Quadrilaterals & Polygons

Study Sheet

Quadrilaterals & Polygons

Vocabulary, angle formulas, special quadrilaterals, and the family tree

Polygon Vocabulary

Concept
What is a polygon?

A polygon is a closed, flat shape made of straight line segments. Each segment is a side, each corner where two sides meet is a vertex, and a segment that connects two non-adjacent vertices is a diagonal.

  • Convex: no vertex is “pushed in”; every diagonal stays inside the shape.
  • Concave: at least one vertex is “pushed in” (it has a dent), and at least one diagonal falls outside.
  • Regular: all sides congruent and all angles congruent.
  • Irregular: not regular (sides or angles differ).

A polygon with nn sides has interior angles summing to (n−2)⋅180∘(n - 2)\cdot180^\circ and exterior angles summing to 360∘360^\circ. The special quadrilaterals form a family tree: parallelograms (opposite sides parallel and equal, diagonals bisect each other), then rectangles, rhombi, and squares, plus trapezoids and kites with their own diagonal facts.

Example
Convex vs. concave
Dconvexconcave
The dashed diagonal of the concave pentagon runs outside it.

The pentagon on the left is convex — it bulges outward at every vertex, and every diagonal stays inside. The notched pentagon on the right is concave: vertex DD is pushed inward, and the diagonal joining its two neighbours (dashed) passes outside the shape.

Side note
Where it goes wrong

Assuming a rhombus is a square or a rectangle is a rhombus. Both are parallelograms with an extra property — equal sides for the rhombus, right angles for the rectangle — and only the square has both.

Concept
Counting diagonals

From a single vertex you can draw n−3\mathbf{n-3} diagonals (you cannot connect a vertex to itself or to its two neighbors). Counting every diagonal of the whole polygon:

number of diagonals=n(n−3)2.\text{number of diagonals} = \frac{n(n-3)}{2}.

Why the formula: each of the nn vertices sends a diagonal to every vertex except itself and its two neighbours — n−3n - 3 of them. That counts n(n−3)n(n - 3) ends of diagonals, but every diagonal has two ends and has been counted once from each, so divide by 22.

Example
Diagonals of a hexagon
A
Solid: the 3 diagonals from A. Dashed: the other 6, for 9 in all.

A hexagon has n=6n=6 sides. From one vertex: 6−3=36-3 = 3 diagonals. In all:

6(6−3)2=6⋅32=9 diagonals.\frac{6(6-3)}{2} = \frac{6\cdot 3}{2} = 9 \text{ diagonals.}
Tip

Memory hook. “Concave” has the word cave in it — a concave polygon has a cave (a dent). Convex shapes have no caves.

Try it
Quick check

How many diagonals does a decagon have? How many start at any one vertex?

Show the answer

From one vertex: 10−3=710 - 3 = \mathbf{7}. In all: 10⋅72=35\tfrac{10 \cdot 7}{2} = \mathbf{35} (the 7070 vertex-counts count each diagonal twice).

Sum of the Interior Angles

Concept
The interior-angle-sum formula

Every polygon can be split into triangles by drawing all diagonals from one vertex. A polygon with nn sides splits into n−2n-2 triangles, and each triangle holds 180∘180^\circ. So:

sum of interior angles=(n−2)⋅180∘.\text{sum of interior angles} = (n-2)\cdot 180^\circ .

Why n−2n - 2 triangles: from one vertex you can draw diagonals to every vertex except itself and its two neighbours, which is n−3n - 3 diagonals, and n−3n - 3 cuts make n−2n - 2 pieces. The angles of those triangles exactly fill the polygon's angles — nothing extra at the vertex you started from, nothing missing elsewhere — so the total is (n−2)⋅180∘(n - 2) \cdot 180^\circ. (For a concave polygon, start from the dented vertex's neighbour so every diagonal stays inside; the total is the same.)

Example
Interior-angle sum of a pentagon
180°180°180°
Diagonals from one vertex: 3 triangles, 3 ×\times 180∘^\circ = 540∘^\circ.

A pentagon has n=5n=5 sides. Drawing diagonals from one vertex makes 5−2=35-2 = 3 triangles.

