Quadrilaterals and Polygons — Free Geometry Review Games.

A hexagon has 6 sides.

Sum = (4-2)*180 = 360 degrees.

In a parallelogram, opposite sides are both parallel and congruent.

A square is a rectangle with all sides equal.

A rectangle (quadrilateral) has 2 diagonals.

Sum = (5-2)*180 = 540 degrees.

Sum = (6-2)*180 = 720. Each angle = 720/6 = 120 degrees.

The diagonals of a rhombus bisect each other at right angles.

A trapezoid has exactly one pair of parallel sides (the bases).

A rectangle's diagonals are congruent but not necessarily perpendicular.

Sum = (8-2)*180 = 1080. Each angle = 1080/8 = 135 degrees.

Diagonals = n(n-3)/2 = 6(3)/2 = 9.

Exterior angles sum to 360. Number of sides = 360/36 = 10.

Consecutive angles in a parallelogram are supplementary: 180-70=110.

The sum of exterior angles is always 360 degrees for any convex polygon.

A rhombus is defined as a parallelogram in which all four sides are congruent. This distinguishes it from a general parallelogram, where only opposite sides must be equal. Choice B describes a general parallelogram. Choice C describes a trapezoid. Choice D is false — adjacent sides of a rhombus are not necessarily perpendicular; it is the diagonals of a rhombus that are perpendicular, not the sides.

The prefix 'hepta-' comes from Greek meaning seven, so a heptagon has \(7\) sides. A pentagon has \(5\) sides, a hexagon has \(6\) sides, and an octagon has \(8\) sides. Memorizing Greek number prefixes (penta-, hexa-, hepta-, octa-) helps identify polygon names quickly.

One fundamental property of a parallelogram is that opposite angles are congruent. Choice A is incorrect — consecutive (adjacent) angles in a parallelogram are supplementary, not opposite ones. Choice D would make every parallelogram a rectangle, which is not generally true.

A rectangle always qualifies as a parallelogram because it has two pairs of opposite parallel sides. A rectangle is a rhombus only when all sides are equal (making it a square), so Choice A is not always true. A rectangle is a square only when all sides are equal, so Choice B is not always true. A kite requires two pairs of consecutive congruent sides, which a non-square rectangle does not have.

The sum of interior angles of a triangle is always \(180°\). This follows from the polygon interior angle sum formula \((n - 2) \times 180°\) with \(n = 3\): \((3 - 2) \times 180° = 180°\). Choice D (\(360°\)) is the interior angle sum for a quadrilateral, a common mix-up. This result for triangles is the foundation for deriving the formula for all polygons.

A kite has two pairs of consecutive (adjacent) sides that are congruent — two sides sharing one vertex are equal in length, and the other two sides sharing the opposite vertex are equal in length. Choice A describes a parallelogram. Choice C describes a rhombus. Choice D describes a trapezoid.

A trapezoid is defined as a quadrilateral with exactly one pair of parallel sides, called the bases. A parallelogram, rectangle, and rhombus all have two pairs of parallel opposite sides. The non-parallel sides of a trapezoid are called legs.

Using the interior angle sum formula \((n - 2) \times 180°\) with \(n = 8\): \((8 - 2) \times 180° = 6 \times 180° = 1080°\). Choice A (\(900°\)) is the sum for a heptagon (\(n = 7\)). Choice C (\(1260°\)) corresponds to a nonagon (\(n = 9\)). Choice D (\(1440°\)) corresponds to a decagon (\(n = 10\)).

First, find the total interior angle sum: \((5 - 2) \times 180° = 3 \times 180° = 540°\). Since a regular pentagon has all angles equal, each interior angle \(= \frac{540°}{5} = 108°\). Choice D (\(120°\)) is the interior angle of a regular hexagon, a frequent mix-up. Choice A (\(100°\)) is incorrect because \(100° \times 5 = 500° \neq 540°\).

