what are the properties of a parallelogram, square, rhombus and rectangle, concerning their diagonals ?

* bisection

*perpendicular

*bisecting the angles of the quadrilateral

etc. as the parameters?

Rectangles

Definition: A rectangle is a quadrilateral with four right angles.

Notice that we use “quadrilateral” in our definition of rectangles. We could have also said that a rectangle is a parallelogram with four right angles, since and quadrilateral with four right angles is also a parallelogram (because their opposite sides would be parallel).

Rectangles have a couple of properties that help distinguish them from other parallelograms. By studying these properties, we will be able to differentiate between various types of parallelograms and classify them more specifically. Keep in mind that all of the figures in this section share properties of parallelograms. That is, they all have

(1) opposite sides that are parallel,

(2) opposite angles that are congruent,

(3) opposite sides that are congruent,

(4) consecutive angles that are supplementary, and

(5) diagonals that bisect each other.

Now, let’s look at the properties that make rectangles a special type of parallelogram.

(1) All four angles of a rectangle are right angles.

(2) The diagonals of a rectangle are congruent.

Rhombuses

Definition: A rhombus is a quadrilateral with four congruent sides.

Similar to the definition of a rectangle, we could have used the word “parallelogram” instead of “quadrilateral” in our definition of rhombus. Thus, rhombuses have all of the properties of parallelograms (stated above), along with a few others. Let’s look at these properties.

(1) Consecutive sides of a rhombus are congruent.

(2) The diagonals of a rhombus bisect pairs of opposite angles.

(3) The diagonals of a rhombus are perpendicular.

Squares

Definition: A square is a parallelogram with four congruent sides and four congruent angles.

Notice that the definition of a square is a combination of the definitions of a rectangle and a rhombus. Therefore, a square is both a rectangle and a rhombus, which means that the properties of parallelograms, rectangles, and rhombuses all apply to squares. Because squares have a combination of all of these different properties, it is a very specific type of quadrilateral.

Look at the hierarchy of quadrilaterals below. This figure shows the progression of our knowledge of polygons, beginning with quadrilaterals, and ending with squares.

Notice that there are two arrows pointing to the square. That is because a square has all the properties of a rectangle and rhombus.

Now that we are aware of the properties of rectangles, rhombuses, and squares, let’s work on a few exercises that will gauge our understanding of this material.

Exercise 1

Identify each parallelogram as a rectangle, rhombus, or a square.

Answer:

First, let’s take a look at Parallelogram A. The figure shows that it has four congruent sides and that its diagonals intersect perpendicularly. Because its sides are congruent, we know that the parallelogram is not a rectangle. The fact that Parallelogram A’s diagonals intersect perpendicularly does not help us because both rhombuses and squares share this characteristic. The angle at the top of Parallelogram A is not a right angle, however. Therefore, we know that it is not a square. Parallelogram A is a rhombus.

In Parallelogram B, we see that there are four right angles and that the pairs of opposite sides are congruent. However, consecutive sides are not congruent, so we can eliminate rhombuses and squares from our options. Thus, Parallelogram B is a rectangle.

Let’s take a look at Parallelogram C now. We note that it has a pair of right angles and four congruent sides. Our inclination leads us to think that this parallelogram is a square, but let’s make sure just in case. We know that the two right angles given to us have a sum of 180°. Because the interior angles of a quadrilateral is 360°, we know that the remaining two angles must have a sum of 180° (because 360-180=180). Opposite angles of parallelograms are congruent, which means that each other the missing angles must have a measure of 90° (since 180÷2=90). This tells us that there are actually four right angles in Parallelogram C, so we know that it is a square (and a rhombus).

Exercise 2

Find the value of x given rectangle ABCD below.

Answer:

We know that ABCD is a rectangle, so let’s use some rectangle properties to help us figure out what x is. It appears as though the focus of this exercise is on the diagonals of the figure. From above, we know that the diagonals of a rectangle are congruent, so let’s set segments AC and BD equal to each other:

So, we get x=12.

Exercise 3

Answer:

Let’s examine the information we have been given from the exercise, in order to deduce more information from it. We know that EKIN is a parallelogram, and that ∠1≌∠2. Since EKIN is a parallelogram, we know that its opposite sides are parallel. Therefore, segments EK and IN are parallel.

Next, we can use the Alternate Interior Angles Theorem to claim that ∠1≌∠4 and ∠2≌∠3. Recall, that the alternate interior angles are congruent if and only if a transversal intersects a pair of parallel lines. In this case, our pair of parallel lines is EK and IN, and our transversal is segment NK.

By transitivity, we can say that ∠1≌∠3 and ∠2≌∠4. Let’s look at our chain of congruences to be assured that the previous statements are true.

The diagonal splits our parallelogram into two triangles. In fact, because two angles of each triangle are congruent, we can say that ∆EKN and ∆INK are isosceles triangles. The converse of the Isosceles Triangle Theorem states that the sides opposite of congruent angles of isosceles triangles are congruent, so we know that segment EK is congruent to segment EN, and that segments IK and IN are congruent.

Now, can say that segments EK and IN are congruent, as are EN and IK because opposite sides of a parallelogram are congruent. By transitivity, we know that EN≌EK≌IN≌IK. Let’s look at our new illustration.

Thus, parallelogram EKIN is a rhombus because it has four congruent sides. Our two-column geometric proof for this exercise is shown below.

Exercise 4

What must the value of y be in order for rhombus PQRS to be a square?

Answer:

Before we can figure our y, we must determine what the value of x is. Ultimately, we want rhombus PQRS to be a square, which means that PQRS should have four right angles.

Let’s begin by figuring out what x is. This is relatively simple because we can just set segment PQ equal to PS:

Now that we know what x is, we can plug it into the measure of the angle given to us. But, first, we need to figure out what the total measure of ∠QSR is. We know that we want ∠PSR to be 90°. Also, we know that the diagonals of a square bisect pairs of opposite angles. Therefore, ∠PSR should be bisected by segment QS, splitting the angle up into two congruent angles of 45° (because 90÷2=45). Now, we can set ∠QSR equal to 45°. We get:

Now, we substitute 7 in for x:

So, the value of y must be 4 in order for rhombus PQRS to also be a square.