Convex Polygons

SAT Questions that focus on Convex Polygons require knowledge of the following topics.


A convex polygon is a simple polygon whose all interior angles are less than 180 degrees.

Sum of the Interior Angles of a Convex Polygon

Consider two convex polygons, one with 4 sides and one with 5 sides (following figures):

The polygon with 4 sides can be divided into 2 triangles
The polygon with 5 sides can be divided into 3 triangles.

It can be demonstrated that:
A polygon with 6 sides can be divided into 4 triangles;
A polygon with 7 sides can be divided into 5 triangles;
.....
A polygon with "n" sides can be divided into "n-2" triangles.

Since the sum of the interior angles of a triangle is 180º, the sum of the interior angles of a convex polygon with "n" sides is given by the formula:

$S_n=180(n-2)$

If the polygon is regular (all sides and all interior angles are congruent), the measure of each of the interior angles is given by the formula:

$a_i=\frac{180(n-2)}{n}$ (only for REGULAR polygons).

Sum of the Exterior Angles of a Convex Polygon

The convex polygon in the following figure has "n" sides.
$a, b, c, d,..., e$ are interior angles, and
$a_{ex}, b_{ex}, c_{ex}, d_{ex},..., e_{ex}$ are exterior angles.

Let's denote the sum of the exterior angles as $S_e$. Thus:
$S_e=a_{ex}+b_{ex}+c_{ex}+d_{ex}+...+e_{ex}$

Now let's add the sum of the interior angles to both sides of the equation:

$S_e+a+b+c+d+...+e=a_{ex}+b_{ex}+c_{ex}+d_{ex}+...+e_{ex}+a+b+c+d+...+e$

Since the sum of the interior angles is given by $180(n-2)$:

$S_e+180(n-2)=(a_{ex}+a)+(b_{ex}+b)+(c_{ex}+c)+(d_{ex}+d)+...+(e_{ex}+e)$

The sum of each interior angle and the corresponding exterior angle is 180º (supplementary angles). Thus:

$S_e+180(n-2)=(180)+(180)+(180)+(180)+...+(180)$

$S_e+180n-180(2)=180n$

$S_e=180(2)=360º$

If the polygon is regular (all sides and all interior angles are congruent), the measure of each of its exterior angles will be given by the formula:

$a_e=\frac{360}{n}$ (only for REGULAR polygons).

Number of Diagonals in a Convex Polygon

Consider two convex polygons, one with 4 sides and one with 5 sides (following figures):

From each vertex in the 4-side polygon we can draw only 1 diagonal;
From each vertex in the 5-side polygon we can draw 2 diagonals;

It can be demonstrated that:
From each vertex in a 6-side polygon we can draw only 3 diagonal;
From each vertex in a 7-side polygon we can draw 4 diagonals;
.....
From each vertex in a n-side polygon we can draw (n-3) diagonals;

Since a polygon with "n" sides has "n" vértices, its total number of diagonals is n(n-3). The problem is that in this formula we are counting twice each diagonal (in the polygon with 4 sides in the figure above, for example, the diagonal that goes from B to D is the same that goes from D to B). Therefore, the number of distinct diagonals in a polygon with "n" sides is given by the formula:

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

Solved SAT Practice Tests

Find Practice Tests in the following link:

SAT Practice Tests - Convex Polygons and

Additional Practice Tests - Convex Polygons