A square combines the diagonal properties of both a rectangle and a rhombus. From the rectangle, its diagonals are congruent. From the rhombus, its diagonals are perpendicular bisectors of each other and bisect the vertex angles. Choice B would describe a non-square rhombus. Choice C would describe a non-square rectangle.

Consecutive angles in a parallelogram are supplementary. Because the opposite sides are parallel, each side acts as a transversal cutting parallel lines, making consecutive interior angles supplementary and summing to \(180°\). Choice A applies to opposite angles, not consecutive ones. Choice D would force every parallelogram to be a rectangle.

The Trapezoid Midsegment Theorem states that the midsegment is parallel to both bases and its length \(= \frac{b_1 + b_2}{2}\), the average of the two bases. For example, if the bases are \(6\) and \(14\), the midsegment \(= \frac{6 + 14}{2} = 10\). Choice A gives the full sum, not the average.

Using the diagonal count formula \(\frac{n(n-3)}{2}\) with \(n = 5\): \(\frac{5(5-3)}{2} = \frac{5 \times 2}{2} = 5\). Each of the \(5\) vertices connects to \(5 - 3 = 2\) non-adjacent vertices, giving \(5 \times 2 = 10\) directed diagonals; dividing by \(2\) removes double-counting. Choice A (\(3\)) is the diagonal count for a quadrilateral, not a pentagon.

In an isosceles trapezoid, the base angles at each base are congruent — if \(\angle A\) and \(\angle B\) are the lower base angles, then \(\angle A = \angle B\). However, a lower base angle and an upper base angle are supplementary (summing to \(180°\)), making Choice A a tempting but incorrect answer when applied to angles at the same base.

A rhombus has all four sides congruent. If all interior angles are also \(90°\), the figure satisfies the definition of both a rhombus and a rectangle simultaneously, making it a square. Choice A (rectangle) requires right angles but not equal sides. Choice D ('regular quadrilateral') is not a standard geometric classification.

Using the interior angle sum formula \((n - 2) \times 180°\) with \(n = 7\): \((7 - 2) \times 180° = 5 \times 180° = 900°\). Choice A (\(720°\)) is the sum for a hexagon (\(n = 6\)). Choice D (\(1080°\)) is the sum for an octagon (\(n = 8\)). Each additional side adds exactly \(180°\) to the sum.

In a parallelogram, opposite angles are congruent, so \(\angle A = \angle C\). Setting the expressions equal: \(3x + 10 = 5x - 20 \Rightarrow 30 = 2x \Rightarrow x = 15\). A common error is treating consecutive angles as equal rather than supplementary, which leads to an incorrect equation and result.

From each of the \(n\) vertices, diagonals can be drawn to every non-adjacent vertex. Each vertex connects to \(n - 3\) others (excluding itself and its two neighbors), giving \(n(n-3)\) directed diagonals. Since each diagonal is counted from both endpoints, divide by \(2\): \(\frac{n(n-3)}{2}\). Choice A, \(\frac{n(n-1)}{2}\), counts all line segments between pairs of vertices including the sides themselves.

The total interior angle sum is \((12 - 2) \times 180° = 10 \times 180° = 1800°\). Each angle of the regular dodecagon \(= \frac{1800°}{12} = 150°\). Alternatively, each exterior angle \(= \frac{360°}{12} = 30°\), so each interior angle \(= 180° - 30° = 150°\). Choice B (\(144°\)) is the interior angle of a regular decagon (\(10\) sides), a common mix-up.

Using the diagonal formula \(\frac{n(n-3)}{2}\) with \(n = 8\): \(\frac{8(8-3)}{2} = \frac{8 \times 5}{2} = \frac{40}{2} = 20\). An alternative check: the total number of line segments between \(8\) vertices is \(\binom{8}{2} = 28\). Subtracting the \(8\) sides leaves \(28 - 8 = 20\) diagonals. Choice A (\(16\)) results from the error \(\frac{8 \times 4}{2}\), using \(n - 4\) instead of \(n - 3\).

