Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
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Chapter 12.3, Problem 14PSC

a.

To determine

To calculate: The total area of a cone frustum if slant height is 5.

a.

Expert Solution
Check Mark

Answer to Problem 14PSC

The total area of frustum is 99π .

Explanation of Solution

Given information:

Radius of top base = 3,

Radius of bottom base = 6,

Angle at base of bottom cone =60 ,

Formula used:

Area of cone =12×C×l

C = Circumference of cone and l = slant height of cone

r = radius of circle

Area of a circle: A=πr2

r = radius of circle

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 14PSC , additional homework tip  1

The surface forms frustum of a cone. Extend the frustum to form a cone’

There are two 306090Δ's formed with parallel bases.

In large triangle, hypotenuse = length of cone =12

The converse of Midline theorem is used. Midline theorem states that the segment that joins the midpoints of two sides of a triangle is parallel to the third side and half as long

In small triangle, hypotenuse =2×3=6

Circumference of top cone: C=2πr=2π×3=6π

Lateral Area of top cone =12×C×h

Lateral Area of top cone =12×6π×6

Lateral Area of top cone =18π

Circumference of bottom cone: C=2πr=2π×6=12π

Lateral Area of bottom cone =12×C×h

Lateral Area of bottom cone =12×12π×12

Lateral Area of bottom cone =72π

Difference in Lateral Area =72π-18π=54π

Area of top base circle =π(6)2=36π

Area of bottom base circle =π(3)2=9π

Total Area = Area of bottom base circle + Area of top base circle + Difference in Lateral Area

Total Area =9π+36π+54π

Total Area =99π

b.

To determine

To find: The total area of a cylindrical shell with height as 5.

b.

Expert Solution
Check Mark

Answer to Problem 14PSC

The total area of cylindrical shell is 60π .

Explanation of Solution

Given information:

Radius of smaller circle = 2,

Height of cylinder = 5.

Formula used:

Area of cylinder =C×h

C = Circumference of cylinder and h = height of cylinder

Area of a circle: A=πr2

r = radius of circle

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 14PSC , additional homework tip  2

The surface forms cylindrical shell.

Radius of smaller circle = 2

Radius of larger circle =2+1=3

Height of cylinder = 5.

Circumference of larger cylinder: C=2πr=2π×3=6π

Lateral Area of larger cylinder C×h

Lateral Area of larger cylinder =6π×5

Lateral Area of larger cylinder =30π

Circumference of smaller cylinder: C=2πr=2π×2=4π

Lateral Area of smaller cylinder =C×h

Lateral Area of smaller cylinder =4π×5

Lateral Area of smaller cylinder =20π

Area of larger circle =π(3)2=9π

Area of smaller circle =π(2)2=4π

Total Area = Area of larger cylinder + Area of smaller cylinder + 2(Area of larger circle - Area of smaller circle)

Total Area =30π+20π+2(9π4π)

Total Area =50π+10π

Total Area =60π

c.

To determine

To calculate: The total area of a solid.

c.

Expert Solution
Check Mark

Answer to Problem 14PSC

The total area of solid is 424π .

Explanation of Solution

Given information:

Radius = 8,

Height of cylinder = 10,

Slant height of cone = 17.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 14PSC , additional homework tip  3

In right angle triangle,

  a2+b2c2

Area of cylinder =C×h

C = Circumference of cylinder and h = height of cylinder

Area of cone =12×C×l

C = Circumference of cone and l = slant height of cone

Area of hemisphere =12×4πr2

r = radius of hemisphere

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 14PSC , additional homework tip  4

  r=8

Side AB can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  (AC)2=(AB)2+(BC)2(17)2=(AB)2+(8)2289=(AB)2+64(AB)2=28964(AB)2=225AB=225AB=15

Height of cone AB=15

Slant height of cone = 17.

Height of cylinder =10 .

Area of cylinder =C×h

  C=2π×8=16π

  h=10

Area of cylinder =16π×10

Area of cylinder =160π

Area of cone =12×C×l

  C=2π×8=16π

  l=17

Area of cone =12×C×l

Area of cone =12×16π×17

Area of cone =136π

Area of hemisphere =12×4πr2

Area of hemisphere =2π(8)2

Area of hemisphere =128π

Total Area = Area of cylinder + Area of cone + Area of hemisphere

Total Area =160π+136π+128π

Total Area =424π

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

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