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Polygon Angles and Tessellations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The total angle

Find the interior angle sum of a convex decagon.
1. Hint 1
  Think of splitting the interior into triangles without adding an interior vertex.
2. Hint 2
  How many triangles does a fan contain compared with the number of sides?
Answer

$1440^\circ$ .

Full solution

A decagon has ten sides.
A fan from one vertex contains
$10-2=8$
triangles.

Each triangle contributes $180^\circ$ , so the total is
$S=8\times180^\circ$ $S=1440^\circ.$

Eight triangles account for the ten interior angles.

Answer

$1440^\circ$ .

Key idea

Counting a convex fan gives the interior angle sum.
Problem 2 The remaining vertex

A convex heptagon has six exterior turns measuring $30^\circ$ , $40^\circ$ , $45^\circ$ , $50^\circ$ , $55^\circ$ , and $65^\circ$ . These are exactly one turn at each of six vertices in the same traversal direction.

Find the exterior turn at the remaining vertex and its neighboring interior angle.
1. Hint 1
  One complete boundary walk accounts for a full turn.
2. Hint 2
  Find the known turns’ total, then use the linear pair at the remaining vertex.
Answer

$\begin{gathered}e=75^\circ\\i=105^\circ\end{gathered}$

Full solution

The six given turns total $285^\circ$ .
The missing turn completes the full turn.
$e=360^\circ-285^\circ$ $e=75^\circ.$

The interior angle and its exterior turn are supplementary.
$i=180^\circ-75^\circ$ $i=105^\circ.$

The remaining interior angle and its turn form a linear pair.

Answer

$\begin{gathered}e=75^\circ\\i=105^\circ\end{gathered}$

Key idea

Exterior turns total a full turn; each interior angle supplements its own turn.
Problem 3 A formula with a reason

For a convex polygon with n sides, where n is a whole number at least three, derive a formula for the sum of its interior angles. Explain why your argument counts every interior angle exactly once.
1. Hint 1
  You know the sum for a triangle. Look for a way to cover the polygon with triangles.
2. Hint 2
  Choose one vertex and examine its connections to nonadjacent vertices.
3. Hint 3
  Relate the triangle count to the boundary sides not ending at the chosen vertex.
Answer

For every convex polygon with n sides, the interior angle sum is $S=(n-2)180^\circ$ .

Full solution

For a convex heptagon, choose vertex A and draw every diagonal from A.
Convexity keeps them inside, and they cannot cross away from A.
The five boundary sides not touching A each belong to one of five fan triangles.
Their angle sums give
$S=5\times180^\circ$ $S=900^\circ.$

For a convex n-sided polygon, choose A in the same way.
The two boundary sides ending at A are excluded.
Each of the other n minus two sides is opposite A in exactly one fan triangle.
The fan fills the polygon without gaps or overlaps.

At each boundary vertex, the triangle angles or angle pieces fill exactly that polygon angle.
No new interior vertex contributes extra angles.
The count works independently of the heptagon example, giving
$S=(n-2)180^\circ.$

The heptagon illustrates the same count used for arbitrary n.

Answer

For every convex polygon with n sides, the interior angle sum is $S=(n-2)180^\circ$ .

Key idea

A derivation must justify both the partition and what its angles count.
Problem 4 Two angles of one polygon

Find one exterior turn and one interior angle of a convex regular dodecagon.

A regular dodecagon, with no angle measurements marked.

Text description of this figure

A convex regular polygon with 12 sides. The diagram has no angle measurements marked.
1. Hint 1
  Equal interior angles have equal supplements.
2. Hint 2
  Share the full turn among twelve vertices, then find the supplement.
Answer

$\begin{gathered}e=30^\circ\\i=150^\circ\end{gathered}$

Full solution

There are twelve equal exterior turns.
$e=\frac{360^\circ}{12}$ $e=30^\circ.$

Take the supplement to obtain the interior angle.
$i=180^\circ-30^\circ$ $i=150^\circ.$

Answer

$\begin{gathered}e=30^\circ\\i=150^\circ\end{gathered}$

Key idea

Regularity lets you share the full turn equally.
Problem 5 Mark a boundary walk

Copy the convex pentagon ABCDE shown in boundary order. Traverse A to B to C to D to E to A. At each vertex, extend the incoming side forward and mark the turn toward the next side.