The diagonals of a rhombus are perpendicular bisectors of each other, so $\angle AEB = 90°$. Triangle $AEB$ is a right triangle with legs \(AE = 6\) and \(BE = 8\). By the Pythagorean theorem: \(AB = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\). Since all four sides of a rhombus are congruent, the perimeter \(= 4 \times 10 = 40\). Choice C (\(48\)) results from mistakenly computing \(4 \times 12 = 48\) using the half-diagonal sum rather than the side length.

Since \(AB \parallel CD\), side \(BC\) is a transversal. Angles \(\angle B\) and \(\angle C\) are co-interior (same-side interior) angles, which are supplementary: \(\angle B + \angle C = 180°\), giving \(\angle C = 180° - 65° = 115°\). Verification: \(\angle D = 180° - \angle A = 110°\), and \(70° + 65° + 115° + 110° = 360°\) confirms the quadrilateral angle sum.

The interior angles of any quadrilateral sum to \(360°\). Setting up the equation: \((2x + 15) + (3x - 5) + (x + 30) + 4x = 360 \Rightarrow 10x + 40 = 360 \Rightarrow 10x = 320 \Rightarrow x = 32\). Therefore \(\angle D = 4(32) = 128°\). A common error is using \(180°\) as the total (the triangle angle sum), which gives \(x = 14\) and an incorrect answer of \(56°\).

A key property of a rhombus is that each diagonal bisects the vertex angles. Since diagonal \(\overline{BD}\) bisects $\angle ABC$, we get $\angle ABD = \frac{\angle ABC}{2} = \frac{60°}{2} = 30°$. Choice D (\(60°\)) is a common error from assuming the diagonal does not bisect the angle. Choice C (\(45°\)) would only apply if the rhombus were a square with $\angle ABC = 90°$.

Using the formula \((n - 2) \times 180° = 2520°\): \(n - 2 = \frac{2520}{180} = 14\), so \(n = 16\). This \(16\)-sided polygon is called a hexadecagon. Choice A (\(14\)) results from forgetting to add \(2\) after dividing — a critical step in solving for \(n\). Verify: \((16 - 2) \times 180° = 14 \times 180° = 2520°\).

By definition, a parallelogram has two pairs of opposite sides that are both parallel and congruent. This distinguishes it from other quadrilaterals. Choice A is incorrect because perpendicular sides are not required in a parallelogram — that property only applies to rectangles, which are a special case.

Using the polygon interior angle sum formula \((n-2) \cdot 180°\), for a quadrilateral \(n = 4\): \((4-2) \cdot 180° = 2 \cdot 180° = 360°\). This applies to all quadrilaterals, convex or concave. Choice D (\(540°\)) is the sum for a pentagon, not a quadrilateral.

A rhombus is a parallelogram in which all four sides are congruent. Choice C describes a rectangle (all right angles). Choice D describes a rectangle or isosceles trapezoid — a rhombus has perpendicular diagonals, but they are generally unequal in length unless the rhombus is also a square.

Regardless of the number of sides, the sum of the exterior angles of any convex polygon (one at each vertex) is always \(360°\). This is because as you traverse the polygon, you complete exactly one full rotation. Choice D is the most tempting distractor, but the sum is constant for all convex polygons.

In a rectangle, the diagonals are always congruent (equal in length) and bisect each other. Choice A describes a rhombus, not all rectangles — in a non-square rectangle the diagonals are not perpendicular. Choice B also describes a rhombus; in a non-square rectangle, the diagonals do not bisect the vertex angles.

By definition, a trapezoid has exactly one pair of parallel sides, called the bases. Choice C describes a parallelogram (two pairs of parallel sides). Choice A describes a rhombus, and Choice D describes a rectangle.

A kite has two pairs of consecutive (adjacent) congruent sides. Choice A describes a rhombus. Choice B describes a parallelogram. Choice D is incorrect because in a kite only one diagonal bisects the other; the diagonals do not mutually bisect each other unless the kite is also a rhombus.

