Missing Angles in Quadrilaterals and Polygons | Grade 8 Math | LevelKite

Suppose you know three of the four angles inside a four-sided figure. Can you find the fourth without measuring it? You can, and this lesson shows you how and why. Then you will stretch the idea to figures with more sides.

A polygon is a flat, closed figure made of straight sides. The sides meet only at their endpoints, and exactly two sides meet at each endpoint. Each meeting point is a vertex (plural: vertices). At every vertex the two sides form an interior angle, the angle that lies inside the polygon. A polygon has as many interior angles as it has sides.

Polygons are named by their number of sides. A triangle has exactly $3$ sides and a quadrilateral has exactly $4$ .

Every polygon in this lesson is convex: each interior angle is greater than $0^\circ$ and less than $180^\circ$ . The angle sums you will learn are also true for polygons that are not convex, but showing that takes a different argument, so this lesson leaves those polygons out.

You will reason from one fact: the three interior angles of any triangle add to $180^\circ$ .

Start with any convex quadrilateral. Sketch one on paper if you like. Name its vertices $A$ , $B$ , $C$ and $D$ in order around the figure. Write $\angle A$ for the measure of the interior angle at $A$ , and do the same for the other vertices.

Now draw the diagonal from $A$ to $C$ . A diagonal is a segment that joins two vertices that are not next to each other. Because the quadrilateral is convex, this diagonal lies inside it and cuts it into two triangles, $ABC$ and $ACD$ .

The diagonal splits the angle at $A$ into two parts and the angle at $C$ into two parts. The angles at $B$ and $D$ stay whole. So the six angles of the two triangles fit together to make exactly the four interior angles of the quadrilateral. Each triangle's angles add to $180^\circ$ , so the four angles of the quadrilateral add to the total for two triangles.

\angle A + \angle B + \angle C + \angle D = 180^\circ + 180^\circ = 360^\circ

So the interior angles of every convex quadrilateral add to $360^\circ$ . This rule turns a missing angle into an equation. Suppose you know three of the interior angles. Let $x$ be the measure of the fourth angle, in degrees. Then the three known measures and $x$ add to $360$ .

To solve the equation, add the three known measures, then subtract that sum from both sides. Subtracting the same number from both sides keeps the two sides equal, so $x$ equals $360$ minus the sum of the three known measures.

Two checks help you catch slips. First, all four measures must add to exactly $360$ . Second, because the quadrilateral is convex, $x$ must be greater than $0$ and less than $180$ .

Worked example

Three known angles in a quadrilateral

A convex quadrilateral has interior angles of $85^\circ$ , $110^\circ$ and $95^\circ$ . Find the measure of its fourth interior angle.

Let x be the measure of the fourth interior angle, in degrees. The four interior angles of a convex quadrilateral add to 360 degrees, so write the sum of all four as an equation.
$85 + 110 + 95 + x = 360$
Add the three known measures to make one number: 85 + 110 = 195, and 195 + 95 = 290. Replacing the three numbers with their sum does not change the value of the left side.
$290 + x = 360$
Subtract 290 from both sides. Taking the same number from both sides keeps them equal, and it leaves x by itself.
$x = 360 - 290 = 70$
Check both conditions. The four measures add to 360, and 70 is greater than 0 and less than 180, as it must be in a convex quadrilateral.
$85 + 110 + 95 + 70 = 360$

Answer

The fourth interior angle measures $70^\circ$ .

Worked example

One known angle in a parallelogram

A parallelogram is a quadrilateral with two pairs of parallel sides. Every parallelogram is convex, and its opposite interior angles have equal measures. Parallelogram $PQRS$ has its vertices named in order around the figure, and its angle at $P$ measures $68^\circ$ . Find the measures of its angles at $Q$ , $R$ and $S$ .

The vertices are named in order, so P and R are not next to each other. That makes them opposite vertices, and Q and S are opposite vertices too. Opposite interior angles of a parallelogram have equal measures, so the angle at R matches the angle at P.
$\angle R = \angle P = 68^\circ$
Let x be the measure of the angle at Q, in degrees. The angle at S is opposite it, so it also measures x degrees. A parallelogram is a convex quadrilateral, so its four interior angles add to 360 degrees.
$68 + x + 68 + x = 360$
Add the two known measures, 68 + 68 = 136, and combine x + x into 2x. Changing the order and grouping of the terms in a sum does not change the sum.
$136 + 2x = 360$
Subtract 136 from both sides, which keeps the two sides equal: 360 - 136 = 224.
$2x = 224$
Divide both sides by 2. Dividing both sides by the same nonzero number keeps them equal, and 224 divided by 2 is 112.
$x = 112$
Check that the four measures add to 360 and that each one is greater than 0 and less than 180.
$68 + 112 + 68 + 112 = 360$

Answer

The angle at $R$ measures $68^\circ$ , and the angles at $Q$ and $S$ each measure $112^\circ$ .

