6-5 Practice Rhombi And Squares Answer Key

6-5 Practice: Rhombi and Squares – A Comprehensive Guide with Answers

This comprehensive guide delves into the intricacies of rhombi and squares, providing detailed explanations, worked examples, and solutions to common practice problems. Understanding the properties of these quadrilaterals is crucial for mastering geometry. We'll explore their unique characteristics, delve into proving theorems, and solidify your understanding with practice problems and solutions. This guide acts as your complete answer key, but more importantly, it helps you understand why the answers are what they are.

Understanding Rhombi

A rhombus is a quadrilateral with all four sides congruent (equal in length). This seemingly simple definition unlocks a wealth of geometric properties. Let's explore some key characteristics:

Key Properties of a Rhombus:

• All sides are congruent: This is the defining characteristic of a rhombus.
• Opposite sides are parallel: This makes a rhombus a parallelogram.
• Opposite angles are congruent: Another property inherited from its parallelogram nature.
• Consecutive angles are supplementary: Meaning their sum is 180 degrees.
• Diagonals bisect each other: The diagonals intersect at a point that divides each diagonal into two equal segments.
• Diagonals are perpendicular bisectors: This means the diagonals intersect at a right angle and bisect each other.

Proving Rhombus Properties:

Many geometric proofs rely on demonstrating these properties. For example, proving that a quadrilateral is a rhombus often involves showing that all four sides are congruent or that the diagonals are perpendicular bisectors. Using postulates and theorems like the Side-Side-Side (SSS) congruence postulate or the Pythagorean theorem is key to constructing these proofs.

Understanding Squares

A square is a special type of rhombus (and also a special type of rectangle). It possesses all the properties of a rhombus, plus some extra defining characteristics.

Key Properties of a Square:

• All sides are congruent: Just like a rhombus.
• All angles are congruent and equal to 90 degrees: This is the defining difference between a square and a rhombus. A rhombus's angles are not necessarily 90 degrees.
• Opposite sides are parallel: Inherited from its rhombus and parallelogram nature.
• Diagonals bisect each other: This property is shared with both rhombi and rectangles.
• Diagonals are perpendicular bisectors: Also shared with rhombi.
• Diagonals are congruent: This is a unique property of squares.

Proving Square Properties:

Proving a quadrilateral is a square often involves showing that it satisfies the properties of both a rhombus and a rectangle. This typically requires demonstrating that all sides are congruent and that all angles are right angles (90 degrees).

Practice Problems and Solutions

Let's tackle some practice problems to reinforce your understanding of rhombi and squares. Remember to always clearly state the properties you're using in your solutions.

Problem 1: Given rhombus ABCD with AB = 6 and angle DAB = 120 degrees. Find the lengths of the diagonals AC and BD.

Solution 1:

1. Draw the rhombus: Sketch rhombus ABCD with the given information.
2. Recognize the properties: In a rhombus, diagonals are perpendicular bisectors. This forms four congruent right-angled triangles.
3. Use trigonometry: Consider triangle ABD. We know AB = 6 and angle DAB = 120 degrees. The diagonals bisect the angles, so angle DAB is bisected into two 60-degree angles. Now we can use trigonometry (SOH CAH TOA). In right-angled triangle AOB (where O is the intersection of the diagonals), we have:
   • AO = AB * cos(60) = 6 * (1/2) = 3
   • BO = AB * sin(60) = 6 * (√3/2) = 3√3
4. Find diagonal lengths: Since the diagonals bisect each other, AC = 2 * AO = 2 * 3 = 6 and BD = 2 * BO = 2 * 3√3 = 6√3.

Therefore, AC = 6 and BD = 6√3.

Problem 2: Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

Solution 2:

1. Start with a rhombus: Begin with rhombus ABCD. Let the diagonals intersect at point O.
2. Congruent triangles: Prove triangles AOB and COD are congruent using the Side-Side-Side (SSS) postulate. Since AB = BC = CD = DA (all sides are congruent in a rhombus), and AO = OC and BO = OD (diagonals bisect each other in a parallelogram), we have SSS congruence.
3. Congruent angles: Since triangles AOB and COD are congruent, their corresponding angles are congruent. Therefore, angle AOB = angle COD.
4. Supplementary angles: Angles AOB and BOC are supplementary (add up to 180 degrees) because they form a straight line.
5. Equal angles: Since angle AOB = angle COD and angle AOB + angle BOC = 180 degrees, we have angle AOB = angle BOC = 90 degrees. This proves that the diagonals are perpendicular.
6. Bisectors: We already established that the diagonals bisect each other from the properties of a parallelogram (which a rhombus is).

Therefore, the diagonals of a rhombus are perpendicular bisectors of each other.

Problem 3: Given square ABCD with side length 8. Find the length of diagonal AC.

Solution 3:

1. Pythagorean theorem: Since a square is a right-angled quadrilateral, we can use the Pythagorean theorem (a² + b² = c²) to find the diagonal length.
2. Apply the theorem: Let AC be the hypotenuse. AB and BC are legs, both with length 8. Therefore:
   • AC² = AB² + BC² = 8² + 8² = 64 + 64 = 128
3. Solve for AC: AC = √128 = 8√2

Therefore, the length of diagonal AC is 8√2.

Problem 4: Prove that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Solution 4:

1. Start with a parallelogram: Begin with parallelogram ABCD. Let the diagonals AC and BD intersect at point O. We are given that the diagonals are perpendicular (AC ⊥ BD).
2. Right angles: This means that angle AOB = angle BOC = angle COD = angle DOA = 90 degrees.
3. Congruent triangles: Consider triangles AOB and BOC. Since AB || CD and AD || BC (opposite sides of a parallelogram are parallel), and AO = OC and BO = OD (diagonals bisect each other in a parallelogram), triangles AOB and BOC are congruent by the Side-Angle-Side (SAS) postulate.
4. Congruent sides: Because triangles AOB and BOC are congruent, their corresponding sides are congruent. Therefore, AB = BC.
5. All sides equal: Similarly, we can show that all other sides are congruent (BC = CD = DA = AB).
6. Rhombus: Since all four sides are congruent, the parallelogram is a rhombus.

Therefore, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Problem 5: A quadrilateral has diagonals that bisect each other and are perpendicular. Is it necessarily a square?

Solution 5: No, it is not necessarily a square. The given conditions describe a rhombus. For it to be a square, we would also need to demonstrate that all angles are 90 degrees (or that the diagonals are equal in length).

These problems and solutions illustrate the crucial interplay between the properties of rhombi and squares and how these properties can be used to prove various geometric relationships. Remember, consistent practice is key to mastering these concepts. Continue to work through various problems, applying the theorems and postulates discussed above. This will build your confidence and understanding, enabling you to tackle more complex geometrical problems with ease. Remember to always clearly state the properties or theorems used in your solutions. This ensures clarity and allows for better understanding of your problem-solving approach.