A polygon has 14 diagonals. How many sides does it have? Also, for which polygon is the sum of the interior angles equal to the sum of the exterior angles?

Step-by-step solution

1. 1

   Derive the diagonal count rather than memorising it. From each of the $n$ vertices you can draw a segment to $n-3$ others (excluding itself and its two neighbours, which give sides, not diagonals). That counts each diagonal twice, once from each end:

   $D=\frac{n(n-3)}{2}.$

2. 2

   Set the count equal to 14 and solve.

   $\frac{n(n-3)}{2}=14\ \Longrightarrow\ n^{2}-3n-28=0\ \Longrightarrow\ (n-7)(n+4)=0.$

   The roots are $n=7$ and $n=-4$; a polygon cannot have a negative number of sides, so

   $n=7\quad(\text{a heptagon}).$

3. 3

   Check it. A heptagon has $\tfrac{7\times 4}{2}=14$ diagonals ✓.

4. 4

   Recall the two angle-sum facts for part (2). For any convex $n$-gon,

   $\text{interior sum}=(n-2)\cdot 180^\circ,\qquad \text{exterior sum}=360^\circ.$

   The exterior sum is the striking one: it is $360^\circ$ for every polygon, no matter how many sides — walking once around the shape turns you through one full revolution.

5. 5

   Set the two sums equal.

   $(n-2)\cdot 180=360\ \Longrightarrow\ n-2=2\ \Longrightarrow\ n=4.$

6. 6

   State both answers with a check. The polygon with $14$ diagonals has $7$ sides, and the polygon whose interior and exterior angle sums agree is the quadrilateral: its interior angles total $(4-2)180=360^\circ$, matching the universal exterior total of $360^\circ$ ✓. Note also that adding one side leaves the exterior sum unchanged while raising the interior sum by $180^\circ$.