Lesson 1.7 · Unit 1 · G-CO.1
The names the rest of the course assumes you already have
This lesson is vocabulary, and it earns its place because almost every later theorem is stated in these terms. A student who is unsure whether a figure counts as a polygon, or what regular means, will misread the hypothesis of a theorem and apply it where it does not hold.
- A polygon is a closed plane figure made of segments that meet only at their endpoints, with exactly two meeting at each endpoint.
- Each condition excludes something. Closed excludes an open path, segments excludes curves, and meeting only at endpoints excludes a figure whose sides cross.
- Name by the number of sides: triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon, and \( n \)-gon in general.
- A polygon is convex when no diagonal falls outside it, equivalently when every interior angle measures less than \( 180^\circ \).
- It is concave when at least one interior angle exceeds \( 180^\circ \), which makes the figure appear dented.
- A regular polygon is both equilateral and equiangular. Either alone is not enough: a rhombus is equilateral but usually not regular, and a rectangle is equiangular but usually not.
- A diagonal joins two nonconsecutive vertices, and an \( n \)-gon has \( \dfrac{n(n-3)}{2} \) of them.
- Assume convex unless told otherwise, since most theorems in the course are stated for convex polygons.
Where students lose marks: calling a figure regular because its sides are equal. Regular requires equal angles too. The distinction matters from unit 7 onward, where a theorem about regular polygons will not apply to a rhombus.
The problem. (a) Decide whether each is a polygon, and if so name it and say whether it is convex: a five-sided closed figure with all interior angles under \( 180^\circ \); a figure made of four segments where two of them cross; a closed figure made of three segments and one arc. (b) How many diagonals does an octagon have? (c) Is a rhombus regular? Is a square?
Step one: test the first figure against the definition. It is closed, made of segments, and the segments meet only at endpoints. So it is a polygon, and with five sides it is a pentagon. Every interior angle is under \( 180^\circ \), so it is convex.
Step two: test the second. It is made of segments and is closed, but two segments cross at a point that is not an endpoint of either. The definition requires the sides to meet only at their endpoints, so this is not a polygon at all. A figure like this is sometimes drawn as a "crossed quadrilateral," and no theorem in this course applies to it.
Step three: test the third. It is closed, but one of its four parts is an arc rather than a segment. A polygon is made of segments, so this fails and is not a polygon.
Step four: set up (b) by counting from one vertex. From any one vertex of an octagon, a diagonal can be drawn to every other vertex except itself and its two neighbors, since a segment to a neighbor is a side rather than a diagonal. That is \( 8 - 3 = 5 \) diagonals from each vertex.
Step five: account for double counting. Multiplying gives \( 8 \times 5 = 40 \), but each diagonal has been counted twice, once from each of its two endpoints. So the number is \( 40 \div 2 = 20 \).
Step six: check against the formula. \( \dfrac{n(n-3)}{2} = \dfrac{8(5)}{2} = 20 \). It agrees, which it must, since the formula is exactly the argument just made written in general. A second check on a smaller case: a quadrilateral should have \( \dfrac{4(1)}{2} = 2 \) diagonals, which matches the two that can be drawn in any four-sided figure.
Step seven: answer (c) for the rhombus. A rhombus has four congruent sides, so it is equilateral. Its angles, however, need not be equal: a typical rhombus has two acute and two obtuse angles. Regular requires both conditions, so a general rhombus is not regular.
Step eight: answer (c) for the square, and note the relationship. A square has four congruent sides and four right angles, so it is both equilateral and equiangular and therefore regular. In fact the square is the only regular quadrilateral. The relationship worth carrying forward: a square is a rhombus that happens to be equiangular, and also a rectangle that happens to be equilateral. That kind of hierarchy is what unit 7 formalizes, and the discipline it requires is naming a figure by what has been proved about it rather than by what it looks like.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- Name a polygon with seven sides.