(5−2)⋅180∘=3⋅180∘=540∘.(5-2)\cdot 180^\circ = 3\cdot 180^\circ = 540^\circ .
Example
Finding a missing angle
100°115°120°90°?
Drawn with the four given angles; the fifth closes the shape at 115∘^\circ.

Four angles of a pentagon are 100∘100^\circ, 115∘115^\circ, 120∘120^\circ, and 90∘90^\circ. Find the fifth. The five angles must total 540∘540^\circ:

100+115+120+90=425,540−425=115.100+115+120+90 = 425,\qquad 540-425 = 115.

The missing angle is 115∘115^\circ.

Side note
Where it goes wrong

Using n⋅180∘n \cdot 180^\circ or (n−1)⋅180∘(n - 1) \cdot 180^\circ. A pentagon's angles total 540∘540^\circ, not 900∘900^\circ or 720∘720^\circ; check the formula on a triangle (n=3n = 3 gives 180∘180^\circ) or a rectangle (n=4n = 4 gives 360∘360^\circ).

Tip

Tip. Always find the total first with (n−2)⋅180∘(n-2)\cdot 180^\circ, then subtract the angles you already know to reveal the missing one.

Try it
Quick check

The angles of a hexagon are xx, xx, 2x2x, 2x2x, 2x2x and 2x2x degrees. Find xx and the largest angle.

Show the answer

A hexagon's angles total (6−2)⋅180∘=720∘(6 - 2) \cdot 180^\circ = 720^\circ. So 10x=72010x = 720 and x=72x = 72; the largest angle is 2x=144∘2x = \mathbf{144^\circ}.

Each Interior Angle of a Regular Polygon

Concept
Sharing the total equally

In a regular polygon all nn interior angles are equal, so each one is the total divided by nn:

each interior angle=(n−2)⋅180∘n.\text{each interior angle} = \frac{(n-2)\cdot 180^\circ}{n}.

As nn grows the angle creeps toward 180∘180^\circ but never reaches it: (n−2)180n=180−360n\tfrac{(n-2)180}{n} = 180 - \tfrac{360}{n}, and 360n\tfrac{360}{n} shrinks without vanishing. A regular polygon with many sides looks more and more like a circle, whose "corners" are flat.

Example
Interior angle of a regular hexagon
120°120°
4 triangles make 720∘^\circ, shared equally by 6 angles.

A regular hexagon has n=6n=6.

(6−2)⋅180∘6=4⋅180∘6=720∘6=120∘.\frac{(6-2)\cdot 180^\circ}{6} = \frac{4\cdot 180^\circ}{6} = \frac{720^\circ}{6} = 120^\circ .

Every interior angle of a regular hexagon is 120∘120^\circ.

Side note
Where it goes wrong

Dividing 360∘360^\circ by nn and calling it the interior angle. For a hexagon, 3606=60∘\tfrac{360}{6} = 60^\circ is the exterior angle (and the central angle); the interior angle is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ.

Tip

Watch out. The formula (n−2)180∘n\frac{(n-2)180^\circ}{n} gives the measure of one angle only when the polygon is regular. In an irregular polygon the angles can be all different.

Try it
Quick check

Find each interior angle of a regular 2020-gon.

Show the answer

(20−2)⋅180∘20=3240∘20=162∘\dfrac{(20 - 2) \cdot 180^\circ}{20} = \dfrac{3240^\circ}{20} = \mathbf{162^\circ} (or 180∘−360∘20=180∘−18∘180^\circ - \tfrac{360^\circ}{20} = 180^\circ - 18^\circ).

Sum of the Exterior Angles

Concept
The exterior angles always total 360∘360^\circ

At each vertex, extend one side; the angle between that extension and the next side is an exterior angle. No matter how many sides a convex polygon has,

sum of exterior angles=360∘.\text{sum of exterior angles} = 360^\circ .

For a regular polygon each exterior angle is the same:

each exterior angle=360∘n.\text{each exterior angle} = \frac{360^\circ}{n}.

An interior angle and its exterior angle form a linear pair, so they add to 180∘180^\circ.