For a regular \(n\)-gon, each interior angle equals \(\frac{(n-2) \cdot 180°}{n}\). With \(n = 4\): \(\frac{(4-2) \cdot 180°}{4} = \frac{360°}{4} = 90°\). Choice A (\(60°\)) is the interior angle of an equilateral triangle (\(n=3\)), and Choice D (\(120°\)) is the interior angle of a regular hexagon (\(n=6\)).

In a parallelogram, the diagonals bisect each other, so \(AE = EC\). Setting the expressions equal: \(3x - 4 = x + 8 \Rightarrow 2x = 12 \Rightarrow x = 6\). Then \(AE = 3(6) - 4 = 14\) and \(EC = 6 + 8 = 14\), so \(AC = 14 + 14 = 28\). Choice B (\(14\)) is only half the diagonal — a common error of forgetting to add both halves.

The interior angle sum of a hexagon is \((6-2) \cdot 180° = 720°\). The five given angles sum to \(110° + 125° + 130° + 120° + 105° = 590°\). The sixth angle is \(720° - 590° = 130°\). Choice A (\(100°\)) results from mistakenly using \(540°\) (the pentagon sum) as the total instead of \(720°\).

Each exterior angle equals \(180° - 144° = 36°\). Since the exterior angles sum to \(360°\), the number of sides is \(n = \frac{360°}{36°} = 10\). Alternatively, \(\frac{(n-2) \cdot 180°}{n} = 144° \Rightarrow 180n - 360 = 144n \Rightarrow 36n = 360 \Rightarrow n = 10\). Choice D (\(12\)) yields interior angles of \(150°\), not \(144°\).

In a rectangle, the diagonals are congruent, so \(AC = BD\): \(5x - 3 = 3x + 11 \Rightarrow 2x = 14 \Rightarrow x = 7\). Then \(AC = 5(7) - 3 = 32\). Check: \(BD = 3(7) + 11 = 32\). ✓ Choice A (\(7\)) is the value of \(x\), not the diagonal length — a classic substitution error.

In a rhombus, the diagonals are perpendicular bisectors of each other — they intersect at right angles and each diagonal bisects the other. Choice A is false in general; a rhombus has equal diagonals only when it is also a square. Choice D is false because the diagonals of a rhombus do bisect the vertex angles.

The midsegment of a trapezoid equals the average of the two bases: \(MN = \frac{AB + CD}{2}\). Substituting: \(14 = \frac{10 + CD}{2} \Rightarrow 28 = 10 + CD \Rightarrow CD = 18\). Choice A (\(4\)) results from subtracting \(14 - 10 = 4\), which ignores the averaging formula entirely.

In a parallelogram, opposite angles are congruent and consecutive angles are supplementary. \(\angle C = \angle A = 72°\) (opposite angles). \(\angle B = 180° - 72° = 108°\) (consecutive to \(\angle A\)). \(\angle D = 108°\) (opposite to \(\angle B\)). So \(\angle B = 108°\), \(\angle C = 72°\), \(\angle D = 108°\).

Using \((n-2) \cdot 180° = 1440°\): \(n - 2 = \frac{1440}{180} = 8\), so \(n = 10\). This polygon is a decagon. Choice A (\(8\)) gives a sum of \((8-2) \cdot 180° = 1080°\), not \(1440°\), confirming that \(n = 8\) is too small.

In an isosceles trapezoid, base angles are congruent. Since \(A\) and \(B\) both lie on base \(\overline{AB}\), they are base angles at the same base: \(\angle A = \angle B = 65°\). Choice C (\(115°\)) is the measure of \(\angle D\) (and \(\angle C\)), since co-interior angles on the same leg sum to \(180°\): \(65° + 115° = 180°\).

In a rectangle, a diagonal is the hypotenuse of the right triangle formed by two adjacent sides. By the Pythagorean theorem: \(AC^2 = AB^2 + BC^2 \Rightarrow 100 = 64 + BC^2 \Rightarrow BC^2 = 36 \Rightarrow BC = 6\). Choice C (\(\sqrt{164}\)) results from incorrectly adding: \(\sqrt{10^2 + 8^2} = \sqrt{164}\), confusing the diagonal for a leg rather than the hypotenuse.

