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 5.4, Problem 1PSA

a.

To determine

To find:The statement “Sides HR, RM, MO and HO are congruent” is true or false.

a.

Expert Solution
Check Mark

Answer to Problem 1PSA

The statement “Sides HR, RM, MO and HO are congruent” is true.

Explanation of Solution

Given information:

A rhombusHOMR with sides HR, RM, MO and HO.

Diagonals of rhombus HOMR are HM and OR which intersect at point B.

  Geometry For Enjoyment And Challenge, Chapter 5.4, Problem 1PSA , additional homework tip  1

One of the two characteristics that make a rhombus unique is that its four sides are equal in length, or congruent. The other identifying property is that opposite sides are parallel.

b.

To determine

To find: The statement “Diagonals HM and RO are perpendicular” is true or false.

b.

Expert Solution
Check Mark

Answer to Problem 1PSA

The statement “Diagonals HM and RO are perpendicular” is true.

Explanation of Solution

Given information:

A rhombus HOMR with sides HR, RM, MO and HO.

Diagonals of rhombus HOMR are HM and OR which intersect at point B.

  Geometry For Enjoyment And Challenge, Chapter 5.4, Problem 1PSA , additional homework tip  2

A wonderful and rare property of a rhombus is that its diagonals are always perpendicular to each other. You can see this for yourself if you lay down your four straight objects to make a rhombus and then draw in diagonals. No matter what angles you have for the rhombus’s four vertices, the diagonals of a rhombus are always at right angles to each other.

c.

To determine

To find: The statement “Diagonals HM and RO bisects the angles” is true or false.

c.

Expert Solution
Check Mark

Answer to Problem 1PSA

The statement “Diagonals HM and RO bisects the angles” is true.

Explanation of Solution

Given information:

A rhombus HOMR with sides HR, RM, MO and HO.

Diagonals of rhombus HOMR are HM and OR which intersect at point B.

  Geometry For Enjoyment And Challenge, Chapter 5.4, Problem 1PSA , additional homework tip  3

A diagonal of a rhombus splits a triangle into two congruent triangles using SAS congruence rule. It is also evident that each of those triangles are isosceles triangles. The two angles formed at each corner of the rhombus must also be equal. Since those equal angles are formed by the diagonal, the diagonal must be a bisector of the corner angles by definition.

Thus, diagonal HM bisects vertex angles at H and M. Diagonal RO bisects vertex angles at R and O.

d.

To determine

To find: The statement “Diagonals HM and RO bisects each other” is true or false.

d.

Expert Solution
Check Mark

Answer to Problem 1PSA

The statement “Diagonals HM and RO bisects each other” is true.

Explanation of Solution

Given information:

A rhombus HOMR with sides HR, RM, MO and HO.

Diagonals of rhombus HOMR are HM and OR which intersect at point B.

  Geometry For Enjoyment And Challenge, Chapter 5.4, Problem 1PSA , additional homework tip  4

The diagonal of a rhombus bisect each other at right angles. In any rhombus, the diagonals bisect each other at right angles, i.e. each diagonal cuts the other into two equal parts, and the angle where they cross is always 90 degrees.

Diagonals HM and RO bisect each other at point Q.

e.

To determine

To find: The statement “Diagonals HM and RO are congruent” is true or false.

e.

Expert Solution
Check Mark

Answer to Problem 1PSA

The statement “Diagonals HM and RO are congruent” is false.

Explanation of Solution

Given information:

A rhombus HOMR with sides HR, RM, MO and HO.

Diagonals of rhombus HOMR are HM and OR which intersect at point B.

  Geometry For Enjoyment And Challenge, Chapter 5.4, Problem 1PSA , additional homework tip  5

If you look in the diagram, the diagonals are not equal, but the sides are equal. One thing you may notice, though, is that the diagonals are perpendicular to each other and they bisect each other.

So, the diagonals are sometimes congruent, but they are always perpendicular bisectors.

