Chapter 12 — Quadrilaterals
Topics in this chapter:
12.1 The Fundamentals
A quadrilateral is a four-sided polygon.
- Convex: all interior angles $< 180°$.
- Concave: one interior angle is a reflex angle ($> 180°$).
A diagonal is a segment from a vertex to a non-adjacent vertex. Every quadrilateral has two diagonals.
A quadrilateral is orthodiagonal if its diagonals are perpendicular.
Angle Sum. The sum of the interior angles of any quadrilateral is $360°$.
Proof: A diagonal splits the quadrilateral into two triangles, each with angle sum $180°$.
Vertices are always named in order when we name a quadrilateral. We would write $ABCD$, never $ACBD$.
Quadrilateral Hierarchy
Each type inherits all properties of its parent types.
12.2 Trapezoids
A trapezoid is a quadrilateral with exactly one pair of parallel sides (bases). The non-parallel sides are legs.
If $AB \parallel CD$:
\[\angle ABC + \angle BCD = \angle BAD + \angle ADC = 180°\]Altitude and Median
- Altitude (height) $h$: the perpendicular distance between the bases.
- Median (midsegment) $XY$: the segment connecting the midpoints of the legs. It is parallel to the bases and its length is the average of the two bases:
Area of a Trapezoid
The proof of the median length uses similar triangles formed by drawing altitudes from the parallel sides.
Isosceles Trapezoids
A trapezoid is isosceles if its legs are equal. Equivalent conditions:
- Base angles are congruent: $\angle ADC = \angle BCD$.
- Diagonals are equal: $AC = BD$.
Example 12-1. An isosceles trapezoid has altitude $4$ and leg length $8$. The smaller base is $5$. Find the area.
Solution: Draw altitudes $AX$ and $BY$. Then $XY = AB = 5$. By the Pythagorean Theorem on $\triangle BYC$: $CY = \sqrt{8^2 - 4^2} = 4\sqrt{3}$. By symmetry $DX = 4\sqrt{3}$.
\[CD = DX + XY + YC = 5 + 8\sqrt{3}\] \[[ABCD] = \frac{AB + CD}{2} \cdot h = \frac{10 + 8\sqrt{3}}{2} \cdot 4 = 20 + 16\sqrt{3}\]Exercise 12-1. Find the area of a trapezoid with height $3$ and bases whose average length is $6$.
Exercise 12-2. One angle of a trapezoid is $20°$. Find $x$ such that another angle must be $x°$.
12.3 Parallelograms
A parallelogram has both pairs of opposite sides parallel. Key properties:
| Property | Statement |
|---|---|
| Opposite sides | Equal: $AB = CD$, $BC = DA$ |
| Opposite angles | Equal: $\angle A = \angle C$, $\angle B = \angle D$ |
| Consecutive angles | Supplementary: $\angle A + \angle B = 180°$ |
| Diagonals | Bisect each other |
Area Formulas
Formula 1 — Base × Height:
\[[ABCD] = \text{base} \times \text{height} = (EF)(XY)\]Formula 2 — Two sides and included angle:
\[[ABCD] = (AB)(BC)\sin B\]Formula 3 — Diagonals and included angle:
\[[ABCD] = \frac{1}{2}d_1 d_2 \sin\theta\]where $d_1$, $d_2$ are the diagonal lengths and $\theta$ is the angle between them.
Example 12-2. The opposite angles of a parallelogram are $3x + 20°$ and $40 - x°$. Find one of the other angles.
Solution: Opposite angles are equal: $3x + 20 = 40 - x \implies x = 5$. Each is $35°$. The other angles are $180° - 35° = 145°$.
Exercise 12-3. Parallelogram with sides $3$ and $6$. The angle opposite the included angle is $30°$. Find the area.
Exercise 12-4. Diagonals of parallelogram $EFGH$ meet at $X$. Find the distance from $X$ to $EF$ if $EF = 8$ and $[EFGH] = 56$.
12.4 Rhombuses
A rhombus is a quadrilateral with all four sides equal.
Properties
Every rhombus is also a parallelogram (proven by SSS on triangles formed by a diagonal). Additionally:
- The diagonals are perpendicular bisectors of each other.
