Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 12.4, Problem 20PSC

a.

To determine

To calculate: The volume of ice in each cube.

a.

Expert Solution
Check Mark

Answer to Problem 20PSC

The volume of ice in each cube is 51.4 cm3 .

Explanation of Solution

Given information:

An ice cube manufacturer makes ice cubes with holes in them. Each cube is 4 cm on a side and hole is 2 cm in diameter.

Formula used:

Volume of cube =s3

Where s = side of cube

Volume of hole =πr2h

r = radius of hole,

h = height of hole.

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.4, Problem 20PSC , additional homework tip  1

Volume of cube =s3

Each cube is 4 cm.

Volume of cube =(4)3

Volume of cube =64 cm3

Volume of hole =πr2h

Volume of hole =π(1)2×4

Volume of hole =4×3.14

Volume of hole =12.56 cm3

Volume of ice = Volume of cube − Volume of hole

Volume of ice =64-12.56

Volume of ice =51.4 cm3

b.

To determine

To find: The volume of water left when 10 cubes will melt.

b.

Expert Solution
Check Mark

Answer to Problem 20PSC

The volume of water left is 457.8 cm3 .

Explanation of Solution

Given information:

An ice cube manufacturer makes ice cubes with holes in them. Each cube is 4 cm on a side and hole is 2 cm in diameter.

Water volume decreases by 11% when it changes from solid to liquid.

Formula used:

Volume of cube =s3

Where s = side of cube Volume of hole =πr2h

r = radius of hole,

h = height of hole.

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.4, Problem 20PSC , additional homework tip  2

Volume of cube =s3

Each cube is 4 cm.

Volume of cube =(4)3

Volume of cube =64 cm3

Volume of hole =πr2h

Volume of hole =π(1)2×4

Volume of hole =4×3.14

Volume of hole =12.56 cm3

Volume of ice = Volume of cube − Volume of hole

Volume of ice =64-12.56

Volume of ice =51.4 cm3

Volume of 10 cubes 10×51.44514.4 cm3

Water volume decreases by 11% when it changes from solid to liquid.

Volume of 10 melted cubes 514.4 -(11%×514.4)

Volume of 10 melted cubes 0.89×514.4457.8 cm3

c.

To determine

To calculate: The total surface area of a single cube.

c.

Expert Solution
Check Mark

Answer to Problem 20PSC

The total surface area of a single cube is 114.8 cm2 .

Explanation of Solution

Given information:

An ice cube manufacturer makes ice cubes with holes in them. Each cube is 4 cm on a side and hole is 2 cm in diameter.

Formula used:

Facial area of the cube =s2

where s = side of cube.

Base area of cylinder =πr2

where r = radius of cylinder

Lateral area of cylinder =Ch

where C=2πr is the circumference of cylinder

r = radius of cylinder

h = height of cylinder.

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.4, Problem 20PSC , additional homework tip  3

Facial area of the cube =s2

Facial area of the cube =(4)2

Facial area of the cube =16

Base area of cylinder =πr2

  d=2r=d2=22=1

Base area of cylinder =π(1)2

Base area of cylinder =π

Lateral area of cylinder =Ch

  C=2πr=2π×1=2π

  h=4

Lateral area of cylinder =2π×4

Lateral area of cylinder =8π

Total Area = 6(Facial area of the cube) − 2(Base area of cylinder) + Lateral area of cylinder

Total Area =(6×16)2π+8π

Total Area =962(3.142)+8(3.142)

Total Area =114.8 cm2

d.

To determine

To verify: The total surface area of a cube is twice the total surface area of cube without holes.

d.

Expert Solution
Check Mark

Explanation of Solution

Given information:

An ice cube manufacturer makes ice cubes with holes in them. Each cube is 4 cm on a side and hole is 2 cm in diameter.

The manufacture claims that these cube cool a drink twice as fast regular cubes of the same size.

Formula used:

Facial area of the cube =s2

where s = side of cube.

Base area of cylinder =πr2

where r = radius of cylinder

Lateral area of cylinder =Ch

where C=2πr is the circumference of cylinder

r = radius of cylinder

h = height of cylinder.