The number of sides is \(n = \frac{360°}{24°} = 15\). The interior angle sum is \((15-2) \cdot 180° = 13 \cdot 180° = 2340°\). Choice B (\(2160°\)) corresponds to a \(14\)-sided polygon: \((14-2) \cdot 180° = 2160°\), but a \(14\)-gon has exterior angles of \(\frac{360°}{14} \approx 25.7°\), not \(24°\).

The diagonals of a rhombus bisect each other at right angles, creating four congruent right triangles. The legs of each right triangle are \(\frac{10}{2} = 5\) and \(\frac{24}{2} = 12\). By the Pythagorean theorem, each side of the rhombus is \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). Perimeter \(= 4 \times 13 = 52\). Choice A (\(34\)) incorrectly computes \(2(5 + 12)\) using the half-diagonals directly.

In a parallelogram, opposite angles are congruent: \(\angle A = \angle C\). Setting equal: \(4x - 20 = 2x + 40 \Rightarrow 2x = 60 \Rightarrow x = 30\). Then \(\angle A = 4(30) - 20 = 100°\). Since consecutive angles in a parallelogram are supplementary: \(\angle B = 180° - 100° = 80°\). Choice C (\(100°\)) is the measure of \(\angle A\) and \(\angle C\), not \(\angle B\).

In kite $ABCD$ with \(AB = AD\) and \(CB = CD\), the line of symmetry through \(A\) and \(C\) makes the non-vertex angles equal: $\angle ABC = \angle ADC$. Using the quadrilateral angle sum: $\angle BAD + \angle BCD + \angle ABC + \angle ADC = 360°$, so $80° + 40° + 2\angle ABC = 360° \Rightarrow 2\angle ABC = 240° \Rightarrow \angle ABC = 120°$. Choice A (\(100°\)) results from incorrectly halving \(200°\) instead of \(240°\).

At any vertex, interior and exterior angles are supplementary: interior \(+\) exterior \(= 180°\). With a \(4:1\) ratio, let interior \(= 4k\) and exterior \(= k\). Then \(5k = 180° \Rightarrow k = 36°\). The exterior angle is \(36°\), so \(n = \frac{360°}{36°} = 10\). Choice D (\(12\)) gives exterior angles of \(30°\) and a ratio of \(\frac{150°}{30°} = 5:1\), not \(4:1\).

Since \(\overline{BD}\) bisects $\angle ABC$: $\angle ABD = \angle DBC = 50°$. Because \(AB \parallel DC\), alternate interior angles formed by transversal \(\overline{BD}\) are congruent: $\angle ADB = \angle DBC = 50°$. This is confirmed in $\triangle ABD$: $\angle DAB = 180° - 100° = 80°$ (consecutive angles in a parallelogram), so $\angle ADB = 180° - 80° - 50° = 50°$.

In a rectangle, the diagonals are congruent and bisect each other, so \(EA = EB\) (each half-diagonal is equal). Triangle $AEB$ is therefore isosceles with \(EA = EB\), giving $\angle EAB = \angle EBA$. Since the angles of a triangle sum to \(180°\): $\angle EAB = \frac{180° - 70°}{2} = \frac{110°}{2} = 55°$. Choice B (\(45°\)) would require $\angle AEB = 90°$, which only occurs in a square where the diagonals are perpendicular.

The interior angle sum of a polygon with \(n\) sides is \((n-2) \times 180°\). For a hexagon, \(n = 6\): \((6-2) \times 180° = 4 \times 180° = 720°\). The choice \(540°\) is the interior angle sum of a pentagon (\(n = 5\)).

A trapezoid is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram — and its special cases rhombus and rectangle — has two pairs of parallel sides, so none of those qualify.

The sum of the exterior angles of any convex polygon is always \(360°\), regardless of the number of sides. Traveling around the polygon completes exactly one full rotation. The value \(180°\) is the sum of angles along a straight line, not the exterior angle sum of a polygon.

