Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
Question
Book Icon
Chapter 12, Problem 10RP

a.

To determine

To calculate: The lateral area and the total area of the pyramid ABCD

a.

Expert Solution
Check Mark

Answer to Problem 10RP

The lateral area and the total area of the solid ABCD are 180 unit 2 and 180+253 unit 2

Explanation of Solution

Given:

The measures of the given lengths are as follows:

AB = 13

AD = 10

Concept used:

Here, we are using the formula:

Lateral surface area of the pyramid = 32×b×h

(Where, “ b ” is base and “ h ” is height)

Total surface area of pyramid = Lateral surface area + 34×a2

(Where, “ a ” is side)

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12, Problem 10RP , additional homework tip  1

Here, In ΔABE

Using Pythagoras theorem

  (hypotenuse)2=(perpendicular)2+(base)2

  (AB)2=(BE)2+(AE)2

  (13)2=(BE)2+(5)2

  BE=16925

  BE=144

  BE=12

Lateral surface area of pyramid = 32×b×h

(Where, “ b ” is base and “ h ” is height)

  → 32×10×12

  → 180

Now, Total surface area of pyramid = Lateral surface area + 34×a2

(Where, “ a ” is side)

  → 180+3×1004

  → 180+253

Conclusion: Hence, the lateral area and the total area of the solid ABCD are 180 unit 2 and 180+253 unit 2

b.

To determine

To calculate: The lateral area and the total area of the cone PQRS

b.

Expert Solution
Check Mark

Answer to Problem 10RP

The lateral area and the total area of the cone PQRS are 60π and 96π

Explanation of Solution

Given:

The measures of the given lengths are as follows:

AD = 8

AC = 10

Concept used:

Here, we are using the formula:

Lateral surface area of the pyramid = 12×circumference×slantheight

Total surface area of pyramid = πr2+πrl

(Where, “ r ” is radius and “ l ” is slant height)

Calculation:

  Geometry For Enjoyment And Challenge, Chapter 12, Problem 10RP , additional homework tip  2

Here, Lateral surface area of the pyramid = 12×circumference×slantheight

  → 12×2πr×l

  → πrl

  → 6π(20)

  → 120π

And, Volume of cylinder = π2×r2×h

(Where, “ r ” is 6 and “ h ” is 20 )

  → π2×(6)2×20

  → π2×36×20

  → 360π

Conclusion: Hence, the volume and the total area of the right cylinder are 360π unit 3 and 156π unit 2

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
Knowledge Booster
Background pattern image
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