- The diagonals bisect the vertex angles.
- Area:
where $d_1, d_2$ are the diagonal lengths.
Example 12-3. Find the side of a rhombus with area $40$ and diagonals $2x$ and $3x - 2$.
Solution:
\[40 = \frac{(2x)(3x - 2)}{2} = 3x^2 - 2x\]Solving $3x^2 - 2x - 40 = 0$: $x = 4$ (rejecting the negative root). Diagonals are $8$ and $10$.
Since the diagonals bisect each other perpendicularly, each quarter-triangle has legs $4$ and $5$:
\[\text{side} = \sqrt{4^2 + 5^2} = \sqrt{41}\]Exercise 12-5. Two sides of a rhombus are $3x + 2$ and $x + 7$. Find the perimeter.
Exercise 12-6. One diagonal of a rhombus is $10$. Find the other if a side is $17$.
12.5 Rectangles and Squares
Rectangles
A rectangle has four right angles. With length $l$ and width $w$:
| Property | Formula |
|---|---|
| Perimeter | $2l + 2w$ |
| Area | $lw$ |
| Diagonal | $\sqrt{l^2 + w^2}$ |
All rectangles are parallelograms, so their diagonals bisect each other. The diagonals are also equal.
Example 12-4. Rectangle with area $40$ and diagonal $10$. Find the perimeter.
Solution: $lw = 40$, $\sqrt{l^2 + w^2} = 10 \implies l^2 + w^2 = 100$.
\[(l + w)^2 = l^2 + w^2 + 2lw = 100 + 80 = 180\] \[l + w = 6\sqrt{5} \implies \text{Perimeter} = 12\sqrt{5}\]Exercise 12-7. The diagonals of a rectangle intersect $5$ units from one side and $3$ units from another. Find the area.
Exercise 12-8. A diagonal forms a $30°$ angle with a side and has length $8$. Find the perimeter.
Squares
A square has all sides equal and all angles equal. With side $s$:
| Property | Formula |
|---|---|
| Perimeter | $4s$ |
| Area | $s^2$ |
| Diagonal | $s\sqrt{2}$ |
A square is both a rectangle and a rhombus.
Example 12-5. Connecting the midpoints of a square’s sides in order forms another square.
Proof: The four corner triangles are isosceles right triangles with $\angle AEH = \angle BEF = 45°$, so $\angle HEF = 90°$. Similarly for the other three angles. By LL congruence the triangles are congruent, so their hypotenuses (sides of $EFGH$) are equal. Equal sides + right angles = square.
Exercise 12-9. One diagonal of a square has length $8$. Find the area.
Exercise 12-10. Square $ABCD$ with side $6$. Point $E$ on $AB$ is twice as far from $A$ as from $B$. $F$ on $CD$ is twice as far from $C$ as from $D$. Find $EF$.
12.6 Hints and Problems
Important. The converse of a true statement is not necessarily true. For example, the diagonals of a rhombus are perpendicular, but a quadrilateral with perpendicular diagonals is not necessarily a rhombus.
Example 12-6. Prove: a quadrilateral whose diagonals are perpendicular and bisect each other is a rhombus.
Proof: By LL congruence: $\triangle ABE \cong \triangle ADE \cong \triangle CDE \cong \triangle CBE$. The hypotenuses (sides of the quadrilateral) are all equal. Hence ABCD is a rhombus.
Example 12-7. Prove: if opposite angles of a quadrilateral are equal, then it is a parallelogram.
Proof: Draw diagonal $BD$ and label angles. From $\triangle BCD$ and $\triangle ABD$: $w + x + y = w + (z-y) + (z-x) = 180°$. Thus $x + y = 2z - x - y$, giving $z = x + y$. Then $z - x = y$ shows $\angle ABD = \angle BDC$, so $AB \parallel CD$. Similarly $AD \parallel BC$.
Example 12-8. Prove: the midpoints of the sides of a parallelogram form a parallelogram.
Proof: Use SAS congruence on the corner triangles to show opposite sides of the midpoint quadrilateral are equal, then apply: opposite sides equal $\implies$ parallelogram.
Exercise 12-11. Show that $ABCD$ is a parallelogram if $AB = CD$ and $AD = BC$.