In a parallelogram, opposite sides are congruent. Since \(AB\) and \(CD\) are opposite sides, \(CD = AB = 9\). Note that \(AD = BC = 6\) — choosing \(6\) would be correct for \(AD\), not \(CD\).

A rectangle is a parallelogram with four right angles. Opposite sides being parallel, opposite sides being congruent, and diagonals bisecting each other are all properties of every parallelogram. Having all angles equal to \(90°\) is the property that specifically distinguishes a rectangle from a general parallelogram.

Using the formula \((n - 2) \times 180° = 540°\): \(n - 2 = 3\), so \(n = 5\). The polygon is a pentagon. Choosing \(4\) would give an interior angle sum of \((4-2) \times 180° = 360°\), which is the sum for a quadrilateral.

Consecutive angles in a parallelogram are supplementary — they sum to \(180°\). This follows because opposite sides are parallel and consecutive angles are co-interior (same-side interior) angles formed by a transversal. Opposite angles in a parallelogram are congruent, but consecutive angles are not.

For a regular \(n\)-gon, each interior angle measures \(\frac{(n-2) \times 180°}{n}\). For a regular octagon (\(n = 8\)): \(\frac{(8-2) \times 180°}{8} = \frac{1080°}{8} = 135°\). The choice \(120°\) is the interior angle of a regular hexagon (\(n = 6\)), and \(150°\) belongs to a regular dodecagon (\(n = 12\)).

Consecutive angles in a parallelogram are supplementary: \(\angle P + \angle Q = 180°\). So \((3x + 15) + (2x + 5) = 180\), giving \(5x + 20 = 180\), then \(5x = 160\) and \(x = 32\). A common error is treating \(\angle P\) and \(\angle Q\) as opposite angles and setting them equal, which gives an invalid result.

The exterior angles of any convex polygon sum to \(360°\). For a regular polygon where all exterior angles are equal: \(n = \frac{360°}{40°} = 9\). The polygon is a regular nonagon. Choosing \(8\) would require each exterior angle to be \(\frac{360°}{8} = 45°\), not \(40°\).

The diagonals of a rhombus bisect each other at right angles, forming four congruent right triangles. Each triangle has legs \(\frac{6}{2} = 3\) and \(\frac{8}{2} = 4\). By the Pythagorean theorem, each side of the rhombus is \(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\). The perimeter is \(4 \times 5 = 20\). Choosing \(14\) adds only the two diagonal lengths.

In a trapezoid, co-interior (same-side interior) angles between the parallel sides are supplementary: \(\angle A + \angle D = 180°\). So \(\angle D = 180° - 65° = 115°\). In an isosceles trapezoid, \(\angle A = \angle B = 65°\) and \(\angle C = \angle D = 115°\). Choosing \(65°\) is the measure of \(\angle B\), not \(\angle D\).

A nonagon has \(n = 9\) sides. The interior angle sum is \((n-2) \times 180° = (9-2) \times 180° = 7 \times 180° = 1260°\). The choice \(1080°\) corresponds to an octagon (\(n = 8\)), while \(1440°\) corresponds to a decagon (\(n = 10\)).

In a kite with \(AB = AD\) and \(CB = CD\), vertices \(B\) and \(D\) are the non-vertex ("wing") angles. By the kite's line of symmetry along diagonal \(\overline{AC}\), reflecting the kite maps \(B\) onto \(D\), so \(\angle B = \angle D\). The vertex angles \(\angle A\) and \(\angle C\) lie on the axis of symmetry and are generally not equal to each other.

In a rectangle, all angles are \(90°\), so triangle $ABC$ is a right triangle with legs \(AB = 6\) and \(BC = 8\). By the Pythagorean theorem: \(AC = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\). Choosing \(7\) might come from incorrectly averaging the two legs, and choosing \(14 = 6 + 8\) adds the legs directly.

The number of diagonals in an \(n\)-sided polygon is \(\frac{n(n-3)}{2}\). For \(n = 12\): \(\frac{12 \times 9}{2} = \frac{108}{2} = 54\). The choice \(66 = \frac{12 \times 11}{2}\) counts all segments connecting any two vertices (including the \(12\) sides themselves), not just the diagonals.