Why 360∘360^\circ: walk around the polygon along its sides. At each corner you turn through the exterior angle, and after the last corner you face exactly the way you started — one full turn, 360∘360^\circ, whatever the number of sides. Algebra agrees: the nn linear pairs total 180n180n, the interior angles take (n−2)180(n - 2)180 of it, and 180n−(n−2)180=360180n - (n - 2)180 = 360.

Example
Exterior angle of a regular octagon
45°45°45°45°
Walk around the octagon: eight 45∘^\circ turns make one full turn.

A regular octagon has n=8n=8, so each exterior angle is

360∘8=45∘.\frac{360^\circ}{8} = 45^\circ .

Then each interior angle is 180∘−45∘=135∘180^\circ - 45^\circ = 135^\circ.

Example
Working backward from an exterior angle

A regular polygon has an exterior angle of 30∘30^\circ. How many sides?

360∘n=30∘  ⇒  n=36030=12.\frac{360^\circ}{n} = 30^\circ \;\Rightarrow\; n = \frac{360}{30} = 12.

It is a regular dodecagon (12 sides).

Side note
Where it goes wrong

Putting an interior angle into n=360En = \tfrac{360}{E}. With an interior angle of 150∘150^\circ, 360150=2.4\tfrac{360}{150} = 2.4 sides is nonsense; first convert, E=180∘−150∘=30∘E = 180^\circ - 150^\circ = 30^\circ, then n=12n = 12.

Tip

Fast trick. To find the number of sides of a regular polygon, it is usually easier to use the exterior angle: n=360∘exterior anglen = \dfrac{360^\circ}{\text{exterior angle}}.

Try it
Quick check

Each interior angle of a regular polygon is 156∘156^\circ. How many sides does it have?

Show the answer

Exterior angle =180∘−156∘=24∘= 180^\circ - 156^\circ = 24^\circ, so n=36024=15n = \tfrac{360}{24} = \mathbf{15}.

Properties of Parallelograms

Concept
What makes a parallelogram?

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. From that one fact, four properties always follow:

  • Opposite sides are congruent.
  • Opposite angles are congruent.
  • Consecutive (next-door) angles are supplementary (add to 180∘180^\circ).
  • The diagonals bisect each other (cut each other in half).

All four properties come from one diagonal. Diagonal AC‾\overline{AC} is a transversal for both pairs of parallel sides, so it makes two pairs of equal alternate interior angles, and it is shared by the triangles ABCABC and CDACDA; they are congruent by ASA. Corresponding parts give equal opposite sides and equal opposite angles. Consecutive angles are same-side interior angles between parallel lines, so they add to 180∘180^\circ. Drawing both diagonals gives four small triangles, and the congruent opposite pairs (ASA again) show the diagonals cut each other in half.

Example
Reading a parallelogram
70°110°70°110°ABCD
Opposite angles equal; neighbours add to 180∘^\circ.

In parallelogram ABCDABCD, side AB∥DCAB \parallel DC and AD∥BCAD \parallel BC. Because opposite angles are equal, m∠A=m∠Cm\angle A = m\angle C and m∠B=m∠Dm\angle B = m\angle D. Because consecutive angles are supplementary, m∠A+m∠B=180∘m\angle A + m\angle B = 180^\circ. If m∠A=70∘m\angle A = 70^\circ, then m∠C=70∘m\angle C = 70^\circ and m∠B=m∠D=110∘m\angle B = m\angle D = 110^\circ.

Example
Diagonals bisect each other
5PQRSM
Each diagonal is cut in half at M.

The diagonals of parallelogram PQRSPQRS meet at MM. Since they bisect each other, PM=MRPM = MR and QM=MSQM = MS. If PM=5PM = 5, then PR=10PR = 10.

Side note
Where it goes wrong

Assuming the diagonals are equal. In parallelogram PQRSPQRS above, PRPR is clearly longer than QSQS; diagonals are equal only in a rectangle. "Bisect each other" says PM=MRPM = MR and QM=MSQM = MS — two separate equalities, not PM=QMPM = QM.

Tip

Remember. “Opposite parts match, next-door angles add to 180∘180^\circ.” Diagonals bisect each other, but in a general parallelogram they are not equal and not perpendicular.

Try it
Quick check

Two consecutive angles of a parallelogram measure (3x+10)∘(3x + 10)^\circ and (5x−30)∘(5x - 30)^\circ. Find all four angles.

