = r\u00b2 + 603\u00b2
\nr\u00b2 = 65\u00b2 – 63\u00b2
\nr = 16 mm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 16\u00b2 \u00d7 63
\n= 16,896 mm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of one green cone is: 16,896 mm\u00b3<\/p>\n

Question 15.
\nReasoning<\/strong> Compare the volumes of two cones. One has a radius of 5 feet and a slant height of 13 feet. The other one has a height of 5 feet and a slant height of 13 feet.
\na. Which cone has the greater volume?
\nAnswer:
\nThe given data is:
\nCone 1: Radius: 5 feet\u00a0 \u00a0 \u00a0 \u00a0Slant height: 13 feet
\nCone 2: Height: 5 feet\u00a0 \u00a0 \u00a0 \u00a0Slant height: 13 feet
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2+ h\u00b2
\nSo,
\nFor cone 1:
\n13\u00b2 = h\u00b2 + 5\u00b2
\nh\u00b2 = 13\u00b2 – 5\u00b2
\nh = 12 feet
\nFor cone 2:
\n13\u00b2 = r\u00b2 + 5\u00b2
\nr\u00b2 = 13\u00b2 – 5\u00b2
\nr = 12 feet
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nFor Cone 1:
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 5\u00b2 \u00d7 12
\n= 314.28 feet\u00b3
\nFor Cone 2:
\nV = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 12\u00b2 \u00d7 5
\n= 754.28 feet\u00b3
\nHence, from the above,
\nWe can conclude that Cone 2 has the greater volume<\/p>\n

b. What is the volume of the larger cone in terms of \u03c0?
\nAnswer:
\nFrom part (a),
\nWe can observe that,
\nCone 2 has a greater volume
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 \u03c0 \u00d7 12\u00b2 \u00d7 5
\n= 240\u03c0 feet\u00b3<\/p>\n

Question 16.
\nAn artist makes a cone-shaped sculpture for an art exhibit. If the sculpture is 7 feet tall and has a base with a circumference of 24.492 feet, what is the volume of the sculpture? Use 3.14 for \u03c0, and round to the nearest hundredth.
\nAnswer:
\nIt is given that
\nAn artist makes a cone-shaped sculpture for an art exhibit and the sculpture is 7 feet tall and has a base with a circumference of 24.492 feet
\nNow,
\nWe know that,
\nCircumference (C) = 2\u03c0r
\nSo,
\n2\u03c0r = 24.492
\nr = \\(\\frac{24.492}{2\u03c0}\\)
\nr = 3.9 feet
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 3.9\u00b2 \u00d7 7
\n= 111.43 feet\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone-shaped sculpture is: 111.43 feet\u00b3<\/p>\n

Question 17.
\nHigher-Order Thinking<\/strong> A cone has a radius of 3 and a height of 11.
\na. Suppose the radius is increased by 4 times its original measure. How many times greater is the volume of the larger cone than the smaller cone?
\nAnswer:
\nIt is given that
\nA cone has a radius of 3 and a height of 11
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 3\u00b2 \u00d7 11
\n= 103.62 units\u00b3
\nNow,
\nIf the radius is increased by 4 times of its original measure, then
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 (3 \u00d7 4)\u00b2 \u00d7 11
\n= 1,657.92 units\u00b3
\nNow,
\nThe number of times the volume of the larger cone than the smaller cone = \\(\\frac{1,657.92}{103.62}\\)
\n= 16
\nHence, from the above,
\nWe can conclude that the volume of the larger cone is 16 times greater than the volume of the smaller cone<\/p>\n

b. How would the volume of the cone change if the radius were divided by four?
\nAnswer:
\nFrom part (a),
\nWe know that,
\nThe volume of the cone with a radius of 3 and a height of 11 units is: 103.62 units\u00b3
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 (3 \u00f7 4)\u00b2 \u00d7 11
\n= 6.476 units\u00b3
\nNow,
\nThe ratio of the volume of the changed cone to the volume of the original cone = \\(\\frac{6.476}{103.62}\\)
\n= 0.06
\nHence, from the above,
\nWe can conclude that the volume of the cone will be 0.06 times of the original cone<\/p>\n

Assessment Practice<\/strong>
\nQuestion 18.
\nList the cones described below in order from least volume to greatest volume.
\n\u2022 Cone 1: radius 6 cm and height 12 cm
\n\u2022 Cone 2: radius 12 cm and height 6 cm
\n\u2022 Cone 3: radius 9 cm and height 8 cm
\nA. Cone 2, Cone 3, Cone 1
\nB. Cone 1, Cone 3, Cone 2
\nC. Cone 2, Cone 1, Cone 3
\nD. Cone 1, Cone 2, Cone 3
\nAnswer:
\nThe given data is:
\na. Cone 1: radius 6 cm and height 12 cm
\nb. Cone 2: radius 12 cm and height 6 cm
\nc.\u00a0 Cone 3: radius 9 cm and height 8 cm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nFor Cone 1:
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 6\u00b2 \u00d7 12
\n= 452.16 cm\u00b3
\nFor Cone 2:
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 12\u00b2 \u00d7 6
\n= 904.32 cm\u00b3
\nFor Cone 3:
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 9\u00b2 \u00d7 8
\n= 678.24 cm\u00b3
\nHence, from the above,
\nWe can conclude that the order of the volumes from the least to the greatest is:
\nCone 1 < Cone 3 < Cone 2<\/p>\n