Label the turns $e_A$ , $e_B$ , $e_C$ , $e_D$ , and $e_E$ . Mark the interior angle $i_B$ . Write an equation relating $i_B$ and $e_B$ .

Vertices are supplied in boundary order.

Text description of this figure

A convex pentagon labeled A, B, C, D, E in clockwise boundary order. No interior angles, extensions, or exterior turns are marked.
1. Hint 1
  A turn starts in the direction in which you arrived, rather than in the reverse direction.
2. Hint 2
  At B, extend AB beyond B. The turn there ends along BC.
3. Hint 3
  The interior angle and the selected exterior turn form a straight angle together.
Answer

Each incoming side is extended forward, with one exterior turn marked at its arrival vertex;
$i_B+e_B=180^\circ.$

Full solution

At A, extend EA beyond A and mark the turn toward AB.
At B, extend AB beyond B and mark the turn toward BC.

At C, extend BC beyond C and mark the turn toward CD.
At D, extend CD beyond D and mark the turn toward DE.
At E, extend DE beyond E and mark the turn toward EA.

The interior angle at B is between BA and BC.
BA and the forward extension of AB are opposite rays, so the two adjacent angles form a linear pair.
$i_B+e_B=180^\circ.$

The arcs select the forward turn once at each vertex.

Answer

Each incoming side is extended forward, with one exterior turn marked at its arrival vertex;
$i_B+e_B=180^\circ.$

Key idea

Choose the incoming extension consistently to mark one turn at every vertex.
Problem 6 Five equal angles

Three interior angles of a convex octagon each measure $125^\circ$ . The other five interior angles all have the same measure. Find that common measure and its exterior turn.
1. Hint 1
  Find the octagon’s total interior angle first.
2. Hint 2
  Remove the three known angles, then share what remains among five vertices.
3. Hint 3
  Take the supplement of the common interior angle.
Answer

$\begin{gathered}i=141^\circ\\e=39^\circ\end{gathered}$

Full solution

A convex octagon has interior sum
$S=6\times180^\circ$ $S=1080^\circ.$

The three known angles total
$3\times125^\circ=375^\circ.$
Let x be each of the other angles.
$5x=1080^\circ-375^\circ$ $5x=705^\circ.$

Divide equally among the five angles.
$x=141^\circ.$
Its exterior turn is
$e=180^\circ-141^\circ$ $e=39^\circ.$

Answer

$\begin{gathered}i=141^\circ\\e=39^\circ\end{gathered}$

Key idea

Equal-angle information is needed before a remaining sum can be shared equally.
Problem 7 Angle specifications

Decide whether each proposed convex regular polygon exists. If it exists, give its side count.

Case A: the interior angle is $168^\circ$ .

Case B: the exterior turn is $28^\circ$ .
1. Hint 1
  A regular polygon has an integer number of equal exterior turns in one full turn.
2. Hint 2
  For Case A, find the exterior turn before calculating sides.
3. Hint 3
  An exact noninteger side count cannot be repaired by rounding.
Answer

Case A exists with $30$ sides. Case B is impossible.

Full solution

Case A: the exterior turn is
$e=180^\circ-168^\circ$ $e=12^\circ.$
The required count is
$n=360\div12$ $n=30.$
This is an integer at least three, so the regular polygon exists.

Case B: dividing the full turn by the proposed turn gives
$n=360\div28$ $n=\frac{90}{7}.$
This is not an integer, so no convex regular polygon has the specified exterior turn.

Answer

Case A exists with $30$ sides. Case B is impossible.

Key idea

Check exact integer feasibility after solving for the side count.
Problem 8 A total independent of shape

For a convex n-sided polygon, select exactly one exterior turn per vertex while traversing the boundary in one direction. Prove that their sum is independent of n. Explain why the selection rule matters.
1. Hint 1
  Compare each exterior turn with its neighboring interior angle.
2. Hint 2
  Total all the vertex linear pairs, then remove the interior-angle total.
3. Hint 3
  If both exterior choices at a vertex were counted, that vertex would be counted twice.
Answer

The exterior sum is always $360^\circ$ , using exactly one consistently selected turn per vertex.