Show the answer

Consecutive angles are supplementary: 8x−20=1808x - 20 = 180, so x=25x = 25. The angles are 3(25)+10=85∘3(25) + 10 = 85^\circ and 5(25)−30=95∘5(25) - 30 = 95^\circ; opposite angles repeat them: 85∘,95∘,85∘,95∘\mathbf{85^\circ, 95^\circ, 85^\circ, 95^\circ}.

Special Parallelograms: Rectangle, Rhombus, Square

Concept
Extra powers

Each special parallelogram keeps all parallelogram properties and adds more:

  • Rectangle: four right angles; diagonals are congruent.
  • Rhombus: four congruent sides; diagonals are perpendicular and bisect the angles.
  • Square: a rectangle and a rhombus — four right angles and four equal sides; diagonals are congruent, perpendicular, and bisect the angles.

Each extra diagonal fact has a one-line reason. Rectangle: triangles ABCABC and BADBAD share AB‾\overline{AB}, have BC=ADBC = AD, and right angles at BB and AA, so they are congruent by SAS and AC=BDAC = BD. Rhombus: both ends of diagonal BD‾\overline{BD} are equally far from AA and CC (all sides are equal), and the points equidistant from AA and CC form the perpendicular bisector of AC‾\overline{AC} — so BD‾\overline{BD} is that perpendicular bisector.

Example
A rhombus and a square with their diagonals
Rhombus: equal sides, perpendicular diagonals. Square: also right angles and equal diagonals.

The rhombus (left) has four equal sides and diagonals that cross at a right angle. The square (right) has everything: equal sides, right angles, and equal, perpendicular diagonals.

Side note
Where it goes wrong

Reversing the diagonal facts. Rectangles have equal diagonals, rhombi have perpendicular ones. A long thin rectangle has diagonals crossing at a sharp angle, far from 90∘90^\circ.

Tip

Diagonal shortcut. Rectangle →\to diagonals are equal (R for Right angles). Rhombus →\to diagonals are perpendicular. Square →\to both at once.

Try it
Quick check

In rectangle ABCDABCD the diagonals are AC=3x+4AC = 3x + 4 and BD=5x−6BD = 5x - 6. Find ACAC.

Show the answer

A rectangle's diagonals are equal: 3x+4=5x−63x + 4 = 5x - 6, so x=5x = 5 and AC=19AC = \mathbf{19}.

Trapezoids and Kites

Concept
Trapezoid parts

A trapezoid has exactly one pair of parallel sides, called the bases. The other two sides are the legs. The midsegment connects the midpoints of the legs; it is parallel to the bases and its length is the average of the two bases:

midsegment=b1+b22.\text{midsegment} = \frac{b_1 + b_2}{2}.

An isosceles trapezoid has congruent legs; its base angles are congruent and its diagonals are congruent.

Why the midsegment is the average: draw a diagonal. It splits the trapezoid into two triangles, and the midsegment splits into two triangle midsegments — one half of the bottom base, one half of the top base. So its length is 12b1+12b2\tfrac12 b_1 + \tfrac12 b_2, and it is parallel to both bases.

Example
Midsegment of a trapezoid
1068
Joining the midpoints of the legs: halfway between 6 and 10.

A trapezoid has bases b1=6b_1 = 6 and b2=10b_2 = 10. The midsegment is

6+102=162=8.\frac{6+10}{2} = \frac{16}{2} = 8.
Side note
Where it goes wrong

Averaging the legs instead of the bases. The midsegment runs between the legs, parallel to the bases, and it is the bases it averages. A trapezoid with bases 66 and 1010 and legs 55 and 55 has midsegment 88, not 55.

Example
A kite
Two pairs of equal neighbouring sides; the long diagonal bisects the short one at right angles.

A kite has two pairs of consecutive congruent sides (not opposite). One pair of opposite angles is congruent, the diagonals are perpendicular, and one diagonal bisects the other. Tick marks show the two pairs of equal sides.

Tip

Don't mix them up. A trapezoid has one pair of parallel sides; a kite has two pairs of equal adjacent sides. Both a rhombus and a kite have perpendicular diagonals, but only the rhombus has all four sides equal.