Question 19.
\nWhat is the volume, in cubic inches, of a cone that has a radius of 8 inches and a height of 12 inches? Use 3.14 for \u03c0, and round to the nearest hundredth.
\nAnswer:
\nIt is given that
\nA cone has a radius of 8 inches and a height of 12 inches
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 8\u00b2 \u00d7 12
\n= 803.84 inches\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 803.84 inches\u00b3<\/p>\n

Lesson 8.4 Find Volume of Spheres<\/h3>\n

Explore It!<\/strong>
\nMarshall uses the beaker to fill the bowl with water.
\n\"Envision
\nI can… find the volume of a sphere and use it to solve problems.<\/p>\n

A. Draw and label three-dimensional figures to represent the beaker and the bowl.
\nAnswer:
\nThe representation of the three-dimensional figures that represent the beaker and bowl are:
\n\"\"<\/p>\n

B. Marshall has to fill the beaker twice to completely fill the bowl with water. How can you use an equation to represent the volume of the bowl?
\nAnswer:
\nThe given figures are:
\n\"Envision
\nNow,
\nIt is given that
\nMarshall has to fill the beaker twice to completely fill the bowl with water
\nNow,
\nFrom part (a),
\nWe can observe that the beaker is in the form of a cone
\nNow,
\nWe know that,
\nThe volume of a cone = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nNow,
\nAccording to the given situation,
\nThe volume of the bowl(V’) = 2 \u00d7 The volume of the beaker(V)
\nV’ = 2 \u00d7 \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nNow,
\nFrom the given figures,
\nWe can observe that
\nThe height of a bowl is 2 times its radius
\nSo,
\nV’ = \\(\\frac{2}{3}\\)\u03c0r\u00b2 (2r)
\nV’ = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the bowl is: \\(\\frac{4}{3}\\)\u03c0r\u00b3<\/p>\n

Focus on math practices
\nReasoning How is the volume of a sphere and the volume of a cone related? What must be true about the radius and height measurements for these relationships to be valid?
\nAnswer:
\nWe know that,
\nThe relationship between the volume of the sphere and the volume of the cone is:
\nThe volume of the sphere = 2 \u00d7 The volume of the cone
\nNow,
\nFor the above relationship to be valid,
\na. The heights of the sphere and the cone must be the same
\nb. The radius of the sphere and the cone must be the same<\/p>\n

Essential Question<\/strong>
\nHow is the volume of a sphere related to the volume of a cone?
\nAnswer:
\nThe volume of a sphere is twice the volume of a cone that has the same circular base and height. i.e.,
\nThe volume of the sphere = 2 \u00d7 The volume of the cone<\/p>\n

Try It!<\/strong><\/p>\n

What is the volume of a ball with a diameter of 6 centimeters? Use 3.14 for \u03c0.
\nV = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>
\n= \\(\\frac{4}{3}\\) \u03c0 _______3<\/sup>
\n\u2248 ________ \u2219 ________
\n= __________
\nThe volume of the ball is about _______ cm3<\/sup>
\nAnswer:
\nIt is given that
\nThe diameter of a ball is: 6 cm
\nNow,
\nWe know that,
\nThe ball is in the form of a sphere
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{6}{2}\\)
\nr = 3 cm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 3\u00b3
\n= 113.04 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given ball is about 113.04 cm\u00b3<\/p>\n

Convince Me!
\nHow is the volume of a sphere related to the volume of a cone that has the same circular base and height?
\nAnswer:
\nThe relationship between the volume of a cone and the volume of a sphere that has the same circular base and height is:
\nThe volume of the sphere = 2 \u00d7 The volume of the cone<\/p>\n

Try It!<\/strong><\/p>\n

What is the volume of the composite figure shown? Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nThe given composite figure is:
\n\"Envision
\nNow,
\nWe know that,
\nA composite figure is made up of 2 or more two-dimensional figures
\nNow,
\nThe given composite figure is made up of Hemisphere and Cone
\nNow,
\nWe know that,
\nThe volume of the given composite figure = The volume of the Hemisphere + The volume of the cone
\nNow,
\nFrom the given composite figure,
\nWe can observe that
\nRadius (r) = 2 in.
\nHeight (h) = 6 in.
\nNow,
\nWe know that,
\nThe volume of the Hemisphere (V) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{2}{3}\\) \u00d7 3.14 \u00d7 2\u00b3
\n= 16.74 in.\u00b3
\nNow,
\nWe know that,
\nThe volume of the cone (V’) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV’ = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 2\u00b2 \u00d7 6
\n= 25.12 in.\u00b3
\nSo,
\nThe volume of the given composite figure = 16.74 + 25.12
\n= 41.84 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given composite figure is: 41.84 in.\u00b3<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