Proof:

  Geometry For Enjoyment And Challenge, Chapter 12.4, Problem 20PSC , additional homework tip  4

Facial area of the cube =s2

Facial area of the cube =(4)2

Facial area of the cube =16

Base area of cylinder =πr2

  d=2r=d2=22=1

Base area of cylinder =π(1)2

Base area of cylinder =π

Lateral area of cylinder =Ch

  C=2πr=2π×1=2π

  h=4

Lateral area of cylinder =2π×4

Lateral area of cylinder =8π

Total Area = 6(Facial area of the cube) − 2(Base area of cylinder) + Lateral area of cylinder

Total Area =(6×16)2π+8π

Total Area =962(3.142)+8(3.142)

Total Area =114.8 cm2

Total area of a cube without hole =6(s)2

Total area of a cube without hole =6(4)2

Total area of a cube without hole =96 cm2

The ratio of area is as follows.

  Total area of a cubeTotal area of a cube without hole=114.896=1.20

The manufacturer claim is not true because the ratio is not two.

Chapter 12 Solutions

Geometry For Enjoyment And Challenge

Ch. 12.1 - Prob. 11PSCCh. 12.2 - Prob. 1PSACh. 12.2 - Prob. 2PSACh. 12.2 - Prob. 3PSACh. 12.2 - Prob. 4PSACh. 12.2 - Prob. 5PSACh. 12.2 - Prob. 6PSBCh. 12.2 - Prob. 7PSBCh. 12.2 - Prob. 8PSBCh. 12.2 - Prob. 9PSBCh. 12.2 - Prob. 10PSCCh. 12.2 - Prob. 11PSCCh. 12.2 - Prob. 12PSCCh. 12.2 - Prob. 13PSCCh. 12.3 - Prob. 1PSACh. 12.3 - Prob. 2PSACh. 12.3 - Prob. 3PSACh. 12.3 - Prob. 4PSACh. 12.3 - Prob. 5PSACh. 12.3 - Prob. 6PSBCh. 12.3 - Prob. 7PSBCh. 12.3 - Prob. 8PSBCh. 12.3 - Prob. 9PSBCh. 12.3 - Prob. 10PSBCh. 12.3 - Prob. 11PSBCh. 12.3 - Prob. 12PSCCh. 12.3 - Prob. 13PSCCh. 12.3 - Prob. 14PSCCh. 12.4 - Prob. 1PSACh. 12.4 - Prob. 2PSACh. 12.4 - Prob. 3PSACh. 12.4 - Prob. 4PSACh. 12.4 - Prob. 5PSACh. 12.4 - Prob. 6PSACh. 12.4 - Prob. 7PSBCh. 12.4 - Prob. 8PSBCh. 12.4 - Prob. 9PSBCh. 12.4 - Prob. 10PSBCh. 12.4 - Prob. 11PSBCh. 12.4 - Prob. 12PSBCh. 12.4 - Prob. 13PSBCh. 12.4 - Prob. 14PSBCh. 12.4 - Prob. 15PSBCh. 12.4 - Prob. 16PSBCh. 12.4 - Prob. 17PSBCh. 12.4 - Prob. 18PSBCh. 12.4 - Prob. 19PSCCh. 12.4 - Prob. 20PSCCh. 12.4 - Prob. 21PSCCh. 12.4 - Prob. 22PSCCh. 12.5 - Prob. 1PSACh. 12.5 - Prob. 2PSACh. 12.5 - Prob. 3PSACh. 12.5 - Prob. 4PSACh. 12.5 - Prob. 5PSACh. 12.5 - Prob. 6PSACh. 12.5 - Prob. 7PSACh. 12.5 - Prob. 8PSBCh. 12.5 - Prob. 9PSBCh. 12.5 - Prob. 10PSBCh. 12.5 - Prob. 11PSBCh. 12.5 - Prob. 12PSBCh. 12.5 - Prob. 13PSBCh. 12.5 - Prob. 14PSBCh. 12.5 - Prob. 15PSBCh. 12.5 - Prob. 16PSBCh. 12.5 - Prob. 17PSCCh. 12.5 - Prob. 18PSCCh. 12.5 - Prob. 19PSCCh. 12.5 - Prob. 20PSCCh. 12.6 - Prob. 1PSACh. 12.6 - Prob. 2PSACh. 12.6 - Prob. 3PSACh. 12.6 - Prob. 4PSACh. 12.6 - Prob. 5PSACh. 