The same cutting idea works for a convex polygon with more sides. Let $n$ be its number of sides, a whole number that is at least $3$ . Choose one vertex and draw a diagonal from it to every vertex that is not next to it. Because the polygon is convex, these diagonals lie inside it and cut it into triangles that do not overlap.

The angles of the triangles fit together to make exactly the interior angles of the polygon, just as they did in the quadrilateral. So the sum of the interior angles is $180^\circ$ times the number of triangles.

How many triangles are there? Each triangle has the chosen vertex as one of its vertices, and the side opposite that vertex is a side of the polygon. Every side of the polygon is used this way except the two sides that meet at the chosen vertex. So $n$ sides give $n - 2$ triangles.

The table names each polygon by its number of sides and counts its triangles. A triangle needs no cutting, so it counts as $1$ triangle.

Convex polygons cut into triangles by the diagonals from one vertex

Polygon	Number of sides, $n$	Number of triangles, $n - 2$	Sum of interior angles
Triangle	$3$	$1$	$1 \times 180^\circ = 180^\circ$
Quadrilateral	$4$	$2$	$2 \times 180^\circ = 360^\circ$
Pentagon	$5$	$3$	$3 \times 180^\circ = 540^\circ$
Hexagon	$6$	$4$	$4 \times 180^\circ = 720^\circ$
Heptagon	$7$	$5$	$5 \times 180^\circ = 900^\circ$
Octagon	$8$	$6$	$6 \times 180^\circ = 1080^\circ$

Key concept

Interior angle sum of a convex polygon

Let $n$ be the number of sides of a convex polygon, a whole number that is at least $3$ , and let $S$ be the sum of its interior angles. The diagonals from one vertex cut the polygon into $n - 2$ triangles, and each triangle's angles add to $180^\circ$ . So $S$ is $n - 2$ times $180^\circ$ .

A quadrilateral has $n = 4$ , so its interior angles add to $2 \times 180^\circ = 360^\circ$ . To find a missing interior angle, write an equation that sets the sum of all the interior angles equal to $S$ , then solve it.

S = (n - 2) \times 180^\circ

Worked example

Two equal angles in a hexagon

A hexagon is a polygon with exactly $6$ sides. A convex hexagon has interior angles of $130^\circ$ , $115^\circ$ , $105^\circ$ and $150^\circ$ . Its other two interior angles are equal to each other. Find the measure of each of those two angles.

Find the angle sum first. A hexagon has n = 6 sides, so the diagonals from one vertex cut it into 6 - 2 = 4 triangles, and each triangle's angles add to 180 degrees.
$S = (6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ$
Let x be the measure, in degrees, of each of the two equal angles. All six interior angles add to 720.
$130 + 115 + 105 + 150 + x + x = 720$
Add the known measures: 130 + 115 = 245, then 245 + 105 = 350, then 350 + 150 = 500. Combine x + x into 2x.
$500 + 2x = 720$
Subtract 500 from both sides to keep them equal: 720 - 500 = 220.
$2x = 220$
Divide both sides by 2, a nonzero number, to keep them equal: 220 divided by 2 is 110.
$x = 110$
Check that all six measures add to 720 and that 110 is greater than 0 and less than 180, as a convex polygon requires.
$130 + 115 + 105 + 150 + 110 + 110 = 720$

Answer

Each of the two equal angles measures $110^\circ$ .

In the hexagon example, two of the angles were equal. In a regular polygon, all the sides are equal in length and all the interior angles are equal in measure. A regular polygon is convex, so the angle-sum rule applies to it.

Suppose a regular polygon has $n$ sides and each interior angle measures $x$ degrees. Its $n$ equal angles add to the angle sum, so $n \times x = (n - 2) \times 180$ . Because $n$ is at least $3$ , it is not zero, so you can divide both sides by $n$ and keep them equal. Each interior angle is the angle sum divided by $n$ .

An octagon has exactly $8$ sides. For a regular octagon, the angle sum is $(8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ$ , and $1080 \div 8 = 135$ . Each interior angle of a regular octagon measures exactly $135^\circ$ .

Divide by $n$ only when you know that all the interior angles are equal, as in a regular polygon. The hexagon in the example was not regular: $720 \div 6 = 120$ , but its two missing angles measure $110^\circ$ each.

Takeaway: the diagonals from one vertex cut a convex polygon with $n$ sides into $n - 2$ triangles, so its interior angles add to $(n - 2) \times 180^\circ$ . For a quadrilateral, that total is $360^\circ$ . To find a missing angle, set the sum of all the interior angles equal to the total, solve the equation, and check your answer. When you know that all $n$ angles are equal, each one is the total divided by $n$ .