The volume of a sphere is twice the volume of a cone that has the same circular base and height. The formula for the volume of a sphere with radius r is V = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How is the volume of a sphere related to the volume of a cone?
\nAnswer:
\nThe volume of a sphere is twice the volume of a cone that has the same circular base and height. i.e.,
\nThe volume of the sphere = 2 \u00d7 The volume of the cone<\/p>\n

Question 2.
\nCritique Reasoning Kristy incorrectly says that the volume of the sphere below is 144\u03c0 cubic units. What mistake might Kristy have made?
\n\"Envision
\nAnswer:
\nThe given sphere is:
\n\"Envision
\nNow,
\nIt is given that
\nKristy incorrectly says that the volume of the sphere below is 144\u03c0 cubic units
\nNow,
\nFrom the given sphere,
\nWe can observe that
\nRadius (r) = 6 units
\nNow,
\nWe know that,
\nThe volume of a spher (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 \u03c0 \u00d7 6\u00b3
\n= 288\u03c0 units\u00b3
\nHence, from the above,
\nWe can conclude that the mistake done by Kristy is:
\nConsidering the sphere as the hemisphere and calculated the volume of the hemisphere instead of the sphere<\/p>\n

Question 3.
\nGeneralize Mehnaj has a set of blocks that are all the same height. The cone-shaped block has a volume of 125 cubic inches. The sphere-shaped block has a volume of 250 cubic inches. What do you know about the radius of the base of the cone-shaped block? Explain.
\nAnswer:
\nIt is given that
\nMehnaj has a set of blocks that are all the same height. The cone-shaped block has a volume of 125 cubic inches. The sphere-shaped block has a volume of 250 cubic inches
\nSo,
\nFrom the given situation,
\nWe can observe that
\na. The heights of the two cone-shaped blocks are the same
\nb. The volumes of the two cone-shaped blocks are different
\nNow,
\nWe know that,
\nThe volume of a cone = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV1<\/sub> =\u00a0<\/sub>\\(\\frac{1}{3}\\)\u03c0r1<\/sub>\u00b2h1<\/sub>
\nV2<\/sub> = \\(\\frac{1}{3}\\)\u03c0r2<\/sub>\u00b2h2<\/sub>
\nSo,
\nV1<\/sub>\/V2<\/sub> = r1<\/sub>\u00b2\/ r2<\/sub>\u00b2
\nr1<\/sub>\u00b2\/ r2<\/sub>\u00b2 = \\(\\frac{125}{250}\\)
\nr1<\/sub>\u00b2 \/ r2<\/sub>\u00b2 = 0.5
\nr1<\/sub> \/ r2<\/sub> = 0.707
\nHence, from the above,
\nWe can conclude that the radius of the first cone-based block is 0.707 times the radius of the second cone-shaped block<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nClarissa has a decorative bulb in the shape of a sphere. If it has a radius of 3 inches, what is its volume? Use 3.14 for \u03c0.
\nAnswer:
\nIt is given that
\nClarissa has a decorative bulb in the shape of a sphere and it has a radius of 3 inches
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 3\u00b3
\n= 113.04 inches\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given decorative bulb is: 113.04 inches\u00b3<\/p>\n

Question 5.
\nA sphere has a surface area of about 803.84 square centimeters. What is the volume of the sphere? Use 3.14 for \u03c0 and round to the nearest whole number.
\nAnswer:
\nIt is given that
\nA sphere has a surface area of about 803.84 square centimeters
\nNow,
\nWe know that,
\nThe surface area of a sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\n4\u03c0r\u00b2 = 803.84
\nr\u00b2 = \\(\\frac{803.84}{4\u03c0}\\)
\nr\u00b2 = 64
\nr = 8 cm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 8\u00b3
\n= 2,143.57 cm\u00b3
\n\u22482,144 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given sphere is about 2,144 cm\u00b3<\/p>\n

Question 6.
\nA water pipe is a cylinder 30 inches long, with a radius of 1 inch. At one end of the cylinder, there is a hemisphere. What is the volume of the water pipe? Explain.
\nAnswer:
\nIt is given that
\nA water pipe is a cylinder 30 inches long, with a radius of 1 inch. At one end of the cylinder, there is a hemisphere.
\nSo,
\nThe volume of the water pipe = The volume of the cylinder + The volume of the hemisphere
\nNow,
\nFrom the given water pipe,
\nWe can observe that
\nRadius (r) = 1 in.
\nHeight (h) = 30 in.
\nNow,
\nWe know that,
\nThe volume of the Hemisphere (V) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{2}{3}\\) \u00d7 3.14 \u00d7 1\u00b3
\n= 2.09 in.\u00b3
\nNow,
\nWe know that,
\nThe volume of the cylinder (V’) = \u03c0r\u00b2h
\nSo,
\nV’ = 3.14 \u00d7 1\u00b2 \u00d7 30
\n= 94.2 in.\u00b3
\nSo,
\nThe volume of the given water pipe = 2.09 + 94.2
\n= 96.29 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given water pipe is: 96.29 in.\u00b3<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Question 7.
\nLeveled Practice What is the amount of air, in cubic centimeters, needed to fill the stability ball? Use 3.14 for \u03c0, and round to the nearest whole number. Use the formula
\n\"Envision
\nThe volume of the stability ball is approximately __________ cubic centimeters.
\nAnswer:
\nFrom the given figure,
\nWe can observe that
\nThe stability ball is in the form of a sphere
\nNow,
\nThe diameter of the stability ball (d) is: 55 cm
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{55}{2}\\)
\nr = 27.5 cm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 (27.5)\u00b3
\n= 87,069.53 cm\u00b3
\n\u224887,070 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the stability ball is approximately 87,070 cm\u00b3<\/p>\n

