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 7PSB

a.

To determine

To calculate: The total area of cone with sides of triangles as 12 and 13.

a.

Expert Solution
Check Mark

Answer to Problem 7PSB

The total area of cone is 90π .

Explanation of Solution

Given information:

Height of cone AB = 12,

Slant height of cone AC = 13.

Formula used:

Lateral Area of cone =12×C×l

C = Circumference of cone.

l = length of cone.

Circumference of cone:

  C=2πr

r = radius of cone

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 7PSB , additional homework tip  1

In right angle triangle,

  a2+b2c2

Area of a circle: A=πr2

r = radius of circle

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 7PSB , additional homework tip  2

The radius r can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  (AC)2=(AB)2+(BC)2(13)2=(12)2+(r)2r2=169144r2=25r=25=5

Lateral Area of cone =12×C×l

Circumference of cone:

  C=2πr=2π×5=10π

Lateral Area of cone =12×10π×13

Lateral Area of cone =65π

Area of base: A=πr2=π(5)2=25π

Total Area = Lateral Area + Area of base

Total Area =65π+25π

Total Area =90π

b.

To determine

To find: The total area of cylinder with height of 8 and slant height of 10.

b.

Expert Solution
Check Mark

Answer to Problem 7PSB

The total area of cylinder is 66π .

Explanation of Solution

Given information:

Height of cylinder PQ = 8,

Slant height of cylinder PR = 10.

Formula used:

Lateral Area of cylinder =C×h

C = Circumference of cylinder.

h = height of cylinder.

Circumference of cylinder:

  C=2πr

r = radius of cylinder

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 7PSB , additional homework tip  3

In right angle triangle,

  a2+b2c2

Area of a circle: A=πr2

r = radius of circle

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 7PSB , additional homework tip  4

The radius r can be calculated by applying Pythagoras Theorem.

In right angled triangle PQR , we get

  (PR)2=(PQ)2+(QR)2(10)2=(8)2+(QR)2(QR)2=10064(QR)2=36QR=36=6r=QR2=62=3

Lateral Area of cylinder =C×h

Circumference of cylinder:

  C=2πr=2π×3=6π

Lateral Area of cylinder =6π×8

Lateral Area of cylinder =48π

Area of base: A=πr2=π(3)2=9π

Total Area = Lateral Area +2( Area of base)

Total Area =48π+(2×9π)

Total Area =48π+18π

Total Area =66π

c.

To determine

To find: The total area of cone with height as 8 and radius as 15.

c.

Expert Solution
Check Mark

Answer to Problem 7PSB

The area of cone is 480π .

Explanation of Solution

Given information:

Height of cone XZ = 8,

Radius of cone r = 15.

Formula used:

The below theorems are used:

Two tangent theorem states that if two tangent segments are drawn to one circle from the same external point, then they are congruent.

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 7PSB , additional homework tip  5

In right angle triangle,

  a2+b2c2

Area of triangle: A=12×b×h

b = base of triangle

h = height of triangle Area of a circle: A=πr2

r = radius of circle

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12.3, Problem 7PSB , additional homework tip  6

The side XY can be calculated by applying Pythagoras Theorem.

In right angled triangle XZY , we get

  (XY)2=(XZ)2+(ZY)2(XY)2=(8)2+(15)2(XY)2=64+225(XY)2=289XY=289=17

Lateral Area of cone =12×C×l

Circumference of cone:

  C=2πr=2π×15=30π

Lateral Area of cone =12×30π×13

Lateral Area of cone =255π

Area of base: A=πr2=π(15)2=225π

Total Area = Lateral Area + Area of base

Total Area =255π+225π

Total Area =480π

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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