Chapter 5 Solutions

Geometry For Enjoyment And Challenge

Ch. 5.1 - Prob. 11PSBCh. 5.1 - Prob. 12PSBCh. 5.1 - Prob. 13PSBCh. 5.1 - Prob. 14PSCCh. 5.1 - Prob. 15PSCCh. 5.2 - Prob. 1PSACh. 5.2 - Prob. 2PSACh. 5.2 - Prob. 3PSACh. 5.2 - Prob. 4PSACh. 5.2 - Prob. 5PSACh. 5.2 - Prob. 6PSACh. 5.2 - Prob. 7PSACh. 5.2 - Prob. 8PSACh. 5.2 - Prob. 9PSACh. 5.2 - Prob. 10PSACh. 5.2 - Prob. 11PSACh. 5.2 - Prob. 12PSACh. 5.2 - Prob. 13PSACh. 5.2 - Prob. 14PSACh. 5.2 - Prob. 15PSACh. 5.2 - Prob. 16PSACh. 5.2 - Prob. 17PSACh. 5.2 - Prob. 18PSACh. 5.2 - Prob. 19PSACh. 5.2 - Prob. 20PSACh. 5.2 - Prob. 21PSACh. 5.2 - Prob. 22PSBCh. 5.2 - Prob. 23PSBCh. 5.2 - Prob. 24PSBCh. 5.2 - Prob. 25PSCCh. 5.2 - Prob. 26PSCCh. 5.2 - Prob. 27PSCCh. 5.2 - Prob. 28PSCCh. 5.3 - Prob. 1PSACh. 5.3 - Prob. 2PSACh. 5.3 - Prob. 3PSACh. 5.3 - Prob. 4PSACh. 5.3 - Prob. 5PSACh. 5.3 - Prob. 6PSACh. 5.3 - Prob. 7PSACh. 5.3 - Prob. 8PSACh. 5.3 - Prob. 9PSACh. 5.3 - Prob. 10PSACh. 5.3 - Prob. 11PSACh. 5.3 - Prob. 12PSACh. 5.3 - Prob. 13PSACh. 5.3 - Prob. 14PSBCh. 5.3 - Prob. 15PSBCh. 5.3 - Prob. 16PSBCh. 5.3 - Prob. 17PSBCh. 5.3 - Prob. 18PSBCh. 5.3 - Prob. 19PSBCh. 5.3 - Prob. 20PSBCh. 5.3 - Prob. 21PSBCh. 5.3 - Prob. 22PSBCh. 5.3 - Prob. 23PSBCh. 5.3 - Prob. 24PSBCh. 5.3 - Prob. 25PSBCh. 5.3 - Prob. 26PSCCh. 5.3 - Prob. 27PSCCh. 5.3 - Prob. 28PSCCh. 5.3 - Prob. 29PSDCh. 5.3 - Prob. 30PSDCh. 5.4 - Prob. 1PSACh. 5.4 - Prob. 2PSACh. 5.4 - Prob. 3PSACh. 5.4 - Prob. 4PSACh. 5.4 - Prob. 5PSACh. 5.4 - Prob. 6PSACh. 5.4 - Prob. 7PSACh. 5.4 - Prob. 8PSACh. 5.4 - Prob. 9PSACh. 5.4 - Prob. 10PSACh. 5.4 - Prob. 11PSACh. 5.4 - Prob. 12PSACh. 5.4 - Prob. 13PSBCh. 5.4 - Prob. 14PSBCh. 5.4 - Prob. 15PSBCh. 5.4 - Prob. 16PSBCh. 5.4 - Prob. 17PSBCh. 5.4 - Prob. 18PSBCh. 5.4 - Prob. 19PSBCh. 5.4 - Prob. 20PSBCh. 5.4 - Prob. 21PSCCh. 5.4 - Prob. 22PSCCh. 5.5 - Prob. 1PSACh. 5.5 - Prob. 2PSACh. 5.5 - Prob. 3PSACh. 5.5 - Prob. 4PSACh. 5.5 - Prob. 5PSACh. 5.5 - Prob. 6PSACh. 5.5 - Prob. 7PSACh. 5.5 - Prob. 8PSACh. 5.5 - Prob. 9PSACh. 5.5 - Prob. 10PSACh. 