Try it
Quick check

A trapezoid's midsegment is 1313 and one base is 99. Find the other base.

Show the answer

9+b2=13\tfrac{9 + b}{2} = 13, so 9+b=269 + b = 26 and b=17b = \mathbf{17}.

The Quadrilateral Family Tree

Concept
Who is who

Every square is a rectangle and a rhombus; every rectangle and every rhombus is a parallelogram; every parallelogram is a quadrilateral. As you climb down the tree, shapes gain more properties; as you climb up, they become more general.

The tree is a statement about definitions, not appearances: a square is a rectangle because it satisfies the rectangle's definition (four right angles), whatever it is called. That is why a theorem proved for all parallelograms — diagonals bisect each other — is automatically true for rectangles, rhombi and squares, and why a property of the square alone cannot be assumed for its ancestors.

Example
The hierarchy at a glance
quadrilateralparallelogramtrapezoidkiterectanglerhombussquare
Each shape inherits every property of the shapes above it on its path.

A square sits at the bottom because it inherits every property above it.

Example
Always, sometimes, or never?
  • A square is always a rectangle (it has four right angles). ✓
  • A rectangle is sometimes a square (only when its sides are all equal).
  • A parallelogram is never a trapezoid under the “exactly one pair” definition, because a parallelogram has two pairs of parallel sides.
Side note
Where it goes wrong

Reading the tree upside down. "Every square is a rectangle" is true; "every rectangle is a square" is false. Arrows point from the general shape to the special one, and "is a" only travels up.

Tip

Reasoning tip. For “always / sometimes / never,” ask: does the lower shape's definition force the upper property? If yes →\to always. If it can happen but isn't required →\to sometimes. If the definitions clash →\to never.

Try it
Quick check

Always, sometimes, or never? (a) A square is a rhombus. (b) A parallelogram has perpendicular diagonals. (c) A trapezoid (exactly one pair of parallel sides) has two pairs of parallel sides.

Show the answer

(a) Always: a square has four equal sides, which is the rhombus definition. (b) Sometimes: only when it is a rhombus. (c) Never: the definition says exactly one pair.

Going Deeper: Advanced Polygon Ideas

Concept
Cyclic quadrilaterals: opposite angles are supplementary

A cyclic quadrilateral is one whose four vertices all lie on a single circle. Its defining property is beautiful and useful:

opposite angles add to 180∘.\text{opposite angles add to } 180^\circ .

So if ABCDABCD is cyclic, then m∠A+m∠C=180∘m\angle A + m\angle C = 180^\circ and m∠B+m∠D=180∘m\angle B + m\angle D = 180^\circ. (The reason: an inscribed angle is half its intercepted arc, and the two opposite angles together intercept the whole circle, or 360∘360^\circ; half of that is 180∘180^\circ.) Every rectangle and every isosceles trapezoid is cyclic; a general parallelogram is cyclic only when it is a rectangle.

Example
Worked example: a missing angle in a cyclic quadrilateral
95°70°85°110°ABCD
Drawn on a circle: 95 + 85 = 70 + 110 = 180.

Quadrilateral ABCDABCD is inscribed in a circle. Its angles are m∠A=95∘m\angle A = 95^\circ and m∠B=70∘m\angle B = 70^\circ. Find m∠Cm\angle C and m∠Dm\angle D. Opposite angles are supplementary, so ∠A\angle A pairs with ∠C\angle C and ∠B\angle B pairs with ∠D\angle D:

m∠C=180∘−95∘=85∘,m∠D=180∘−70∘=110∘.m\angle C = 180^\circ - 95^\circ = 85^\circ, \qquad m\angle D = 180^\circ - 70^\circ = 110^\circ .

Check: all four angles total 95+70+85+110=360∘95 + 70 + 85 + 110 = 360^\circ, exactly the interior-angle sum of any quadrilateral.

Concept
Varignon's theorem: the midpoint quadrilateral
ABCD
Two sides of the shaded parallelogram are parallel to the dashed diagonal AC.

Take any quadrilateral and mark the midpoint of each of its four sides. Connect those midpoints in order. The result — called the Varignon parallelogram — is always a parallelogram, no matter how lopsided the original shape is.

