Honors Geometry Unit 6 Review: Quadrilaterals

A trapezoid has a pair of parallel sides called bases. The other two sides are legs. Many textbooks say exactly one pair of parallel sides, but some use at least one pair, which makes every parallelogram a trapezoid. Check which definition your class uses.

Because the bases are parallel, the two angles along each leg are same-side interior angles, so they add to $180^\circ$. Here, an isosceles trapezoid has exactly one pair of parallel sides and congruent legs. Both pairs of base angles are congruent, and the diagonals are congruent.

The midsegment, or median, connects the midpoints of the legs. It is parallel to both bases, and its length is the average of the bases, $\frac{b_1 + b_2}{2}$.

Here, a kite has two pairs of adjacent congruent sides, with the pairs having different lengths. Some definitions also include rhombuses as kites. Its diagonals are perpendicular. The diagonal between the vertices where congruent sides meet bisects the other diagonal and bisects those two vertex angles. The congruent pair of opposite angles is the other pair, where each angle sits between two sides of different lengths. The area of a kite is $\frac{1}{2}d_1d_2$.

Worked example. A trapezoid has bases $2x + 4$ and $3x$, and its midsegment is 17. Set up $\frac{(2x + 4) + 3x}{2} = 17$. Multiply by 2 to get $5x + 4 = 34$, so $x = 6$. The bases are 16 and 18, and their average is 17, which checks.

A quadrilateral is a polygon with four sides. Its interior angles always add to $360^\circ$, because an interior diagonal splits it into two triangles of $180^\circ$ each. If three angles are $85^\circ$, $95^\circ$, and $110^\circ$, the fourth is $360 - 290 = 70^\circ$.

Classifying a quadrilateral means finding its most specific name. Parallelograms have both pairs of opposite sides parallel. Rectangles are parallelograms with four right angles. Rhombuses are parallelograms with four congruent sides. Squares are both rectangles and rhombuses. Whether trapezoids and kites overlap the parallelogram branch depends on the definitions your class uses.

To prove a quadrilateral is a parallelogram, show any one of these:

• both pairs of opposite sides are parallel
• both pairs of opposite sides are congruent
• one pair of opposite sides is both parallel and congruent
• both pairs of opposite angles are congruent
• the diagonals bisect each other

In coordinate problems, slopes test for parallel and perpendicular sides, and the distance formula tests for congruent sides. The midpoint formula shows whether diagonals bisect each other, since bisecting diagonals share a midpoint. Honors courses often ask you to classify a shape from its four vertices this way.

The table below compares the diagonals, since they are often the quickest test.

In parallelogram $ABCD$, opposite sides are parallel by definition. From that, four more facts follow. Opposite sides are congruent, opposite angles are congruent, consecutive angles are supplementary, and the diagonals bisect each other.

The proofs all use triangles. Draw diagonal $\overline{AC}$. Since $\overline{AB} \parallel \overline{CD}$, alternate interior angles give $\angle BAC \cong \angle DCA$. Since $\overline{BC} \parallel \overline{AD}$, you also get $\angle BCA \cong \angle DAC$. The shared side $\overline{AC}$ completes ASA, so $\triangle ABC \cong \triangle CDA$, and CPCTC gives the congruent opposite sides and angles.

A common mix-up is thinking parallelogram diagonals are equal in length. They only cut each other in half. Congruent diagonals belong to rectangles.

Worked example. In parallelogram $ABCD$, $\angle A = (2x + 10)^\circ$ and $\angle B = (3x)^\circ$. These are consecutive, so $2x + 10 + 3x = 180$. Then $5x = 170$ and $x = 34$. So $\angle A = 78^\circ$ and $\angle B = 102^\circ$. Opposite angles match, so $\angle C = 78^\circ$ and $\angle D = 102^\circ$.

Worked example. The diagonals meet at $M$, with $AM = 2y + 1$ and $MC = 4y - 7$. Since the diagonals bisect each other, $2y + 1 = 4y - 7$, so $y = 4$. Each half is 9, and $AC = 18$.

Rectangles, rhombuses, and squares keep every parallelogram property and add more. A rectangle has four right angles, and its diagonals are congruent. A rhombus has four congruent sides. Its diagonals are perpendicular, and each diagonal bisects a pair of opposite angles. A square has everything from both lists, so each corner splits into two $45^\circ$ angles.

To prove these shapes, first show the figure is a parallelogram, then add one condition. A parallelogram with one right angle or with congruent diagonals is a rectangle. A parallelogram with perpendicular diagonals, two congruent consecutive sides, or a diagonal that bisects an angle is a rhombus. A shape that is both a rectangle and a rhombus is a square.

The rhombus diagonals make four congruent right triangles. Each triangle's legs are half-diagonals and its hypotenuse is a side, so the Pythagorean theorem shows up constantly.

Worked example. A rhombus has side 13 and one diagonal of 24. Half of that diagonal is 12. In one right triangle, $12^2 + b^2 = 13^2$, so $b^2 = 169 - 144 = 25$ and $b = 5$. That is half the other diagonal, so the other diagonal is 10. The area is $\frac{1}{2}(24)(10) = 120$.