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Heptagon
- What makes a polygon regular?
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All sides congruent and all angles congruent
- Is a circle a polygon?
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It is not made of segments. No
- How many diagonals does a pentagon have?
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\( \dfrac{5(2)}{2} = 5 \). 5
- What does concave mean?
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At least one interior angle exceeds \( 180^\circ \)
- How many diagonals does a decagon have?
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\( \dfrac{n(n-3)}{2} \) with \( n = 10 \): \( \dfrac{10(7)}{2} = 35 \). Checking the reasoning directly: from each of the 10 vertices, diagonals go to \( 10 - 3 = 7 \) others, giving 70 endpoints, and each diagonal has two endpoints, so \( 70 \div 2 = 35 \). 35
- A polygon has 27 diagonals. How many sides does it have?
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Set the formula equal to 27: \( \dfrac{n(n-3)}{2} = 27 \), so \( n(n - 3) = 54 \) and \( n^2 - 3n - 54 = 0 \). Factoring: two numbers multiplying to \( -54 \) and adding to \( -3 \) are \( -9 \) and 6, so \( (n - 9)(n + 6) = 0 \), giving \( n = 9 \) or \( n = -6 \). A polygon cannot have a negative number of sides, so \( n = 9 \). Check: \( \dfrac{9(6)}{2} = 27 \). Correct. The figure is a nonagon. 9 sides
- Is an equilateral polygon necessarily regular? Is an equiangular one?
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Neither implication holds for polygons in general, and a counterexample settles each. Equilateral but not regular: a rhombus that is not a square. All four sides are congruent, but the angles come in two different sizes, so it is not equiangular. Equiangular but not regular: a rectangle that is not a square. All four angles are right angles, but the sides come in two different lengths. So both conditions are genuinely needed in the definition of regular. The triangle is the exception worth knowing: an equilateral triangle is necessarily equiangular and the reverse also holds, which lesson 5.7 proves. Triangles are rigid in a way other polygons are not, and that rigidity is why the triangle congruence criteria of unit 5 exist at all. No to both; a rhombus and a rectangle are the counterexamples, though triangles are a special case where each does imply the other
- Explain why the definition of a polygon requires the sides to meet only at their
endpoints.
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Without that condition, a figure whose sides cross in the middle would count as a polygon, and almost every theorem in the course would become false. Consider a four-segment closed path where two sides cross, sometimes drawn as a bowtie. Its interior is not well defined, so "interior angle" has no clear meaning and the angle sum formula of lesson 7.1 does not apply. Its area cannot be computed by the usual methods because the two loops would count with opposite orientation. The condition is doing real work: it guarantees that the figure separates the plane into a well-defined inside and outside, which is what makes interior angles, area, and the whole apparatus of unit 7 and unit 11 meaningful. This is a general feature of geometric definitions. Each clause excludes a specific pathological case, and reading a definition carefully means asking what each clause is there to rule out. It guarantees a well-defined interior, without which interior angles and area would be meaningless
- A regular polygon has a diagonal count equal to twice its number of sides. Find the
polygon.
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Let \( n \) be the number of sides. The condition is \( \dfrac{n(n-3)}{2} = 2n \). Multiply both sides by 2: \( n(n - 3) = 4n \), so \( n^2 - 3n = 4n \) and \( n^2 - 7n = 0 \). Factor: \( n(n - 7) = 0 \), giving \( n = 0 \) or \( n = 7 \). A polygon must have at least three sides, so \( n = 0 \) is rejected on those grounds and \( n = 7 \). Check: a heptagon has \( \dfrac{7(4)}{2} = 14 \) diagonals, and twice its number of sides is \( 2(7) = 14 \). They match. Note that the factoring had to avoid dividing both sides by \( n \), which would have discarded the root \( n = 0 \). It happens to be a root that gets rejected anyway, but relying on that is luck rather than method. A heptagon, with 7 sides and 14 diagonals