Quadrilaterals

SAT Questions that focus on Quadrilaterals require knowledge of the following topics.

Trapezoid

Trapezoid is a quadrilateral that has at least one pair of parallel opposite sides, called bases (the other two sides are called legs).

Types of trapezoids:

Isosceles Trapezoid: the base angles are congruent, and, as a consequence, the two legs are also congruent;
Right Trapezoid: has two adjacent right angles;
Acute trapezoid: the two angles on its longer base are acute;
Obtuse trapezoid: one angle on each base is acute, the other is obtuse.

Trapezoid Properties

P1: $\hat{a}+\hat{d}=\hat{b}+\hat{c}=180º$ (figure below).
Since AB and DC are parallel, $\hat{a}$ and $\hat{d}$ are colateral interior angles, and, therefore, are supplementary. For the same reason, angles $\hat{b}$ and $\hat{c}$ are also supplementary.

P2: The bisectors of angles $\hat{b}$ and $\hat{c}$ intersect forming a right angle. The bisectors of angles $\hat{a}$ and $\hat{d}$ also intersect forming a right angle.
In the figure below, BO is the bisector of angle $\hat{b}$, and CO is the bisector of angle $\hat{c}$. Thus, the interior angles of triangle BOC are $\frac{\hat{b}}{2}$, $\frac{\hat{c}}{2}$ and $\hat{O}$.

The sum of the three interior angles of any triangle is 180º. Thus,
$\frac{\hat{b}}{2}+\frac{\hat{c}}{2}+\hat{O}=180$

Multiplying both sides of the equation by 2:
$\hat{b}+\hat{c}+2\hat{O}=360$

By the previous property (P1), we know that $\hat{b}+\hat{c}=180$. Thus,
$180+2\hat{O}=360$
$2\hat{O}=180$
$\hat{O}=90º$

Parallelogram

A parallelogram is a quadrilateral whose opposite sides are parallel. In the figure below, ABCD is a parallelogram: AB is parallel to DC; AD is parallel to BC:

Since AD and BC are parallel, $\hat{a}$ and $\hat{b}$ are colateral interior angles, as well as $\hat{d}$ and $\hat{c}$. Thus,
$\hat{a}+\hat{b}=180$ (equation I) and
$\hat{d}+\hat{c}=180$. (equation II)

Since AB and DC are also parallel, $\hat{b}$ and $\hat{c}$ are colateral interior angles, as well as $\hat{a}$ and $\hat{d}$. Thus,
$\hat{b}+\hat{c}=180$ (equation III) and
$\hat{a}+\hat{d}=180$. (equation IV)

Let's subtract equation IV from equation I:
$\hat{a}+\hat{d}-\hat{a}-\hat{b}=180-180$
$\hat{d}-\hat{b}=0$
$\hat{d}=\hat{b}$

Now let's subtract equation III from equation I:
$\hat{a}+\hat{b}-\hat{b}-\hat{c}=180-180$
$\hat{a}-\hat{c}=0$
$\hat{a}=\hat{c}$

In other words, opposite angles in a parallelogram are congruent.

Parallelogram Properties

P1: The two diagonals bisect each other.
In parallelogram ABCD (figure below), the red lines are the diagonals, that intersect in point M, and split the interior angles of the parallelogram, forming angles $\hat{x}$, $\hat{y}$, $\hat{r}$ and $\hat{s}$.

Note that triangles AMD and CMB are similar, since they have all three interior angles congruent, and sides AD and BC measure the same (opposite sides of the parallelogram). Thus, MB=MD and MA=MC. The diagonals bisect each other.

P2: If we connect the midpoint of each side of any quadrilateral, the segments form a parallelogram.
In the figure below, ABCD is a quadrilateral, with sides and interior angles that are incongruous. $M_{ab}$ is the midpoint of side $AB$;
$M_{bc}$ is the midpoint of side $BC$;
$M_{cd}$ is the midpoint of side $CD$;
$M_{da}$ is the midpoint of side $DA$:

$M_{cd}M_{da}$ (the segment that connects the midpoints of sides $AD$ e $CD$) is parallel to the diagonal $AC$.
$M_{ab}M_{bc}$ (the segment that connects the midpoints of sides $AB$ e $BC$) is parallel to the diagonal $AC$.
Therefore, $M_{cd}M_{da}$ is parallel to $M_{ab}M_{bc}$.

$M_{da}M_{ab}$ (the segment that connects the midpoints of sides $AD$ e $AB$) is parallel to the diagonal $DB$.
$M_{cd}M_{bc}$ (the segment that connects the midpoints of sides $DC$ e $BC$) is parallel to the diagonal $DB$.
Therefore, $M_{da}M_{ab}$ is parallel to $M_{cd}M_{bc}$

Conclusion: $M_{ab}M_{bc}M_{cd}M_{da}$ is a parallelogram.

Rectangle

A rectangle is a quadrilateral in which the four interior angles measure 90º.

All the properties that apply to parallelograms, also apply to rectangles, since a rectangle is just a special parallelogram.

One property that applies to rectangles, but does not apply to parallelograms, is the congruence of the diagonals. ABCD (figure below) is rectangle, and AC and BD are its diagonals:

Diagonal AC forms a right triangle ADC. And diagonal BD forms a right triangle BCD.
These two right triangles are similar, since cathetus AD measures the same as cathetus BC (opposite sides of the rectangle), and cathetus CD is the same in both triangles. Therefore, hypotenuse AC is the same length as BD.

Rhombus

Rhombus is a special parallelogram, whose four sides are the same length. All the properties that apply to parallelograms, also apply to rhombus.

The diagonals of any rhombus bisect the interior angles, and intersect each other forming right angles. In the figure below, ABCD is a rhombus, and the red lines are its diagonals:

Diagonal DB divides rhombus ABCD into two isosceles triangles BCD and BAD. These two triangles are similar, because they have all three sides congruent. Thus, angle $C\hat{B}D$ is congruent to angle $A\hat{B}D$, that is, diagonal DB bisects angle $A\hat{B}C$.

Since in any parallelogram adjacent angles are supplementary:
$D\hat{A}B+A\hat{B}C=180º$

Dividing both sides of the equation by 2:
$\frac{D\hat{A}B}{2}+\frac{A\hat{B}C}{2}=90º$

$\frac{D\hat{A}B}{2}$ and $\frac{A\hat{B}C}{2}$ are the two interior angles of triangle ABM. Therefore, angle $A\hat{M}B=90º$ (the two diagonals are perpendicular to each other).

Square

A square is a rectangle with four sides that are the same length; it is also a rhombus with four interior angles that measure 90º. Therefore, a square share all the properties of rectangles, rhombus and parallelograms. For example:

P1: Diagonals are interior angle bisectors;
P2: Diagonals intersect each other in their midpoint;
P3: Diagonals are perpendicular to each other.

Solved SAT Practice Tests

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