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 6.3, Problem 3PSA

a

To determine

To prove: The two planes ‘r’ and ‘t’ are parallel

a

Expert Solution
Check Mark

Explanation of Solution

Given information:

Plane 1 is parallel to given plane 2

Plane 2 is parallel to given plane 3

Two given lines are parallel

Formula used:

Associative property

Proof:

Using associative property

If plane 1 is parallel to plane 2

And plane 2 is parallel to plane 3

Then

Plane 3 will be parallel to plane 1.

Hence

  rt

b

To determine

To prove: The ABFE is a plane figure

b

Expert Solution
Check Mark

Explanation of Solution

Given information:

Plane 1 is parallel to given plane 2

Plane 2 is parallel to given plane 3

Two given lines are parallel

Formula used:

Plane figures have only two dimensions and covers area

Proof:

Since figure ABFE have

only two dimensions namely, length and breadth

and area of the figure can be calculated using length and breadth only

Hence, figure ABFE is a plane figure

c

To determine

To prove: The lines AB and EF are parallel

c

Expert Solution
Check Mark

Explanation of Solution

Given information:

Plane 1 is parallel to given plane 2

Plane 2 is parallel to given plane 3

Two given lines are parallel

Formula used:

Parallel lines never meet even on extending them to infinity.

Proof:

Given lines are AB and EF

They are parallel because

They will never meet even on extending to infinity

d

To determine

To prove: The line segments AB and EF are congruent

d

Expert Solution
Check Mark

Explanation of Solution

Given information:

Plane 1 is parallel to given plane 2

Plane 2 is parallel to given plane 3

Two given lines are parallel

Formula used:

Congruent objects are always equal or have same shape and size

Proof:

Since AB and EF are parallel and a part of simple figure,

Thus, they are equal in magnitude

By the definition of congruency,

  AB¯=EF¯AB¯EF¯

Hence proved

Chapter 6 Solutions

Geometry For Enjoyment And Challenge

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