6-5 Additional Practice Properties Of Special Parallelograms

6 + 5 Additional Practice Properties of Special Parallelograms

Parallelograms, quadrilaterals with opposite sides parallel, form the foundation for understanding many other geometric shapes. Within the parallelogram family lie special types: rectangles, rhombuses, and squares, each possessing unique properties beyond the basic parallelogram characteristics. This article delves into six fundamental properties of parallelograms, followed by five additional properties that are particularly useful when dealing with special parallelograms (rectangles, rhombuses, and squares). We'll then explore practice problems to solidify your understanding.

Six Fundamental Properties of Parallelograms

Before we delve into the special properties, let's refresh our understanding of the basic characteristics of parallelograms:

1. Opposite sides are parallel: This is the defining characteristic. Lines AB and CD are parallel, as are lines BC and DA. This parallelism is denoted by AB || CD and BC || DA.

2. Opposite sides are congruent: The lengths of opposite sides are equal. AB = CD and BC = DA.

3. Opposite angles are congruent: Opposite angles within the parallelogram have the same measure. ∠A = ∠C and ∠B = ∠D.

4. Consecutive angles are supplementary: Any two angles that share a side are supplementary; their measures add up to 180 degrees. ∠A + ∠B = 180°, ∠B + ∠C = 180°, ∠C + ∠D = 180°, and ∠D + ∠A = 180°.

5. Diagonals bisect each other: The diagonals of a parallelogram intersect at a point that divides each diagonal into two equal segments. If the diagonals intersect at point E, then AE = EC and BE = ED.

6. Each diagonal divides the parallelogram into two congruent triangles: Diagonal AC divides parallelogram ABCD into two congruent triangles, ΔABC and ΔADC. Similarly, diagonal BD divides the parallelogram into congruent triangles ΔABD and ΔBCD. This congruence is based on Side-Angle-Side (SAS) or Side-Side-Side (SSS) congruency rules.

Five Additional Properties, Especially Useful for Special Parallelograms

These five properties, while sometimes implied in the fundamental properties, become particularly important when distinguishing and working with rectangles, rhombuses, and squares:

1. Rectangles: Right Angles: A rectangle is a parallelogram where all four angles are right angles (90°). This property immediately distinguishes it from other parallelograms. This implies that consecutive sides are perpendicular. The diagonals of a rectangle are congruent and bisect each other.

2. Rhombuses: Congruent Sides: A rhombus is a parallelogram with all four sides congruent. This is a key differentiating factor. The diagonals of a rhombus are perpendicular bisectors of each other. They also bisect the angles of the rhombus.

3. Squares: Right Angles and Congruent Sides: A square is a parallelogram that is both a rectangle and a rhombus. It possesses all the properties of both shapes: four right angles and four congruent sides. The diagonals of a square are congruent, perpendicular bisectors of each other, and they bisect the angles of the square.

4. Diagonals and Area: The area of a parallelogram can be calculated using the formula: Area = base * height. However, knowing the lengths of the diagonals can also be useful, particularly in special cases. For example, in a rhombus, the area can be calculated as: Area = (1/2)d₁d₂, where d₁ and d₂ are the lengths of the diagonals.

5. Midpoint Theorem and Parallelograms: The midpoint theorem provides a useful link between parallelograms and line segments. If you connect the midpoints of consecutive sides of any quadrilateral, the resulting quadrilateral is always a parallelogram. This is true even if the original quadrilateral isn't a parallelogram. This property is particularly helpful in proving or constructing parallelograms.

Practice Problems

Let's test your understanding with some practice problems that utilize both fundamental and additional properties:

Problem 1: In parallelogram ABCD, ∠A = 110°. Find the measure of ∠B, ∠C, and ∠D.

Solution: Consecutive angles are supplementary, so ∠B = 180° - 110° = 70°. Opposite angles are congruent, so ∠C = ∠A = 110° and ∠D = ∠B = 70°.

Problem 2: ABCD is a rhombus with AB = 6 cm and diagonal AC = 8 cm. Find the area of the rhombus.

Solution: In a rhombus, diagonals are perpendicular bisectors. Therefore, we can divide the rhombus into four congruent right-angled triangles. Let's consider triangle ABC. We know AB = 6 cm and AC = 8 cm. We can use the Pythagorean theorem to find BC: BC² = AB² + (AC/2)² = 6² + 4² = 52. So BC = √52 cm. However, a simpler approach is to use the diagonal area formula for a rhombus. First, we need to find the length of the other diagonal (BD). Since the diagonals are perpendicular bisectors, they form four right-angled triangles. Using Pythagoras on one of those triangles we would need another side length. Alternatively, if we are given one diagonal, we cannot find the area without another piece of information. We are given that AC = 8cm. The other diagonal is unknown. We'll need additional information to solve this problem completely. We need the length of BD to apply the formula Area = (1/2)d₁d₂.

Problem 3: Prove that the diagonals of a rectangle are congruent.

Solution: Let ABCD be a rectangle. We need to prove that AC = BD. In a rectangle, all angles are 90°. Consider right-angled triangles ΔABC and ΔBAD. AB is common to both triangles. BC = AD (opposite sides of a rectangle are congruent). Therefore, by the Pythagorean theorem or the RHS congruence theorem, ΔABC ≅ ΔBAD. Thus, AC = BD (corresponding sides of congruent triangles).

Problem 4: A parallelogram has vertices at A(1, 2), B(4, 3), C(6, 7), and D(x, y). Find the coordinates of D.

Solution: In a parallelogram, the diagonals bisect each other. Let M be the midpoint of AC and also the midpoint of BD. The midpoint formula states that the coordinates of the midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) are ((x₁ + x₂)/2, (y₁ + y₂)/2). Therefore, the midpoint of AC is M = ((1 + 6)/2, (2 + 7)/2) = (3.5, 4.5). Since M is also the midpoint of BD, we have (3.5, 4.5) = ((4 + x)/2, (3 + y)/2). Solving for x and y, we get x = 3 and y = 6. Therefore, D = (3, 6).

Problem 5: Show that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Solution: Let ABCD be a parallelogram with perpendicular diagonals AC and BD intersecting at point O. In a parallelogram, diagonals bisect each other, so AO = OC and BO = OD. Since the diagonals are perpendicular, we have four right-angled triangles: ΔAOB, ΔBOC, ΔCOD, and ΔDOA. In ΔAOB and ΔBOC, AO = OC and BO is common. Since ∠AOB = ∠BOC = 90°, by the RHS congruence rule, ΔAOB ≅ ΔBOC. Therefore, AB = BC. Similarly, we can show that BC = CD and CD = DA. Hence, all sides are congruent, proving that the parallelogram is a rhombus.

These problems illustrate the application of various properties of parallelograms, emphasizing the distinctions and relationships between parallelograms, rectangles, rhombuses, and squares. By practicing more problems, you'll build a strong understanding of these geometric shapes and their properties. Remember to always start by identifying the type of parallelogram you're working with, as this dictates which specific properties you can apply.