Exercise 12-12. $ABCD$ is a trapezoid with $AB \parallel CD$. Prove: if $\angle A = \angle B$, then $ABCD$ is isosceles.
Exercise 12-13. Prove: if the diagonals of a quadrilateral are equal and bisect each other, it is a rectangle.
Exercise 12-14. Prove: the sum of the squares of the sides of a parallelogram equals the sum of the squares of its diagonals.
Exercise 12-15. Use the previous exercise to find medial length $BM$ in $\triangle ABC$ with $AB = 5$, $BC = 6$, $AC = 7$.
End-of-Chapter Problems
197. Prove: if a quadrilateral is orthodiagonal, its area equals half the product of its diagonals. (M&IQ 1991)
198. A rhombus is inscribed in a circle. One diagonal has length $8x$. Find the other. (MAΘ 1990)
199. Find the area of trapezoid $DUCK$ (given: parallel bases $15$ and legs $13$). (MATHCOUNTS 1992)
200. Find the area of a rhombus with side $13$ and one diagonal $24$. (MAΘ 1990)
201. Rectangular lot diagonal $= 37$. Length is $1$ less than $3$ times the width. Find the perimeter. (MAΘ 1987)
202. Trapezoid $ABCD$: $AB \parallel DC$, $AB = 5$, $BC = 3\sqrt{2}$, $\angle BCD = 45°$, $\angle CDA = 60°$. Find $DC$. (AHSME 1984)
203. Prove: if a quadrilateral is orthodiagonal, the midpoints of its sides form a rectangle. (M&IQ 1991)
204. Picture length is $3\times$ width; frame is $4$ in. wide; outer perimeter is $96$ in. Find the picture length. (MATHCOUNTS 1985)
205. Rectangle $ABCD$: $M$ midpoint of $AB$, $AB = 24$, $BC = 18$, $x = DE$. Find $x$ so that $[AMED] = 2[MBCE]$. (MATHCOUNTS 1984)
206. Find the length of the external tangent of two externally tangent circles with radii $8$ and $11$. (MAΘ 1990)
207. The segment joining the midpoints of the diagonals of a trapezoid has length $3$. Longer base is $97$. Find the shorter base. (AHSME 1959)
208. Trapezoid $ABCD$: $AB = 2 \cdot DC$, diagonals intersect at $E$, $AC = 11$. Find $EC$. (AHSME 1972)
209. Prove: connecting the midpoints of any quadrilateral’s sides in order forms a parallelogram.
210. Squares $ABCD$ and $EFGH$ with $AB = 1$. Find the area of $EFGH$. (Mandelbrot #1)
211. Prove: if $ABCD$ is orthodiagonal, then $AB^2 + CD^2 = BC^2 + DA^2$. (M&IQ 1992)
212. Prove: if trapezoid $ABCD$ ($AB \parallel CD$) is orthodiagonal, then $AC^2 + BD^2 = (AB + CD)^2$. (M&IQ 1991)
213. $AB \parallel CD$, $\angle B = 2\angle D$, $CB = a$, $AB = b$. Find $CD$ in terms of $a$ and $b$. (AHSME 1970)
214. Rectangle $ABCD$, $P$ on $AB$, $PS \perp BD$, $PR \perp AC$, $AF \perp BD$, $PQ \perp AF$. Which always equals $PR + PS$? (AHSME 1958)
215. Parallelogram $ABCD$ with area $10$, $AB = 3$, $BC = 5$. $E$, $F$, $G$ on $AB$, $BC$, $AD$ with $AE = BF = AG = 2$. Line through $G$ parallel to $EF$ meets $CD$ at $H$. Find $[EFHG]$. (AHSME 1992)
216. Prove: if isosceles trapezoid $ABCD$ ($AB \parallel CD$) is orthodiagonal, its altitude equals $\frac{AB + CD}{2}$. (M&IQ 1991)
217. $E$ is the intersection of diagonals of convex $ABCD$. $P$, $Q$, $R$, $S$ are circumcenters of $\triangle ABE$, $\triangle BCE$, $\triangle CDE$, $\triangle ADE$. Prove $PQRS$ is a parallelogram. (AHSME 1977)