Intermediate Geometry : How to find if rhombuses are similar

Study concepts, example questions & explanations for Intermediate Geometry

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Example Questions

Example Question #1 : Rhombuses

Given: Rhombuses \(\displaystyle ABCD\) and \(\displaystyle EFGH\).

\(\displaystyle AB = EF\)

True, false, or undetermined: Rhombus \(\displaystyle ABCD \cong\) Rhombus \(\displaystyle EFGH\).

Possible Answers:

False

Undetermined

True

Correct answer:

Undetermined

Explanation:

Two figures are congruent by definition if all of their corresponding sides are congruent and all of their corresponding angles are congruent.

By definition, a rhombus has four sides of equal length.  If we let \(\displaystyle s_{1}\) be the common sidelength of Rhombus \(\displaystyle ABCD\) and \(\displaystyle s_{2}\) be the common sidelength of Rhombus \(\displaystyle EFGH\), then, since \(\displaystyle AB = EF\), it follows that \(\displaystyle s_{1} = s_{2}\), so corresponding sides are congruent. However, no information is given about their angle measures. Therefore, it cannot be determined whether or not the two rhombuses are congruent.

 

Example Question #1 : How To Find If Rhombuses Are Similar

Given: Rhombuses \(\displaystyle ABCD\) and \(\displaystyle EFGH\).

\(\displaystyle m \angle A = 60 ^{\circ }\) and \(\displaystyle m \angle F = 120 ^{\circ }\)

True, false, or undetermined: Rhombus \(\displaystyle ABCD \sim\) Rhombus \(\displaystyle EFGH\).

Possible Answers:

False

Undetermined

True

Correct answer:

True

Explanation:

Two figures are similar by definition if all of their corresponding sides are proportional and all of their corresponding angles are congruent.

By definition, a rhombus has four sides that are congruent. If we let \(\displaystyle s_{1}\) be the common sidelength of Rhombus \(\displaystyle ABCD\) and \(\displaystyle s_{2}\) be the common sidelength of Rhombus \(\displaystyle EFGH\), it can easily be seen that the ratio of the length of each side of the former to that of the latter is the same ratio, namely, \(\displaystyle \frac{s_{1}}{s_{2}}\).

Also, a rhombus being a parallelogram, its opposite angles are congruent, and its consecutive angles are supplementary. Therefore, since \(\displaystyle m \angle A = 60 ^{\circ }\), it follows that \(\displaystyle m\angle C = 60 ^{\circ }\), and \(\displaystyle m \angle B = m \angle D = 180 ^{\circ } - 60^{\circ } = 120 ^{\circ }\). By a similar argument, \(\displaystyle m \angle H = m \angle F = 120 ^{\circ }\) and \(\displaystyle m \angle E = m \angle G= 180 ^{\circ } - 120 ^{\circ } = 60^{\circ }\). Therefore, 

\(\displaystyle \angle A \cong \angle E\)

\(\displaystyle \angle B \cong \angle F\)

\(\displaystyle \angle C \cong \angle F\)

\(\displaystyle \angle D \cong \angle H\)

Since all corresponding sides are proportional and all corresponding angles are congruent, it holds that Rhombus \(\displaystyle ABCD \sim\) Rhombus \(\displaystyle EFGH\).

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