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Geometry - Angle Properties (Polygons and Parallel Lines)

Grade 12A Level

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The sum of the exterior angles of any convex polygon is always 360∘360^\circ. For a regular polygon with nn sides, each exterior angle is calculated as 360∘n\frac{360^\circ}{n}.

Diagram showing an exterior angle of a pentagon formed by extending one side.
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When a transversal intersects two parallel lines, several angle relationships are formed: Corresponding angles are equal (F-shape), Alternate angles are equal (Z-shape), and Co-interior angles sum to 180∘180^\circ (C-shape).

Parallel lines L1 and L2 intersected by a transversal, showing alternate interior angles a and b.
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Interior and exterior angles at any vertex of a polygon are supplementary, meaning they add up to 180∘180^\circ. This is because they lie on a straight line.

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The sum of interior angles of a polygon depends on the number of triangles it can be divided into from one vertex, which is (n−2)(n-2). Hence, the sum is (n−2)×180∘(n-2) \times 180^\circ.

📐Formulae

Sum of interior angles = (n−2)×180∘(n - 2) \times 180^\circ

Each interior angle (regular polygon) = (n−2)×180∘n\frac{(n - 2) \times 180^\circ}{n}

Sum of exterior angles = 360∘360^\circ

Each exterior angle (regular polygon) = 360∘n\frac{360^\circ}{n}

Number of sides (n) = 360∘Exterior Angle\frac{360^\circ}{\text{Exterior Angle}}

Interior Angle + Exterior Angle = 180∘180^\circ

💡Examples

Problem 1:

A regular polygon has an interior angle of 150∘150^\circ. Calculate the number of sides (n) of this polygon.

Solution:

180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ. n=360∘/30∘=12n = 360^\circ / 30^\circ = 12.

Explanation:

First, find the exterior angle using the supplementary rule (Interior + Exterior = 180°). Then, use the property that the sum of exterior angles is 360° divided by the measure of one exterior angle to find the number of sides.

Problem 2:

In a pentagon, four of the interior angles are 110∘,90∘,120∘,110^\circ, 90^\circ, 120^\circ, and 100∘100^\circ. Find the size of the fifth angle.

Solution:

Sum = (5−2)×180∘=540∘(5-2) \times 180^\circ = 540^\circ. Fifth angle = 540∘−(110∘+90∘+120∘+100∘)=540∘−420∘=120∘540^\circ - (110^\circ + 90^\circ + 120^\circ + 100^\circ) = 540^\circ - 420^\circ = 120^\circ.

Explanation:

Calculate the total sum of interior angles for a pentagon (n=5). Subtract the sum of the known four angles from the total sum to find the remaining angle.

Problem 3:

Line L1L_1 and L2L_2 are parallel. A transversal cuts them. If a pair of co-interior angles are represented by (2x+10)∘(2x + 10)^\circ and (3x+20)∘(3x + 20)^\circ, find the value of xx.

Solution:

(2x+10)+(3x+20)=180⇒5x+30=180⇒5x=150⇒x=30(2x + 10) + (3x + 20) = 180 \Rightarrow 5x + 30 = 180 \Rightarrow 5x = 150 \Rightarrow x = 30.

Explanation:

Co-interior angles between parallel lines are supplementary, meaning they add up to 180°. Set up an algebraic equation summing the two expressions to 180 and solve for x.

Problem 4:

In the diagram provided, ABAB is parallel to CDCD. Given that ∠AGH=125∘\angle AGH = 125^\circ, find the value of the angle xx and the angle yy.

Two parallel lines intersected by a transversal with angles labeled x, y, and 125 degrees.

Solution:

y=∠AGH=125∘y = \angle AGH = 125^\circ (Vertically opposite angles) Since AB∥CDAB \parallel CD, ∠GHD+∠AGH=180∘\angle GHD + \angle AGH = 180^\circ is not the direct path. Instead, use alternate interior angles: ∠GHC=∠AGH=125∘\angle GHC = \angle AGH = 125^\circ Since ∠GHC\angle GHC and xx are on a straight line: x=180∘−125∘=55∘x = 180^\circ - 125^\circ = 55^\circ

Explanation:

We identify yy as vertically opposite to the given angle. Then we use the property that alternate interior angles are equal to find the relationship between the given angle and the angles on the parallel line CDCD.

Problem 5:

The diagram shows a regular hexagon. Calculate the size of the interior angle marked aa and the exterior angle marked bb.

A regular hexagon with an interior angle labeled 'a' and an exterior angle labeled 'b'.

Solution:

For a regular hexagon, n=6n = 6. Sum of interior angles = (6−2)×180∘=4×180∘=720∘(6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ Each interior angle aa = 720∘6=120∘\frac{720^\circ}{6} = 120^\circ Each exterior angle bb = 180∘−120∘=60∘180^\circ - 120^\circ = 60^\circ Alternatively, b=360∘6=60∘b = \frac{360^\circ}{6} = 60^\circ.

Explanation:

We use the formula for the sum of interior angles of a polygon and divide by the number of sides for a regular polygon. The exterior angle is then found using the supplementary rule.

Angle Properties (Polygons and Parallel Lines) Grade 12 Notes & Examples