Finding Angles in Regular Polygons Worksheet | Geometry PDF Worksheets - Free Printable

Here are the solutions to the problems on the worksheet, broken down by section.

Section A: Formulas and Basic Calculations

1) Find the size of each interior angle in a regular hexagon.
* Step 1: A hexagon has 6 sides ($n=6$).
* Step 2: The sum of interior angles is $(n-2) \times 180^\circ$. So, $(6-2) \times 180 = 4 \times 180 = 720^\circ$.
* Step 3: Since it is regular, divide the sum by the number of angles (6).
* Calculation: $720 \div 6 = 120$.
* Answer: $120^\circ$

2) Find the size of each interior angle in a regular nonagon.
* Step 1: A nonagon has 9 sides ($n=9$).
* Step 2: Sum of interior angles = $(9-2) \times 180 = 7 \times 180 = 1260^\circ$.
* Step 3: Divide by 9.
* Calculation: $1260 \div 9 = 140$.
* Answer: $140^\circ$

3) Find the size of each exterior angle in a regular dodecagon.
* Step 1: A dodecagon has 12 sides ($n=12$).
* Step 2: The sum of exterior angles for any convex polygon is always $360^\circ$.
* Step 3: Divide by the number of sides.
* Calculation: $360 \div 12 = 30$.
* Answer: $30^\circ$

4) Write an expression for the sum of the interior angles in a polygon with n sides.
* Logic: You can split any polygon into triangles. The number of triangles is always 2 less than the number of sides. Each triangle is $180^\circ$.
* Answer: $(n - 2) \times 180^\circ$

5) Write an expression for the size of each interior angle in a polygon with n sides.
* Logic: Take the total sum from question 4 and divide it by the number of angles ($n$).
* Answer: $\frac{(n - 2) \times 180}{n}$

6) Write an expression for the size of each exterior angle in a polygon with n sides.
* Logic: The total exterior sum is $360^\circ$. Divide by the number of sides ($n$).
* Answer: $\frac{360}{n}$

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Section B: Finding Angles in Diagrams

1) Regular Pentagon
* Angle a (Interior):
* Sides ($n$) = 5.
* Interior Angle = $\frac{(5-2) \times 180}{5} = \frac{540}{5} = 108^\circ$.
* a = $108^\circ$
* Angle b (Exterior):
* Exterior Angle = $\frac{360}{5} = 72^\circ$. (Or $180 - 108 = 72$).
* b = $72^\circ$

2) Regular Pentagon with a diagonal
* Angle c:
* This angle is inside an isosceles triangle formed at the top vertex.
* The top vertex angle is the interior angle of the pentagon: $108^\circ$.
* The two base angles are equal. Let's call them $x$.
* $108 + x + x = 180 \rightarrow 2x = 72 \rightarrow x = 36^\circ$.
* c = $36^\circ$
* Angle d:
* Angle $d$ is part of the bottom-right interior angle ($108^\circ$).
* The diagonal creates symmetry. The angle adjacent to $d$ (inside the triangle we just looked at) is also $36^\circ$ due to alternate interior angles or symmetry of the trapezoid formed below.
* Actually, simpler view: The shape below the diagonal is an isosceles trapezoid. The base angles are equal. The full interior angle is $108^\circ$. The "top" part of that corner angle corresponds to angle $c$ via alternate interior angles if we draw a parallel line, but let's stick to the triangle method.
* In the triangle containing angle $c$, the other two angles are $36^\circ$. By symmetry, the angle adjacent to $d$ inside the pentagon is also $36^\circ$.
* Therefore, $d = 108^\circ - 36^\circ = 72^\circ$.
* d = $72^\circ$

3) Regular Hexagon with diagonals
* Angle e:
* This is simply the interior angle of a regular hexagon.
* Interior Angle = $\frac{(6-2) \times 180}{6} = 120^\circ$.
* e = $120^\circ$
* Angle f:
* Look at the triangle on the left side formed by the short diagonal. It uses two sides of the hexagon and one diagonal.
* This is an isosceles triangle. The vertex angle is the interior angle of the hexagon ($120^\circ$).
* The two base angles are equal. $(180 - 120) \div 2 = 30^\circ$.
* Angle $f$ is one of these base angles.
* f = $30^\circ$

4) Regular Hexagon divided into triangles
* Angle g (Exterior):
* This is the exterior angle of a regular hexagon.
* Calculation: $360 \div 6 = 60^\circ$.
* g = $60^\circ$
* Angle h:
* The hexagon is divided into 6 equilateral triangles meeting at the center.
* Angle $h$ is the angle of one of these equilateral triangles at the center? No, looking closely, $h$ is the interior angle of one of the small triangles at the center vertex.
* The angles at the center add up to $360^\circ$. There are 6 identical triangles.
* $360 \div 6 = 60^\circ$.
* h = $60^\circ$