Full solution

For a convex quadrilateral, four linear pairs total $720^\circ$ .
The interior sum is $360^\circ$ , leaving
$E=720^\circ-360^\circ$ $E=360^\circ.$

For a convex n-sided polygon, the n linear pairs total 180n degrees.
Subtract the interior sum.
With all angle measures in degrees,
$\begin{aligned}E&=180n-180(n-2)\\E&=360.\end{aligned}$
The n terms cancel, so the sum does not depend on the side count.

The same subtraction accounts for one supplement at each vertex.
Counting both exterior choices would count each supplement twice.
Consistent traversal also ensures each selected angle describes the actual boundary turn.

A square illustrates one consistent selection; the proof applies to every convex polygon.

Answer

The exterior sum is always $360^\circ$ , using exactly one consistently selected turn per vertex.

Key idea

Pairing each exterior turn with its interior angle gives a complete count.
Problem 9 Compare two tile choices

Compare regular tessellations made from congruent squares with those proposed from congruent regular octagons. Tiles must meet whole side to whole side.

For each choice, test the possible number of corners at a vertex. Explain which choice can cover the plane, supporting any positive conclusion with a repeating arrangement.
1. Hint 1
  An angle total must fill a full turn exactly.
2. Hint 2
  Find a regular octagon’s interior angle, then compare two and three of its corners.
3. Hint 3
  A successful angle test needs a pattern that continues along matching edges.
Answer

Squares tile in a grid, with four corners at each vertex. Regular octagons alone cannot form such a tessellation.

Full solution

Each square corner is $90^\circ$ .
Four corners total
$4\times90^\circ=360^\circ.$
Rows and columns of congruent squares pair every side with a neighbor and continue in both directions, so they tile the plane.

A regular octagon has interior angle
$i=\frac{6\times180^\circ}{8}$ $i=135^\circ.$
Two corners total
$2\times135^\circ=270^\circ,$
leaving a $90^\circ$ gap.
Three total
$3\times135^\circ=405^\circ,$
requiring $45^\circ$ of overlap.

No whole number of these octagon corners totals a full turn: two are too few and three are too many.
Thus octagons alone fail the required local condition.
This does not rule out a tiling mixing octagons with another polygon.

Two octagon corners leave space; adding a third would exceed a full turn.

Answer

Squares tile in a grid, with four corners at each vertex. Regular octagons alone cannot form such a tessellation.

Key idea

Use angles to rule out a tile choice and a repeating construction to establish a successful one.
Problem 10 A repeating design

A tessellation must use congruent copies of one convex regular polygon, meeting whole side to whole side, with exactly three tiles at every vertex.

Identify the polygon. Draw a patch of at least five tiles and give a repeating placement rule that justifies covering the plane without gaps or overlaps. Explain what this adds beyond the angle calculation.
1. Hint 1
  Work backward from the equal corners needed to fill a full turn.
2. Hint 2
  Convert the required interior angle to an exterior turn and a side count.
3. Hint 3
  After identifying the tile, arrange matching staggered columns and explain how they continue.
Answer

Use regular hexagons. A patch of matching staggered columns extends throughout the plane, with three corners at every vertex.

Full solution

Three equal corners must fill the full turn.
$i=360^\circ\div3$ $i=120^\circ.$
The exterior turn is
$e=180^\circ-120^\circ$ $e=60^\circ.$
Therefore
$n=360\div60$ $n=6.$
The required tile is a regular hexagon.

Draw flat-topped regular hexagons in a column, with each horizontal bottom side matching the horizontal top side of the next hexagon.
Place the next column halfway down, with its slanted sides matching the exposed slanted sides of the first column.
Repeat this placement in both directions.

Every column continues upward and downward through matching horizontal sides.
Each adjacent column fills the strip beside it through matching slanted sides.
The same edge matches repeat, with three corners totaling a full turn at each vertex, so no unfilled sector or overlap appears.

The angle calculation identifies a feasible local fit.
The continuing column rule supplies the additional evidence that these local fits cover the plane.
Other patches following the same complete edge matches are acceptable.

Repeat the staggered columns above, below, left, and right.

Answer

Use regular hexagons. A patch of matching staggered columns extends throughout the plane, with three corners at every vertex.

Key idea

A tiling argument needs both feasible vertex angles and a construction that continues.

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