Question 8.
\nThe spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon. Assume that the balloon remains a sphere. Keep all answers in terms of \u03c0.
\na. Find the volume of the fully-inflated balloon.
\nAnswer:
\nIt is given that
\nThe spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon.
\nNow,
\nFrom the given situation,
\nThe diameter of the fully inflated balloon (d) is: 22 inches
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{22}{2}\\)
\nr = 11 cm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 \u03c0 \u00d7 11\u00b3
\n= 1,774.6\u03c0 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the fully inflated balloon is: 1,774.6\u03c0 cm\u00b3<\/p>\n

b. Find the volume of the half-inflated balloon.
\nAnswer:
\nIt is given that
\nThe spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon.
\nSo,
\nThe diameter of half-inflated balloon (d) = \\(\\frac{The diameter of fully inflated balloon}{2}\\)
\nd = \\(\\frac{22}{2}\\)
\nd = 11 cm
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{11}{2}\\)
\nr = 5.5 cm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 \u03c0 \u00d7 (5.5)\u00b3
\n= 221.8\u03c0 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the half-inflated balloon is: 221.8\u03c0 cm\u00b3<\/p>\n

c. What is the radius of the half-inflated balloon? Round to the nearest tenth.
\nAnswer:
\nFrom part (b),
\nWe can observe that
\nThe diameter of half-inflated balloon (d) = \\(\\frac{The diameter of fully inflated balloon}{2}\\)
\nd = \\(\\frac{22}{2}\\)
\nd = 11 cm
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{11}{2}\\)
\nr = 5.5 cm
\nHence, from the above,
\nWe can conclude that the radius of the half-inflated balloon is: 5.5 cm<\/p>\n

Question 9.
\nFind the volume of the figure. Use 3.14 for \u03c0, and round to the nearest whole number.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe figure is a combination of a hemisphere and a cone
\nSo,
\nFrom the given figure,
\nDiameter (d) =14 cm
\nHeight (h) = 17 cm
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{14}{2}\\)
\nr = 7 cm
\nNow,
\nWe know that,
\nThe volume of the figure (V) = The volume of a hemisphere + The volume of a cone
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 7\u00b2 \u00d7 17
\n= 871.87 cm\u00b3
\nNow,
\nWe know that,
\nThe volume of a hemisphere (V’) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV’ = \\(\\frac{2}{3}\\) \u00d7 3.14 \u00d7 7\u00b3
\n= 718.01 cm\u00b3
\nSo,
\nThe volume of the given figure = 871.87 + 718.01
\n= 1,589.88 cm\u00b3
\n\u22481,590 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given figure is: 1,590 cm\u00b3<\/p>\n

Question 10.
\nThe surface area of a sphere is about 2,826 square millimeters. What is the volume of the sphere? Use 3.14 for \u03c0, and round to the nearest whole number.
\nAnswer:
\nIt is given that
\nThe surface area of a sphere is about 2,826 square millimeters
\nNow,
\nWe know that,
\nThe surface area of a sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\n4\u03c0r\u00b2 = 2,826
\nr\u00b2 = \\(\\frac{2,826}{4\u03c0}\\)
\nr\u00b2 = 225
\nr = 15 mm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 15\u00b3
\n= 14,130 mm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given sphere is: 14,130 mm\u00b3<\/p>\n

Question 11.
\nA sphere has a volume of 1,837.35 cubic centimeters. What is the radius of the sphere? Use 3.14 for \u03c0, and round to the nearest tenth.
\nAnswer:
\nIt is given that
\nA sphere has a volume of 1,837.35 cubic centimeters
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\n1,837.35 = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\n\u03c0r\u00b3 = \\(\\frac{1,837.35 \u00d7 3}{4}\\)
\n\u03c0r\u00b3 = 1,378.01
\nr\u00b3 = \\(\\frac{1,378.01}{\u03c0}\\)
\nr\u00b3 = 438.85
\nr = 0.333 cm
\nHence, from the above,
\nWe can conclude that the radius of the given sphere is: 0.333 cm<\/p>\n