5) Regular Octagon
* Angle i:
* First, find the interior angle of a regular octagon ($n=8$).
* Interior Angle = $\frac{(8-2) \times 180}{8} = \frac{1080}{8} = 135^\circ$.
* Angle $i$ is part of this interior angle. We need to subtract the angle belonging to the triangle formed by the diagonal.
* Consider the isosceles triangle formed by the top-left vertex and the two adjacent vertices connected by the chord. The vertex angle is $135^\circ$. The base angles are $(180-135)/2 = 22.5^\circ$.
* Wait, let's look at the triangle containing angle $j$ first, it might be easier.
* Let's look at the triangle defined by the long diagonal.
* Actually, let's look at the quadrilateral or triangle properties directly.
* Angle $i$ is the angle between a side and a diagonal that skips one vertex.
* Let's use the property of the circle circumscribing the octagon. Each side subtends $360/8 = 45^\circ$ at the center.
* Angle $i$ is an inscribed angle subtending 3 sides (from the vertex where $i$ is, across to the other end of the diagonal). Wait, the diagonal for $i$ goes to the vertex 2 steps away.
* Let's stick to simple geometry.
* Triangle at the top left corner: Isosceles with vertex angle $135^\circ$. Base angles = $22.5^\circ$.
* Angle $i$ is the remainder of the interior angle after removing that base angle? No, the diagram shows the diagonal going from the vertex *below* the top-left one, to the top-left one? No, it looks like the diagonal starts at the vertex with angle $i$.
* Let's trace the lines. Angle $i$ is inside a triangle formed by three vertices of the octagon.
* Let's calculate angle $j$ first. Angle $j$ is at the bottom. The triangle containing $j$ has its third vertex at the top. This is a large isosceles triangle.
* Let's try a different approach for $i$. The diagonal cuts off a triangle with 2 sides of the octagon. The angle at the vertex between those two sides is $135^\circ$. The other two angles are $22.5^\circ$. Angle $i$ is adjacent to one of these $22.5^\circ$ angles? No, angle $i$ IS the base angle of the triangle formed by connecting vertex 1 to vertex 3.
* Yes, Angle $i$ is the base angle of an isosceles triangle with vertex angle $135^\circ$.
* So, $i = (180 - 135) / 2 = 22.5^\circ$.
* i = $22.5^\circ$
* Angle j:
* Angle $j$ is part of the interior angle at the bottom vertex.
* The diagonal forming the right side of angle $j$ connects to the top vertex. This divides the octagon in half.
* The diagonal forming the left side of angle $j$ connects to the vertex 2 spots away (same as angle $i$'s setup but mirrored).
* Let's look at the triangle containing angle $j$. It is formed by the bottom vertex and two others.
* Actually, notice the symmetry. The line dividing the octagon vertically is a line of symmetry.
* Angle $j$ is bounded by a vertical line of symmetry and a diagonal.
* The interior angle is $135^\circ$.
* The vertical line bisects the interior angle? No, only if it connects to the opposite vertex. It does. So the angle between the side and the vertical diameter is $135 / 2 = 67.5^\circ$.
* Now, what is the other line bounding $j$? It connects to the vertex "2 steps" away.
* Let's calculate the angle between the side and the diagonal that skips 1 vertex. We already found that angle is $22.5^\circ$ (like angle $i$).
* So, angle $j$ is the difference between the half-interior angle ($67.5^\circ$) and the small slice ($22.5^\circ$)? Or the sum?
* Looking at the diagram, $j$ is the angle between the diagonal skipping 1 vertex and the main vertical diagonal.
* Angle between Side and Diagonal(skip 1) = $22.5^\circ$.
* Angle between Side and Vertical Diagonal = $67.5^\circ$.
* Therefore, $j = 67.5 - 22.5 = 45^\circ$.
* j = $45^\circ$