5.5 - Prob. 11PSACh. 5.5 - Prob. 12PSACh. 5.5 - Prob. 13PSACh. 5.5 - Prob. 14PSACh. 5.5 - Prob. 15PSBCh. 5.5 - Prob. 16PSBCh. 5.5 - Prob. 17PSBCh. 5.5 - Prob. 18PSBCh. 5.5 - Prob. 19PSBCh. 5.5 - Prob. 20PSBCh. 5.5 - Prob. 21PSBCh. 5.5 - Prob. 22PSBCh. 5.5 - Prob. 23PSBCh. 5.5 - Prob. 24PSBCh. 5.5 - Prob. 25PSBCh. 5.5 - Prob. 26PSBCh. 5.5 - Prob. 27PSBCh. 5.5 - Prob. 28PSCCh. 5.5 - Prob. 29PSCCh. 5.5 - Prob. 30PSCCh. 5.6 - Prob. 1PSACh. 5.6 - Prob. 2PSACh. 5.6 - Prob. 3PSACh. 5.6 - Prob. 4PSACh. 5.6 - Prob. 5PSACh. 5.6 - Prob. 6PSACh. 5.6 - Prob. 7PSACh. 5.6 - Prob. 8PSACh. 5.6 - Prob. 9PSACh. 5.6 - Prob. 10PSACh. 5.6 - Prob. 11PSBCh. 5.6 - Prob. 12PSBCh. 5.6 - Prob. 13PSBCh. 5.6 - Prob. 14PSBCh. 5.6 - Prob. 15PSBCh. 5.6 - Prob. 16PSBCh. 5.6 - Prob. 17PSBCh. 5.6 - Prob. 18PSBCh. 5.6 - Prob. 19PSCCh. 5.6 - Prob. 20PSCCh. 5.6 - Prob. 21PSDCh. 5.7 - Prob. 1PSACh. 5.7 - Prob. 2PSACh. 5.7 - Prob. 3PSACh. 5.7 - Prob. 4PSACh. 5.7 - Prob. 5PSACh. 5.7 - Prob. 6PSACh. 5.7 - Prob. 7PSACh. 5.7 - Prob. 8PSACh. 5.7 - Prob. 9PSACh. 5.7 - Prob. 10PSACh. 5.7 - Prob. 11PSBCh. 5.7 - Prob. 12PSBCh. 5.7 - Prob. 13PSBCh. 5.7 - Prob. 14PSBCh. 5.7 - Prob. 15PSBCh. 5.7 - Prob. 16PSBCh. 5.7 - Prob. 17PSBCh. 5.7 - Prob. 18PSBCh. 5.7 - Prob. 19PSBCh. 5.7 - Prob. 20PSBCh. 5.7 - Prob. 21PSBCh. 5.7 - Prob. 22PSBCh. 5.7 - Prob. 23PSBCh. 5.7 - Prob. 24PSBCh. 5.7 - Prob. 25PSBCh. 5.7 - Prob. 26PSCCh. 5.7 - Prob. 27PSCCh. 5.7 - Prob. 28PSCCh. 5.7 - Prob. 29PSCCh. 5 - Prob. 1RPCh. 5 - Prob. 2RPCh. 5 - Prob. 3RPCh. 5 - Prob. 4RPCh. 5 - Prob. 5RPCh. 5 - Prob. 6RPCh. 5 - Prob. 7RPCh. 5 - Prob. 8RPCh. 5 - Prob. 9RPCh. 5 - Prob. 10RPCh. 5 - Prob. 11RPCh. 5 - Prob. 12RPCh. 5 - Prob. 13RPCh. 5 - Prob. 14RPCh. 5 - Prob. 15RPCh. 5 - Prob. 16RPCh. 5 - Prob. 17RPCh. 5 - Prob. 18RPCh. 5 - Prob. 19RPCh. 5 - Prob. 20RPCh. 5 - Prob. 21RPCh. 5 - Prob. 22RPCh. 5 - Prob. 23RPCh. 5 - Prob. 24RPCh. 5 - Prob. 25RPCh. 5 - Prob. 26RPCh. 5 - Prob. 27RPCh. 5 - Prob. 28RPCh. 5 - Prob. 29RPCh. 5 - Prob. 30RP
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