Question 12.
\nFind the volume of the solid. Use 3.14 for \u03c0, and round to the nearest whole number.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the above,
\nWe can observe that the given figure is a composite figure
\nNow,
\nFrom the given figure,
\nWe can observe that
\nRadius (r) = 4 m
\nHeight (h) = 17 m
\nNow,
\nWe know that,
\nThe volume of the given figure = The volume of a hemisphere + The volume of a cylinder
\nNow,
\nWe know that,
\nThe volume of the Hemisphere (V) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{2}{3}\\) \u00d7 3.14 \u00d7 4\u00b3
\n= 133.97 m\u00b3
\nNow,
\nWe know that,
\nThe volume of the cylinder (V’) = \u03c0r\u00b2h
\nSo,
\nV’ = 3.14 \u00d7 4\u00b2 \u00d7 17
\n= 854.08 m\u00b3
\nSo,
\nThe volume of the given figure = 133.9 + 854.08
\n= 987.98 m\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given figure is: 987.98 m\u00b3<\/p>\n

Question 13.
\nYour friend says that the volume of a sphere with a diameter of 3.4 meters is 164.55 cubic meters. What mistake might your friend have made? Find the correct volume. Use 3.14 for \u03c0 and round to the nearest hundredth.
\nAnswer:
\nIt is given that
\nYour friend says that the volume of a sphere with a diameter of 3.4 meters is 164.55 cubic meters
\nNow,
\nFrom the given information,
\nDiameter (d) = 3.4 m
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{3.4}{2}\\)
\nr = 1.7 m
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 (1.7)\u00b3
\n= 20.56 m\u00b3
\nHence, from the above,
\nWe can conclude that
\nThe mistake made by your friend is:
\nConsideration of diameter as radius and find the volume of the given sphere<\/p>\n

Question 14.
\nA solid figure has a cone and hemisphere hollowed out of it. What is the volume of the remaining part of the solid? Use 3.14 for \u03c0, and round to the nearest whole number.
\n\"Envision
\nAnswer:
\nIt is given that
\nA solid figure has a cone and hemisphere hollowed out of it
\nSo,
\nThe volume of the remaining part of the solid = | The volume of a hemisphere – The volume of a cone |
\nNow,
\nFrom the given solid,
\nWe can observe that
\nRadius (r) = 6 in.
\nHeight = 23 in.
\nNow,
\nWe know that,
\nThe volume of a hemisphere (V) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{2}{3}\\)\u03c0 \u00d7 6\u00b3
\n= 144\u03c0 in.\u00b3
\nNow,
\nWe know that,
\nThe volume of a cone (V’) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV’ = \\(\\frac{1}{3}\\)\u03c0 \u00d7 6\u00b2 \u00d7 23
\n= 276\u03c0 in.\u00b3
\nSo,
\nThe volume of the remaining part of the solid = |144\u03c0 – 276\u03c0|
\n= 132\u03c0
\n= 132 \u00d7 3.14
\n= 414.48 in.\u00b3
\n\u2248415 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the remaining part of the given solid is about 415 in.\u00b3<\/p>\n

Question 15.
\nHigher-Order Thinking<\/strong> A student was asked to find the volume of a solid where the inner cylinder is hollow. She incorrectly said the volume is 2,034.72 cubic inches.
\n\"Envision
\nAnswer:
\nIt is given that
\nA student was asked to find the volume of a solid where the inner cylinder is hollow. She incorrectly said the volume is 2,034.72 cubic inches.<\/p>\n

a. Find the volume of the solid. Use 3.14 for \u03c0. Round to the nearest whole number.
\nAnswer:
\nThe given solid is:
\n\"Envision
\nNow,
\nFrom the given solid,
\nWe can observe that it is a combination of a cone and a cylinder
\nSo,
\nThe volume of the given solid = The volume of a cylinder + The volume of a cone
\nNow,
\nFrom the given solid,
\nWe can observe that
\nThe radius of the cylinder is 3 in.
\nThe height of the cylinder is 15 in.
\nThe diameter of the cone is 12 in.
\nThe slant height of the cone is 9 in.
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = 3.14 \u00d7 3\u00b2 \u00d7 15
\n= 423.9 in.\u00b3
\nNow,
\nThe volume of a cone:
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{12}{2}\\)
\nr = 6 in.
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2 + h\u00b2
\nSo,
\n9\u00b2 = 6\u00b2 + h\u00b2
\nh\u00b2 = 9\u00b2 – 6\u00b2
\nh\u00b2 = 45
\nh = 6.70 in.
\nNow,
\nWe know that,
\nThe volume of a cone (V’) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV’ = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 6\u00b2 \u00d7 6.70
\n= 252.45 in.\u00b3
\nSo,
\nThe volume of the given solid = 423.9 + 252.45
\n= 676.3 in.\u00b3
\n\u2248676 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given solid is about 676 in.\u00b3<\/p>\n

b. What mistake might the student have made?
\nAnswer:<\/p>\n

Assessment Practice<\/strong>
\nQuestion 16.
\nThe spherical boulder is 20 feet in diameter and weighs almost 8 tons. Find its volume. Use 3.14 for \u03c0. Round to the nearest cubic foot.
\nAnswer:
\nIt is given that
\nThe spherical boulder is 20 feet in diameter and weighs almost 8 tons.
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{20}{2}\\)
\nr = 10 feet
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 3.14 \u00d7 10\u00b3
\n= 4,186.6 feet\u00b3
\n\u2248 4,187 feet\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given spherical boulder is: 4,187 feet\u00b3<\/p>\n