The diagonals of a parallelogram bisect each other, so \(AE = EC\). Setting \(3x - 2 = x + 6\) gives \(2x = 8\), so \(x = 4\). Then \(AE = 3(4) - 2 = 10\) and \(EC = 4 + 6 = 10\). The full diagonal \(AC = AE + EC = 10 + 10 = 20\). Choosing \(10\) is a common error of reporting the half-diagonal \(AE\) rather than the full length.

Using \((n - 2) \times 180° = 2340°\): \(n - 2 = \frac{2340}{180} = 13\), so \(n = 15\). The polygon is a pentadecagon. Choosing \(13\) is a common error of stopping after finding \(n - 2\) instead of adding \(2\) to solve for \(n\).

Consecutive angles in a rhombus are supplementary: $\angle ABC = 180° - 70° = 110°$. The diagonals of a rhombus bisect each vertex angle, so diagonal \(\overline{BD}\) bisects $\angle ABC$: $\angle ABD = \frac{110°}{2} = 55°$. Choosing \(35°\) results from mistakenly bisecting $\angle BCD = 70°$ instead of $\angle ABC$.

The area of a trapezoid is \(A = \frac{1}{2}(b_1 + b_2) \cdot h = \frac{1}{2}(18 + 10)(8) = \frac{1}{2}(28)(8) = 112\) square units. Choosing \(80 = 10 \times 8\) uses only the smaller base. Choosing \(144 = 18 \times 8\) uses only the larger base. Choosing \(224 = (18 + 10) \times 8\) omits the required factor of \(\frac{1}{2}\).

Opposite sides of a parallelogram are congruent: \(AB = CD\). Setting \(2x + 3 = 3x - 5\) gives \(x = 8\). Then \(AB = CD = 2(8) + 3 = 19\) and \(BC = AD = 8 + 7 = 15\). The perimeter is \(2(AB + BC) = 2(19 + 15) = 2(34) = 68\). A common error is equating \(AB = BC\) as if the figure were a rhombus, yielding an incorrect value of \(x\).

Let each exterior angle equal \(x\). Then each interior angle is \(5x\). Since interior and exterior angles at each vertex are supplementary: \(x + 5x = 180°\), giving \(6x = 180°\) and \(x = 30°\). The number of sides is \(n = \frac{360°}{30°} = 12\). Choosing \(10\) corresponds to exterior angle \(36°\) and interior angle \(144°\), a ratio of \(4\), not \(5\).

First find \(AB\): since $ABCD$ is a rectangle, \(AC^2 = AB^2 + BC^2\), so \(400 = AB^2 + 144\), giving \(AB = 16\). Since \(E\) is the midpoint of \(\overline{AB}\), \(AE = 8\). In right triangle $DAE$: \(DA = BC = 12\) and \(AE = 8\), so \(DE = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}\). Choosing \(4\sqrt{10}\) results from incorrectly using \(AE = 4\) (one-quarter of \(AB\)) rather than the midpoint value \(AE = 8\).

Since \(AB = AD = 5\) and $\angle BAD = 90°$, triangle $ABD$ is an isosceles right triangle. By the Pythagorean theorem: \(BD = \sqrt{AB^2 + AD^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\). Choosing \(10 = 2 \times 5\) incorrectly doubles one leg. Choosing \(13\) confuses the diagonal with a value derived from the longer side \(CB = 12\).

Drop perpendiculars from \(D\) and \(C\) to \(\overline{AB}\). Since \(AB - CD = 16 - 8 = 8\), symmetry places each side overhang at \(\frac{8}{2} = 4\). The height is a leg of a right triangle with hypotenuse \(AD = 5\) and other leg \(4\): \(h = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\). Choosing \(4\) mistakes the horizontal overhang for the height. Choosing \(\sqrt{41} = \sqrt{25 + 16}\) adds instead of subtracts under the radical.