12.6 - Prob. 6PSBCh. 12.6 - Prob. 7PSBCh. 12.6 - Prob. 8PSBCh. 12.6 - Prob. 9PSBCh. 12.6 - Prob. 10PSBCh. 12.6 - Prob. 11PSBCh. 12.6 - Prob. 12PSCCh. 12.6 - Prob. 13PSCCh. 12.6 - Prob. 14PSCCh. 12.6 - Prob. 15PSCCh. 12.6 - Prob. 16PSCCh. 12.6 - Prob. 17PSDCh. 12.6 - Prob. 18PSDCh. 12 - Prob. 1RPCh. 12 - Prob. 2RPCh. 12 - Prob. 3RPCh. 12 - Prob. 4RPCh. 12 - Prob. 5RPCh. 12 - Prob. 6RPCh. 12 - Prob. 7RPCh. 12 - Prob. 8RPCh. 12 - Prob. 9RPCh. 12 - Prob. 10RPCh. 12 - Prob. 11RPCh. 12 - Prob. 12RPCh. 12 - Prob. 13RPCh. 12 - Prob. 14RPCh. 12 - Prob. 15RPCh. 12 - Prob. 16RPCh. 12 - Prob. 17RPCh. 12 - Prob. 18RPCh. 12 - Prob. 19RPCh. 12 - Prob. 20RPCh. 12 - Prob. 21RPCh. 12 - Prob. 22RPCh. 12 - Prob. 1CRCh. 12 - Prob. 2CRCh. 12 - Prob. 3CRCh. 12 - Prob. 4CRCh. 12 - Prob. 5CRCh. 12 - Prob. 6CRCh. 12 - Prob. 7CRCh. 12 - Prob. 8CRCh. 12 - Prob. 9CRCh. 12 - Prob. 10CRCh. 12 - Prob. 11CRCh. 12 - Prob. 12CRCh. 12 - Prob. 13CRCh. 12 - Prob. 14CRCh. 12 - Prob. 15CRCh. 12 - Prob. 16CRCh. 12 - Prob. 17CRCh. 12 - Prob. 18CRCh. 12 - Prob. 19CRCh. 12 - Prob. 20CRCh. 12 - Prob. 21CRCh. 12 - Prob. 22CRCh. 12 - Prob. 23CRCh. 12 - Prob. 24CRCh. 12 - Prob. 25CRCh. 12 - Prob. 26CRCh. 12 - Prob. 27CRCh. 12 - Prob. 28CRCh. 12 - Prob. 29CRCh. 12 - Prob. 30CRCh. 12 - Prob. 31CRCh. 12 - Prob. 32CRCh. 12 - Prob. 33CRCh. 12 - Prob. 34CRCh. 12 - Prob. 35CRCh. 12 - Prob. 36CRCh. 12 - Prob. 37CRCh. 12 - Prob. 38CRCh. 12 - Prob. 39CRCh. 12 - Prob. 40CRCh. 12 - Prob. 41CRCh. 12 - Prob. 42CR

Additional Math Textbook Solutions

Find more solutions based on key concepts
Knowledge Booster
Background pattern image
Geometry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Elementary Geometry For College Students, 7e
Geometry
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Cengage,
Text book image
Elementary Geometry for College Students
Geometry
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Cengage Learning
ALGEBRAIC EXPRESSIONS & EQUATIONS | GRADE 6; Author: SheenaDoria;https://www.youtube.com/watch?v=fUOdon3y1hU;License: Standard YouTube License, CC-BY
Algebraic Expression And Manipulation For O Level; Author: Maths Solution;https://www.youtube.com/watch?v=MhTyodgnzNM;License: Standard YouTube License, CC-BY
Algebra for Beginners | Basics of Algebra; Author: Geek's Lesson;https://www.youtube.com/watch?v=PVoTRu3p6ug;License: Standard YouTube License, CC-BY
Introduction to Algebra | Algebra for Beginners | Math | LetsTute; Author: Let'stute;https://www.youtube.com/watch?v=VqfeXMinM0U;License: Standard YouTube License, CC-BY