6) Regular Nonagon (9 sides)
* First, Interior Angle of Nonagon:
* Sum = $(9-2) \times 180 = 1260^\circ$.
* One Interior Angle = $1260 / 9 = 140^\circ$.
* Angle k:
* Angle $k$ is the vertex angle of an isosceles triangle formed by two sides of the nonagon and a short diagonal.
* Wait, looking at the diagram, $k$ is at the top vertex. The triangle legs are sides of the nonagon.
* So $k$ is simply the interior angle of the regular nonagon.
* k = $140^\circ$
* Angle l:
* Angle $l$ is an interior angle of the triangle formed by the center? No, there is no center point marked.
* Let's look at the triangle containing $l$. The vertices are: Top vertex, Bottom-Right vertex, and... another vertex?
* Let's trace the lines defining $l$.
* Line 1: Connects Top Vertex to Bottom-Right Vertex.
* Line 2: Connects Top-Left Vertex (approx) to Bottom-Right Vertex.
* This is getting complex. Let's use the "angles in a circle" method (inscribed angles), which is very reliable for regular polygons.
* Imagine a circle passing through all vertices.
* The entire circle is $360^\circ$.
* Each side of the nonagon subtends an arc of $360 / 9 = 40^\circ$.
* An inscribed angle is half the measure of the intercepted arc.
* For Angle k: It intercepts the arc covering the rest of the polygon (8 sides). Arc = $8 \times 40 = 320^\circ$. Angle = $320 / 2 = 160^\circ$? No, that's the reflex angle. The interior angle intercepts the arc of the *other* way? No.
* Let's stick to the triangle calculation for $k$ to be safe. We established $k = 140^\circ$.
* For Angle l:
* Vertex is at the bottom right.
* One ray goes to the Top vertex. How many sides away is the Top vertex? Let's count from Bottom-Right clockwise. 1, 2, 3, 4 steps to the top. So it spans 4 sides. Arc = $4 \times 40 = 160^\circ$.
* The other ray goes to the vertex on the left (let's say "Top-Left"). From Bottom-Right counter-clockwise: 1, 2, 3 steps. So it spans 3 sides. Arc = $3 \times 40 = 120^\circ$.
* Wait, angle $l$ is the angle *between* these two chords.
* The angle subtended by an arc at the circumference is half the arc.
* Let's define the position of vertices 1 to 9. Let Bottom-Right be Vertex 1.
* Ray 1 goes to Vertex 5 (Top). The arc from 1 to 5 is 4 steps ($160^\circ$). The inscribed angle subtending this arc would be at Vertex 9 or something. This is confusing.
* Let's use the triangle sum method.
* Consider the triangle formed by: Top Vertex (V5), Bottom-Right Vertex (V1), and the vertex to the left of V1 (V9)? No, the line goes to V3 or V4?
* Let's look at the diagram 6 again carefully.
* The angle $l$ is at a vertex. Let's call the vertex with $l$: $A$.
* The two lines forming $l$ go to two other vertices.
* Line 1 goes to the top peak.
* Line 2 goes to a vertex on the left side.
* Let's assume standard orientation. Top is V1. Clockwise V2... V9.
* Angle $k$ is at V1. It is the interior angle. $140^\circ$. Correct.
* Angle $l$ is at V4 (bottom right-ish).
* One line connects V4 to V1 (Top).
* The other line connects V4 to V8 (left side).
* We need the angle $\angle V1-V4-V8$.
* This is an inscribed angle subtending the arc from V1 to V8.
* Path from V1 to V8 not containing V4: V1 -> V9 -> V8. That is 2 sides.
* Arc measure = $2 \times 40^\circ = 80^\circ$.
* Inscribed Angle = Half the arc = $80 / 2 = 40^\circ$.
* Let's double check.
* Triangle V1-V4-V8.
* Angle at V4 ($l$) subtends arc V1-V8 (2 sides). Angle = $40^\circ$.
* Angle at V1 subtends arc V4-V8 (4 sides: V4,V5,V6,V7,V8 is 4 steps? V4->V5->V6->V7->V8 is 4 edges). Arc = $160^\circ$. Angle = $80^\circ$.
* Angle at V8 subtends arc V4-V1 (3 sides: V4,V3,V2,V1 is 3 edges). Arc = $120^\circ$. Angle = $60^\circ$.
* Sum: $40 + 80 + 60 = 180^\circ$. Perfect.
* l = $40^\circ$

Final Answer:
Section A
1) $120^\circ$
2) $140^\circ$
3) $30^\circ$
4) $(n - 2) \times 180^\circ$
5) $\frac{(n - 2) \times 180}{n}$
6) $\frac{360}{n}$

Section B
1) a = $108^\circ$, b = $72^\circ$
2) c = $36^\circ$, d = $72^\circ$
3) e = $120^\circ$, f = $30^\circ$
4) g = $60^\circ$, h = $60^\circ$
5) i = $22.5^\circ$, j = $45^\circ$
6) k = $140^\circ$, l = $40^\circ$