Question 17.
\nA bowl is in the shape of a hemisphere (half a sphere) with a diameter of 13 inches. Find the volume of the bowl. Use 3.14 for \u03c0, and round to the nearest cubic inch.
\nAnswer:
\nIt is given that
\nA bowl is in the shape of a hemisphere (half a sphere) with a diameter of 13 inches. Find the volume of the bowl
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{13}{2}\\)
\nr = 6.5 inches
\nNow,
\nWe know that,
\nThe volume of a hemisphere (V) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{2}{3}\\) \u00d7 3.14 \u00d7 (6.5)\u00b3
\n= 574.88 inches\u00b3
\n\u2248 575 inches\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given bowl is about 575 inches\u00b3<\/p>\n

3-ACT MATH<\/strong><\/p>\n

3-Act Mathematical Modeling:
\nMeasure Up
\n\"Envision<\/p>\n

ACT 1<\/strong>
\n1. After watching the video, what is the first question that comes to mind?
\n\"Envision
\nAnswer:<\/p>\n

Question 2.
\nWrite the Main Question you will answer.
\nAnswer:<\/p>\n

Question 3.
\nConstruct Arguments Predict an answer to this Main Question. Explain your prediction.
\nAnswer:<\/p>\n

Question 4.
\nOn the number line below, write a number that is too small to be the answer. Write a number that is too large.
\n\"Envision
\nAnswer:<\/p>\n

Question 5.
\nPlot your prediction on the same number line.
\nAnswer:<\/p>\n

ACT 2<\/strong>
\nQuestion 6.
\nWhat information in this situation would be helpful to know? How would you use that information?
\n\"Envision
\nAnswer:<\/p>\n

Question 7.
\nUse Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
\nAnswer:<\/p>\n

Question 8.
\nModel with Math
\nRepresent the situation using mathematics. Use your representation to answer the Main Question.
\nAnswer:<\/p>\n

Question 9.
\nWhat is your answer to the Main Question? Is it higher or lower than your prediction? Explain why.
\n\"Envision
\nAnswer:<\/p>\n

ACT 3<\/strong>
\nQuestion 10.
\nWrite the answer you saw in the video.
\n\"Envision
\nAnswer:<\/p>\n

Question 11.
\nReasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
\nAnswer:<\/p>\n

Question 12.
\nMake Sense and Persevere Would you change your model now that you know the answer? Explain.
\n\"Envision
\nAnswer:<\/p>\n

ACT 3 Extension<\/strong>
\nReflect
\nQuestion 13.
\nModel with Math
\nExplain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?
\nAnswer:<\/p>\n

Question 14.
\nMake Sense and Persevere When did you struggle most while solving the problem? How did you overcome that obstacle?
\n\"Envision
\nAnswer:<\/p>\n

SEQUEL<\/strong>
\nQuestion 15.
\nGeneralize Suppose you have a graduated cylinder half the height of the one in the video. How wide does the cylinder need to be to hold the liquid in the flask?
\nAnswer:<\/p>\n

Topic 8 REVIEW<\/h3>\n

Topic Essential Question
\nHow can you find volumes and surface areas of three-dimensional figures?
\nAnswer:
\nThe “Surface area” is the sum of the areas of all faces (or surfaces) on a 3D shape.
\nEx:
\nA cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces
\nWe know that,
\nThe volume of a three-dimensional figure = <\/b>Cross-sectional area \u00d7 length<\/p>\n

Vocabulary Review<\/strong>
\nComplete each definition and then provide an example of each vocabulary word.
\nVocabulary
\ncomposite figure
\ncone
\ncylinder
\nsphere<\/p>\n

\"Envision
\nAnswer:
\n\"\"<\/p>\n

Use Vocabulary in Writing
\nDraw a composite figure that includes any two of the following: a cylinder, a cone, a sphere, and a hemisphere. Label each part of your drawing. Then describe each part of your composite figure. Use vocabulary terms in your description.<\/p>\n

Concepts and Skills Review<\/strong><\/p>\n

Lesson 8.1 Find Surface Area of Three-Dimensional Figures<\/strong><\/p>\n

Quick Review<\/strong>
\nSurface area is the total area of the surfaces of a three-dimensional figure. The chart gives formulas for finding the surface area of a cylinder, a cone, and a sphere.
\n\"Envision<\/p>\n

Example<\/strong>
\nWhat is the surface area of the cylinder? Use 3.14 for \u03c0.
\nAnswer:
\n\"Envision<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nWhat is the surface area of the cone? Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nWe can observe that
\nThe slant height of the cone (l) is: 13 m
\nThe radius of the cone (r) is: 5 m
\nNow,
\nWe know that,
\nThe surface area of a cone (S.A) = \u03c0r\u00b2 + \u03c0rl
\nSo,
\nS.A = 3.14 \u00d7 5\u00b2 + 3.14 \u00d7 5 \u00d7 13
\n= 78.5 + 204.1
\n= 282.6 m\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given cone is: 282.6 m\u00b2<\/p>\n