The key idea is the triangle midsegment: in triangle ABCABC, the segment joining the midpoints of ABAB and BCBC is parallel to diagonal ACAC and half its length. The same holds on the other side, so both pairs of opposite midpoint-sides are parallel — the definition of a parallelogram.

Tip

Consequences of Varignon. The Varignon parallelogram's sides are half the lengths of the original diagonals, so its perimeter equals the sum of the diagonals of the outer quadrilateral, and its area is exactly half the area of the original quadrilateral.

Concept
The apothem and the area of a regular polygon

The apothem aa of a regular polygon is the perpendicular distance from the center to the middle of any side. Slice the polygon into nn congruent triangles, each with base equal to a side ss and height equal to the apothem. Adding the triangle areas gives

Area=12 a P,\text{Area} = \tfrac{1}{2}\,a\,P,

where P=nsP = n s is the perimeter. In words: half the apothem times the perimeter.

Example
Worked example: area of a regular hexagon from its apothem
63√3
One of 6 triangles, each with base 6 and height the apothem 333\sqrt{3}.

A regular hexagon has side length s=6s = 6 and apothem a=33≈5.196a = 3\sqrt{3} \approx 5.196. Find its area. The perimeter is P=ns=6⋅6=36P = n s = 6 \cdot 6 = 36. Then

Area=12 a P=12 (33)(36)=543≈93.5.\text{Area} = \tfrac{1}{2}\,a\,P = \tfrac{1}{2}\,(3\sqrt{3})(36) = 54\sqrt{3} \approx 93.5 .

So the hexagon covers about 93.593.5 square units.

Concept
Area from coordinates: the shoelace formula

When you know the corner coordinates of a polygon, you can find its area directly — no base or height needed. List the vertices in order (all the way around) as (x1,y1),(x2,y2),…,(xn,yn)(x_1,y_1), (x_2,y_2), \dots, (x_n,y_n), then

Area=12 ∣(x1y2−x2y1)+(x2y3−x3y2)+⋯+(xny1−x1yn)∣.\text{Area} = \tfrac{1}{2}\,\bigl| (x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + \cdots + (x_n y_1 - x_1 y_n) \bigr| .

It is called the shoelace formula because the cross-multiplications lace diagonally like shoelaces. Go around once in a consistent direction and take the absolute value at the end.

Example
Worked example: quadrilateral area by the shoelace method
xy1234512345ABCD
A(1,1), B(5,2), C(4,5), D(2,4): area 9.

Find the area of quadrilateral with vertices A(1,1)A(1,1), B(5,2)B(5,2), C(4,5)C(4,5), and D(2,4)D(2,4), taken in order. Set up the “down-then-up” products and subtract:

down products (xiyi+1):1⋅2+5⋅5+4⋅4+2⋅1=2+25+16+2=45,up products (yixi+1):1⋅5+2⋅4+5⋅2+4⋅1=5+8+10+4=27.\begin{array}{rcl} \text{down products } (x_i y_{i+1}): & & 1\cdot2 + 5\cdot5 + 4\cdot4 + 2\cdot1 = 2+25+16+2 = 45,\\[2pt] \text{up products } (y_i x_{i+1}): & & 1\cdot5 + 2\cdot4 + 5\cdot2 + 4\cdot1 = 5+8+10+4 = 27. \end{array}

Then

Area=12 ∣45−27∣=12 (18)=9 square units.\text{Area} = \tfrac{1}{2}\,|45 - 27| = \tfrac{1}{2}\,(18) = 9 \text{ square units.}
Side note
Where it goes wrong

Listing the vertices out of order. Going A(1,1)A(1,1), C(4,5)C(4,5), B(5,2)B(5,2), D(2,4)D(2,4) traces a self-crossing bow-tie, and the shoelace sum gives 12∣−2∣=1\tfrac12|{-2}| = 1 instead of 99. The vertices must be taken in order around the boundary.

Concept
Finding nn from an angle condition

Angle formulas run both directions: given a single interior or exterior angle of a regular polygon, you can solve for the number of sides nn.

