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 1PSA

a.

To determine

To calculate: The total surface area of a sphere having radius 7 units.

a.

Expert Solution
Check Mark

Answer to Problem 1PSA

The total surface area of a sphere is 616units2. .

Explanation of Solution

Given Information:

Radius of a sphere (r)=6units

Formula used:

Total surface area of a sphere (TSA)=4πr2

  "r" is the radius of a sphere

Calculation:

We know that, Total surface area of a sphere (TSA)=4πr2

Where "r" is the radius of a sphere

  TSA=4×227×72...(r=7)TSA=4×22×7TSA=616units2

Hence, total surface area of a sphere is 616units2.

b.

To determine

To find: The total surface area of a sphere having radius of 3 units.

b.

Expert Solution
Check Mark

Answer to Problem 1PSA

The total surface area of a sphere is 113.14units2. .

Explanation of Solution

Given Information:

Radius of a sphere (r)=3units

Formula used:

Total surface area of a sphere (TSA)=4πr2

  "r" is the radius of a sphere

Calculation:

We know that, Total surface area of a sphere (TSA)=4πr2

Where "r" is the radius of a sphere

  TSA=4×227×32...(r=7)TSA=4×227×9TSA=113.14units2

Hence, total surface area of a sphere is 113.14units2.

c.

To determine

To find: The total surface area of a sphere having diameter of 6 units.

c.

Expert Solution
Check Mark

Answer to Problem 1PSA

The total surface area of a sphere with diameter 6 is 113.14units2.

Explanation of Solution

Given Information:

Diameter of a sphere (d)=6units

Formula used:

Total surface area of a sphere (TSA)=4πr2

  "r" is the radius of a sphere

Calculation:

As diameter of a sphere (d)=6units

Therefore, radius (r)=d2=3units

We know that, Total surface area of a sphere (TSA)=4πr2

Where "r" is the radius of a sphere

  TSA=4×227×32...(r=3)TSA=4×227×9TSA=113.14units2

Hence, total surface area of a sphere is 113.14units2.

d.

To determine

To find: The total surface area of a sphere having diameter of 5 units.

d.

Expert Solution
Check Mark

Answer to Problem 1PSA

The total surface area of a sphere is 78.57units2.

Explanation of Solution

Given Information:

Diameter of a sphere (d)=5units

Formula used:

Total surface area of a sphere (TSA)=4πr2

  "r" is the radius of a sphere

Calculation:

As diameter of a sphere (d)=5units

Therefore, radius (r)=d2=52units

We know that, Total surface area of a sphere (TSA)=4πr2

Where "r" is the radius of a sphere

  TSA=4×227×(52)2...(r=52)TSA=4×227×254TSA=227×25TSA=78.57units2

Hence, total surface area of a sphere is 78.57units2.

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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