Question 2.
\nWhat is the surface area of the sphere in terms of \u03c0?
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nWe can observe that
\nThe diameter of the sphere (d) is: 10 cm
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{10}{2}\\)
\nr = 5 cm
\nNow,
\nWe know that,
\nThe surface area of a sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\nS.A = 4 \u00d7 3.14 \u00d7 5\u00b2
\n= 314 cm\u00b2
\nHence, from the above,
\nWe can conclude that the surface area of the given sphere is: 314 cm\u00b2<\/p>\n

Question 3.
\nWhat is the surface area of the cylinder in terms of \u03c0?
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that,
\nThe diameter of the cylinder (d) is 12 in.
\nThe height of the cylinder (h) is 15 in.
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{12}{2}\\)
\nr = 6 in.
\nNow,
\nWe know that,
\nThe surface area of a cylinder (S.A) = 2\u03c0r\u00b2 + 2\u03c0rh
\nSo,
\nS.A = 2 \u00d7 3.14 \u00d7 6\u00b2 + 2 \u00d7 3.14 \u00d7 6 \u00d7 15
\n= 226.08 + 565.2
\n= 791.28 in.\u00b3
\nHence, from the above,
\nWe can conclude that the surface area of the given cylinder is: 791.28 in.\u00b3<\/p>\n

Lesson 8.2 Find Volume of Cylinders<\/strong><\/p>\n

Quick Review<\/strong>
\nThe volume of a cylinder is equal to the area of its base times its height.
\nV = area of base \u00b7 height, or V = \u03c0r\u00b2h
\n\"Envision<\/p>\n

Example<\/strong>
\nWhat is the volume of the cylinder? Use 3.14 for \u03c0.
\nAnswer:
\n\"Envision<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nWhat is the volume of the cylinder in terms of \u03c0?
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nWe can observe that
\nThe diameter of the cylinder (d) is: 2 m
\nThe height of the cylinder (h) is: 6 m
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{2}{2}\\)
\nr = 1 m
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\nV = 3.14 \u00d7 1\u00b2 \u00d7 6
\n= 18.84 m\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cylinder is: 18.84 m\u00b3<\/p>\n

Question 2.
\nThe volume of the cylinder is 141.3 cubic centimeters. What is the radius of the cylinder? Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nIt is given that
\nThe volume of the cylinder is 141.3 cubic centimeters
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe height of the cylinder (h) is: 5 cm
\nNow,
\nWe know that,
\nThe volume of a cylinder (V) = \u03c0r\u00b2h
\nSo,
\n141.3 = 3.14 \u00d7 r\u00b2 \u00d7 5
\nr\u00b2 = \\(\\frac{141.3}{3.14 \u00d7 5}\\)
\nr\u00b2= 9
\nr = 3 cm
\nHence, from the above,
\nWe can conclude that the radius of the given cylinder is: 3 cm<\/p>\n

Lesson 8.3 Find Volume of Cones<\/strong><\/p>\n

Quick Review<\/strong>
\nTo find the volume of a cone, use the formula V = \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h.<\/p>\n

Example<\/strong>
\nWhat is the volume of the cone? Use 3.14 for \u03c0.
\nAnswer:
\n\"Envision<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nWhat is the volume of the cone in terms of \u03c0?
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe radius of the cone (r) is 3 in.
\nThe height of the cone (h) is 8 in.
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 3\u00b2 \u00d7 8
\n= 75.36 in.\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 75.36 in.\u00b3<\/p>\n

Question 2.
\nWhat is the volume of the cone? Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe radius of the cone (r) is: 4 cm
\nThe slant height of the cone (h) is: 5 cm
\nNow,
\nWe know that,
\nl\u00b2 = r\u00b2 + h\u00b2
\n5\u00b2 = 4\u00b2 + h\u00b2
\nh\u00b2 = 5\u00b2 – 4\u00b2
\nh\u00b2 = 9
\nh = 3 cm
\nNow,
\nWe know that,
\nThe volume of a cone (V) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV = \\(\\frac{1}{3}\\) \u00d7 3.14 \u00d7 4\u00b2 \u00d7 3
\n= 50.24 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given cone is: 50.24 cm\u00b3<\/p>\n

Lesson 8.4 Find Volume of Spheres<\/strong><\/p>\n

Quick Review<\/strong>
\nTo find the volume of a sphere, use the formula V = \\(\\frac{4}{3}\\)\u03c0r3<\/sup>
\n\"Envision<\/p>\n

