6.4 Trapezoids and kites

Trapezoids

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Properties of Trapezoids

A trapezoid is a quadrilateral with exactly one pair of parallel sides. The parallel sides are called bases, and the non-parallel sides are called legs.

An isosceles trapezoid is a trapezoid whose legs are congruent. It comes with bonus properties:

• Base angles are congruent (each pair of angles sharing the same base)
• Diagonals are congruent

The median (also called the midsegment) is the segment connecting the midpoints of the two legs. It has two key properties:

• It's parallel to both bases
• Its length equals the average of the two base lengths: $\text{median} = \frac{b_1 + b_2}{2}$

Trapezoid Median Theorem

The trapezoid median theorem states that the median of a trapezoid is parallel to both bases and has a length equal to the average of the base lengths.

The standard proof uses coordinate geometry or the triangle midsegment theorem. Here's the approach using the midsegment theorem:

1. Given trapezoid $ABCD$ with $AB \parallel CD$, let $M$ be the midpoint of leg $AD$ and $N$ be the midpoint of leg $BC$.
2. Draw diagonal $AC$. This creates triangles $ACD$ and $ABC$.
3. In $\triangle ACD$, the segment from $M$ (midpoint of $AD$) to the point where it meets $AC$ is a midsegment, so it's parallel to $CD$.
4. Similarly, working with $\triangle ABC$ and midpoint $N$, you can show the midsegment is parallel to $AB$.
5. Since both bases are parallel to the median, $MN \parallel AB \parallel CD$.
6. The length relationship $MN = \frac{AB + CD}{2}$ follows from combining the two midsegment lengths.

Problem-Solving with Trapezoids

Example: Find the length of the median in a trapezoid with bases 12 cm and 20 cm.

$\text{median} = \frac{12 + 20}{2} = \frac{32}{2} = 16 \text{ cm}$

Example: The median of a trapezoid is 15 cm and one base is 10 cm. Find the other base.

$15 = \frac{10 + b}{2}$ $30 = 10 + b$ $b = 20 \text{ cm}$

This second type shows up on tests frequently. Just rearrange the median formula to solve for the unknown base.

Constructing an Isosceles Trapezoid

Given the lengths of both bases and the leg length:

1. Draw base $AB$ with the longer base length.
2. Find the difference between the bases and divide by 2. This tells you how far inward each top vertex sits from the endpoints of $AB$.
3. From point $A$, mark a point along $AB$ that distance inward. Do the same from point $B$.
4. At each of those interior marks, construct perpendicular lines upward.
5. From $A$ and $B$, swing arcs with radius equal to the leg length. Where each arc intersects the corresponding perpendicular gives you points $D$ and $C$.
6. Connect $D$ to $C$ to complete the trapezoid.

This method ensures the trapezoid is symmetric about the perpendicular bisector of the bases, which is what makes it isosceles.

Kites

Properties of Kites

A kite is a quadrilateral with two pairs of consecutive (adjacent) congruent sides. If you label it $ABCD$, then $AB \cong AD$ and $CB \cong CD$. Notice the congruent sides share a vertex rather than being opposite each other.

Key properties:

• The diagonals are perpendicular to each other
• The diagonal connecting the vertices where unequal sides meet (the "main diagonal," $AC$ in our labeling) bisects the other diagonal ($BD$)
• One pair of opposite angles is congruent: the angles at $B$ and $D$ (the vertices between the two different pairs of congruent sides)
• The other pair of opposite angles (at $A$ and $C$) are generally not congruent

Kite Diagonal Theorem

The kite diagonal theorem states that the diagonals of a kite are perpendicular, and the diagonal connecting the vertices of the unequal-side pairs bisects the other diagonal.

Proof outline:

1. Consider kite $ABCD$ with $AB \cong AD$ and $CB \cong CD$.
2. Since $AB \cong AD$ and $CB \cong CD$, point $A$ and point $C$ are each equidistant from $B$ and $D$.
3. Any point equidistant from two points lies on the perpendicular bisector of the segment joining them. So both $A$ and $C$ lie on the perpendicular bisector of $BD$.
4. Since $A$ and $C$ both lie on the perpendicular bisector of $BD$, diagonal $AC$ is the perpendicular bisector of $BD$.
5. Therefore $AC \perp BD$, and $AC$ bisects $BD$ at their intersection point $E$.

Problem-Solving with Kites

Example: In kite $ABCD$, diagonal $AC = 24$ cm and diagonal $BD = 18$ cm. $E$ is the intersection of the diagonals. Find $BE$.

Since $AC$ bisects $BD$ (the main diagonal bisects the other):

$BE = ED = \frac{BD}{2} = \frac{18}{2} = 9 \text{ cm}$

Note that $AC$ is not bisected by $BD$ in a general kite. So $AE \neq EC$ unless you're given additional information. Be careful with which diagonal gets bisected.

Example: In the same kite, if $AE = 7$ cm, find the area.

Since the diagonals are perpendicular, the area of a kite is:

$\text{Area} = \frac{d_1 \cdot d_2}{2} = \frac{24 \cdot 18}{2} = 216 \text{ cm}^2$

The perpendicular diagonals property makes area calculations straightforward.

Constructing a Kite

Given two different side lengths (say $a$ and $b$) and the length of the main diagonal:

1. Draw diagonal $AC$ with the given diagonal length.
2. Construct the perpendicular bisector of... actually, you need a different approach. Draw $AC$ first.
3. Construct a circle centered at $A$ with radius $a$ and another circle centered at $C$ with radius $b$.
4. These two circles intersect at two points. Label them $B$ and $D$ (they'll be on opposite sides of $AC$).
5. Connect $A$ to $B$, $B$ to $C$, $C$ to $D$, and $D$ to $A$.

This works because $B$ and $D$ are each at distance $a$ from $A$ and distance $b$ from $C$, giving you the two pairs of adjacent congruent sides.