  • From an exterior angle EE:   n=360∘En = \dfrac{360^\circ}{E}.
  • From an interior angle II: first get the exterior angle E=180∘−IE = 180^\circ - I, then n=360∘En = \dfrac{360^\circ}{E}; or solve (n−2)180∘n=I\dfrac{(n-2)180^\circ}{n} = I directly.

A valid answer must be a whole number n≥3n \ge 3 — if the division is not exact, no regular polygon has that angle.

Example
Worked example: which regular polygon has a 150∘150^\circ interior angle?

Each interior angle is I=150∘I = 150^\circ, so each exterior angle is

E=180∘−150∘=30∘,n=360∘30∘=12.E = 180^\circ - 150^\circ = 30^\circ, \qquad n = \frac{360^\circ}{30^\circ} = 12 .

It is a regular dodecagon. As a check, try the direct equation:

(n−2)180n=150  ⇒  180n−360=150n  ⇒  30n=360  ⇒  n=12.✓\frac{(n-2)180}{n} = 150 \;\Rightarrow\; 180n - 360 = 150n \;\Rightarrow\; 30n = 360 \;\Rightarrow\; n = 12 . \checkmark

By contrast, an interior angle of 140∘140^\circ gives E=40∘E = 40^\circ and n=9n = 9 (a nonagon), while 155∘155^\circ gives E=25∘E = 25^\circ and n=14.4n = 14.4 — not a whole number, so no regular polygon has a 155∘155^\circ interior angle.

Concept
Justifying always / sometimes / never

The family tree turns every classification claim into a logic question. Compare the definitions of the two shapes:

  • Always — the first shape's definition forces the second's properties. Every rhombus is a parallelogram: having four equal sides guarantees both pairs of opposite sides are parallel.
  • Sometimes — the property can hold but is not required. A rhombus is sometimes a square: only when its angles are also right angles.
  • Never — the definitions contradict. A square is never a (strict) trapezoid: a square has two pairs of parallel sides, but the strict trapezoid definition demands exactly one.

The habit that never fails: state each definition, then ask whether the first must, may, or cannot satisfy the second.

Tip

Big picture. These advanced tools all come from the same two engines you already know: (1) angles built from 180∘180^\circ and 360∘360^\circ, and (2) breaking a polygon into triangles. Cyclic angles, the apothem area, and Varignon's theorem are just those engines pushed one step further.

Try it
Quick check

Quadrilateral ABCDABCD is inscribed in a circle with m∠A=(2x+10)∘m\angle A = (2x + 10)^\circ and m∠C=(3x−5)∘m\angle C = (3x - 5)^\circ. Find m∠Am\angle A.

Show the answer

∠A\angle A and ∠C\angle C are opposite, so they are supplementary: 5x+5=1805x + 5 = 180, x=35x = 35, and m∠A=2(35)+10=80∘m\angle A = 2(35) + 10 = \mathbf{80^\circ} (with m∠C=100∘m\angle C = 100^\circ).

Why It Works

Interior angle sum: pick one vertex and draw diagonals to the others, cutting the polygon into n−2n - 2 triangles; each contributes 180∘180^\circ. Exterior angles sum to 360∘360^\circ because walking around the polygon you turn through each exterior angle in turn and end up facing the way you started — one full turn. A parallelogram's diagonals bisect each other because the two triangles they form with a pair of opposite sides are congruent by ASA (alternate interior angles from the parallel sides).

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Polygon angle sums
interior sum=(n−2)⋅180∘,exterior sum=360∘\text{interior sum}=(n-2)\cdot 180^\circ,\qquad \text{exterior sum}=360^\circ

What it means. Interior angles grow with the number of sides; exterior angles always total one full turn.

Example. A hexagon (n=6n=6) has interior angle sum (6−2)⋅180∘=720∘(6-2)\cdot180^\circ=720^\circ.

Why it works. Cutting an nn-gon from one vertex to all the others makes n−2n-2 triangles, each contributing 180∘180^\circ. Walking once around the polygon turns you through every exterior angle and returns you to your starting direction — one full 360∘360^\circ rotation.

Tip. For a regular polygon divide by nn: each interior angle is (n−2)180∘n\tfrac{(n-2)180^\circ}{n} and each exterior angle is 360∘n\tfrac{360^\circ}{n}.