Example<\/strong>
\nFind the volume of the composite figure. Use 3.14 for \u03c0.
\n\"Envision
\nAnswer:
\nFirst, find the volume of the sphere. Use 3.14 for \u03c0.
\nV = \\(\\frac{4}{3}\\)\u03c0r3<\/sup>
\n= \\(\\frac{4}{3}\\)\u03c0(3.5)3<\/sup> \u2192 Substitute 3.5 for r.
\n= 57.17\u03c0 \u2248 179.5 cm3<\/sup>
\nDivide by 2 to find the volume of the hemisphere: 179.5 \u00f7 2 \u2248 89.75 cubic centimeters.
\nThen, find the volume of the cone. Use 3.14 for \u03c0.
\nV = \\(\\frac{1}{3}\\)\u03c0r2<\/sup> h
\n= \\(\\frac{1}{3}\\)\u03c0(3.5)2<\/sup>(14) \u2192 Substitute 3.5 for r and 14 for h.
\n= 57.171 \u2248 179.5 cm3<\/sup>
\nThe volume of the composite figure is approximately 89.75 + 179.5 \u2248 269.25 cubic centimeters.<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nWhat is the volume of the sphere? Use \\(\\frac{22}{7}\\) for \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe radius of the sphere (r) is: 14 cm
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 14\u00b3
\n= 11,498.6 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given sphere is: 11,498.6 cm\u00b3<\/p>\n

Question 2.
\nThe surface area of a sphere is 1,017.36 square inches. What is the volume of the sphere? Use \\(\\frac{22}{7}\\) for \u03c0.
\nAnswer:
\nIt is given that
\nThe surface area of a sphere is 1,017.36 square inches.
\nNow,
\nWe know that,
\nThe surface area of a sphere (S.A) = 4\u03c0r\u00b2
\nSo,
\n4\u03c0r\u00b2 = 1,017.36
\nr\u00b2 = \\(\\frac{1,017.36}{4\u03c0}\\)
\nr\u00b2 = 80.92
\nr = 8.99 inches
\nNow,
\nWe know that,
\nThe volume of a sphere (V) = \\(\\frac{4}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 (8.99)\u00b3
\n= 3,044.68 inches\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given sphere is: 3,044.68 inches\u00b3<\/p>\n

Question 3.
\nWhat is the volume of the composite figure? Use \\(\\frac{22}{7}\\) for \u03c0.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nFrom the above,
\nWe can observe that the given figure is a combination of a hemisphere and a cone
\nSo,
\nThe volume of the given figure = The volume of a hemisphere + The volume of a cone
\nNow,
\nFrom the given figure,
\nWe can observe that
\nDiameter (d) = 4 cm
\nHeight (h) = 10 cm
\nNow,
\nWe know that,
\nRadius (r) = \\(\\frac{Diameter}{2}\\)
\nr = \\(\\frac{4}{2}\\)
\nr = 2 cm
\nNow,
\nWe know that,
\nThe volume of a hemisphere (V) = \\(\\frac{2}{3}\\)\u03c0r\u00b3
\nSo,
\nV = \\(\\frac{2}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 2\u00b3
\n= 16.76 cm\u00b3
\nNow,
\nWe know that,
\nThe volume of a cone (V’) = \\(\\frac{1}{3}\\)\u03c0r\u00b2h
\nSo,
\nV’ = \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 2\u00b2 \u00d7 10
\n= 41.90 cm\u00b3
\nSo,
\nThe volume of the given figure = 41.90 + 16.76
\n= 58.66 cm\u00b3
\nHence, from the above,
\nWe can conclude that the volume of the given figure is: 58.66 cm\u00b3<\/p>\n

Topic 8 Fluency Practice<\/h3>\n

Hidden Clue<\/strong>
\nFor each ordered pair, solve the equation to find the unknown coordinate. Then locate and label the corresponding point on the graph. Draw line segments to connect the points in alphabetical order. Use the completed picture to help answer the riddle below.
\n\"Envision<\/p>\n

What do squares, triangles, pentagons, and octagons have in common?
\n\"Envision<\/p>\n

\"Envision
\nAnswer:
\nStep 1:
\nFor each ordered pair, solve the equation to find the unknown coordinate.
\n\"\"
\nStep 2:
\nThen locate and label the corresponding point on the graph.
\nStep 3:
\nDraw line segments to connect the points in alphabetical order.
\n\"\"
\nStep 4:
\nUse the completed picture to help answer the riddle below.
\nAll the squares, triangles, pentagons, and octagons have in common the sum of all the total angles i.e., 360\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

Go through the\u00a0enVision Math Common Core Grade 8 Answer Key\u00a0Topic 8 Solve Problems Involving Surface Area And Volume\u00a0regularly and improve your accuracy in solving questions. enVision Math Common Core 8th Grade Answers Key Topic 8 Solve Problems Involving Surface Area And Volume Topic 8 Essential Question How can you find volumes and surface areas of …<\/p>\n","protected":false},"author":1,"featured_media":25553,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/envisionmathanswerkey.com\/wp-content\/uploads\/2021\/09\/Envision-Math-Common-Core-Grade-8-Answer-Key-Topic-8-Solve-Problems-Involving-Surface-Area-And-Volume.png","_links":{"self":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts\/4249"}],"collection":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/comments?post=4249"}],"version-history":[{"count":4,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts\/4249\/revisions"}],"predecessor-version":[{"id":27947,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts\/4249\/revisions\/27947"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/media\/25553"}],"wp:attachment":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/media?parent=4249"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/categories?post=4249"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/tags?post=4249"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}