Homework 4 Rhombi And Squares Answers

Homework 4: Rhombi and Squares - Answers and Explanations

This comprehensive guide provides detailed answers and explanations for homework problems focusing on rhombi and squares. We'll cover key properties, theorems, and problem-solving strategies to solidify your understanding of these quadrilaterals. Remember that understanding the why behind the solutions is just as crucial as getting the correct answer.

Understanding Rhombi and Squares

Before diving into the specific homework problems, let's review the defining characteristics of rhombi and squares:

Rhombi:

A rhombus is a quadrilateral with all four sides congruent. This fundamental property leads to several other important characteristics:

• Opposite angles are congruent: ∠A ≅ ∠C and ∠B ≅ ∠D.
• Consecutive angles are supplementary: ∠A + ∠B = 180°, ∠B + ∠C = 180°, etc.
• Diagonals bisect each other: The diagonals intersect at a point that divides each diagonal into two equal segments.
• Diagonals are perpendicular: The diagonals intersect at a right angle (90°).
• Diagonals bisect opposite angles: Each diagonal bisects a pair of opposite angles.

Squares:

A square is a special type of rhombus (and also a special rectangle). It possesses all the properties of a rhombus and a rectangle. Therefore:

• All four sides are congruent.
• All four angles are congruent and equal to 90°.
• Opposite sides are parallel.
• Diagonals bisect each other.
• Diagonals are perpendicular.
• Diagonals bisect opposite angles.
• Diagonals are congruent. This is a key difference between a rhombus and a square.

Sample Homework Problems and Solutions

Let's tackle some common homework problems involving rhombi and squares. Remember to always draw a diagram to visualize the problem.

Problem 1: Given rhombus ABCD with AB = 6 and ∠A = 120°. Find the lengths of the diagonals AC and BD.

Solution:

1. Draw the rhombus: Sketch rhombus ABCD, labeling the given information.
2. Use properties of rhombi: Since the diagonals of a rhombus bisect each other at a right angle, we can form four congruent right-angled triangles within the rhombus.
3. Focus on one triangle: Consider triangle ABC. We know AB = 6 and ∠A = 120°. Since the diagonals bisect the angles, ∠BAC = 60°.
4. Trigonometry: We can use trigonometry (specifically, the sine and cosine rules) to find the lengths of the diagonals. In triangle ABC, we have:
   • AC = 2 * AB * sin(60°) = 2 * 6 * (√3/2) = 6√3
   • BD can be found using the area of the rhombus. The area can be calculated using the formula Area = AB * BC * sin(∠A) = 6 * 6 * sin(120°) = 18√3. Alternatively, the area can be calculated as (1/2) * d1 * d2, where d1 and d2 are the lengths of the diagonals. Equating the two, we get 18√3 = (1/2) * AC * BD, leading to BD = 6.

Therefore, AC = 6√3 and BD = 6.

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

Solution:

This problem requires a formal proof. We'll use coordinate geometry to demonstrate this.

1. Establish coordinates: Let's place the rhombus in the coordinate plane. Let A = (0, a), B = (b, 0), C = (0, -a), and D = (-b, 0). Note that we're assuming the center of the rhombus is the origin (0,0). This simplifies our calculations without loss of generality.
2. Find midpoints of diagonals: The midpoint of AC is ((0+0)/2, (a-a)/2) = (0,0). The midpoint of BD is ((b-b)/2, (0+0)/2) = (0,0). Both diagonals have the same midpoint. Hence, they bisect each other.
3. Find slopes of diagonals: The slope of AC is undefined (vertical line). The slope of BD is 0 (horizontal line). Since a horizontal and a vertical line are perpendicular, the diagonals are perpendicular.
4. Conclusion: Since the diagonals bisect each other and are perpendicular, they are perpendicular bisectors of each other.

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

Solution:

1. Pythagorean Theorem: A square is a special case of a rectangle and a rhombus. Since it has right angles, we can use the Pythagorean theorem directly.
2. Application: Consider right-angled triangle ABC. We have AB = BC = 8. Therefore, by Pythagorean theorem, AC² = AB² + BC² = 8² + 8² = 128.
3. Solution: AC = √128 = 8√2.

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

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

Solution:

This requires demonstrating that if the diagonals of a parallelogram are perpendicular, all four sides are congruent.

1. Properties of Parallelograms: Recall that in a parallelogram, opposite sides are equal in length and parallel. Diagonals bisect each other.
2. Perpendicular Diagonals: We're given that the diagonals are perpendicular.
3. Consider Triangles: The diagonals divide the parallelogram into four triangles. Since the diagonals bisect each other, we have four triangles with congruent pairs of sides.
4. Right-angled Triangles: The perpendicular diagonals create four right-angled triangles.
5. Congruence: By the Pythagorean theorem or hypotenuse-leg congruence, we can demonstrate that all four sides of the parallelogram are congruent. Therefore, the parallelogram is a rhombus.

Advanced Problems and Concepts

Let's explore more complex problems involving rhombi and squares:

Problem 5: In rhombus ABCD, the diagonals intersect at point O. If AO = 5 and BO = 12, find the area of the rhombus.

Solution:

1. Diagonals and Area: The area of a rhombus is half the product of its diagonals. Since the diagonals are perpendicular bisectors, AO and BO are half the lengths of the diagonals AC and BD respectively.
2. Calculate Full Diagonals: AC = 2 * AO = 10 and BD = 2 * BO = 24.
3. Area: Area = (1/2) * AC * BD = (1/2) * 10 * 24 = 120.

Therefore, the area of the rhombus is 120 square units.

Problem 6: Prove that the area of a rhombus is equal to half the product of its diagonals.

Solution:

This problem can be solved using several methods, including coordinate geometry (as shown in a previous problem) or using the formula for the area of a triangle (Area = (1/2) * base * height) and splitting the rhombus into four congruent triangles.

Problem 7: A square has a perimeter of 36 cm. Find the area of the square.

Solution:

1. Side Length: The perimeter of a square is 4 times the side length (P = 4s). Therefore, the side length is 36/4 = 9 cm.
2. Area: The area of a square is the side length squared (A = s²). Therefore, the area is 9² = 81 cm².

Practice and Further Exploration

This guide provided solutions and explanations for several homework problems related to rhombi and squares. Remember to thoroughly understand the underlying principles and properties of these quadrilaterals. The more you practice, the more comfortable you'll become with solving a wide variety of problems. Consider searching for additional practice problems online or in your textbook. Focus on understanding the "why" behind each solution, not just memorizing formulas. This will help you develop a strong foundation in geometry and problem-solving. Remember to always draw diagrams; visualization significantly aids in solving geometry problems.