Angle sum of quadrilaterals

Take any four-sided closed figure — a kite-shaped park boundary, a tilted picture frame, even a lopsided sheet of paper torn into four straight edges — and measure its four interior angles with a protractor. Add them up. No matter how stretched, skewed, or irregular the shape is, the total always comes to exactly $360^\circ$ . That single fixed number is one of the most reliable facts in plane geometry, and it turns "find the missing angle" problems into simple arithmetic. This section explains where the $360^\circ$ comes from — a two-triangle split that needs nothing beyond the angle-sum property of a triangle — and shows how the same reasoning scales up to any convex polygon, giving the general rule $(n-2)\times180^\circ$ . You will also see how exterior angles fit into the picture, work through a full ratio-based problem, and collect the traps examiners like to set around this topic.

The core insight

The four interior angles of any simple quadrilateral always sum to $360^\circ$ , because a single diagonal splits it into two triangles whose $180^\circ$ angle sums add together without overlap.

The core idea

A quadrilateral is a closed plane figure bounded by four straight sides, with four vertices and four interior angles. The angle-sum property says these four angles — however unequal — always total the same fixed amount:

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

where: $\angle A, \angle B, \angle C, \angle D$ are the four interior angles of quadrilateral $ABCD$ , taken in order around the shape.

Why a triangle-split works. Any triangle's three interior angles add to $180^\circ$ — a fact from the previous chapter that this whole result leans on. A quadrilateral is not a triangle, but one diagonal turns it into two of them, and $180^\circ + 180^\circ$ lands exactly on $360^\circ$ .

It scales to any polygon. The same diagonal-splitting idea, repeated from one vertex, gives the interior angle sum of any convex polygon with $n$ sides:

\text{Interior angle sum} = (n-2)\times180^\circ

where: $n$ is the number of sides (and vertices) of the convex polygon. A quadrilateral is simply the case $n = 4$ , giving $(4-2)\times180^\circ = 360^\circ$ — consistent with the direct triangle-split above. A pentagon ( $n=5$ ) gives $540^\circ$ , a hexagon ( $n=6$ ) gives $720^\circ$ , and so on.

Exterior angles behave differently. At each vertex, the exterior angle and the interior angle form a straight line (a linear pair), so exterior $= 180^\circ -$ interior. Remarkably, for any convex polygon — triangle, quadrilateral, decagon, whatever $n$ is — the sum of all exterior angles, one at each vertex, is always:

\text{Exterior angle sum} = 360^\circ

This does not depend on $n$ at all, unlike the interior sum. For a regular $n$ -gon (all sides and angles equal), each interior angle is $\dfrac{(n-2)\times180^\circ}{n}$ and each exterior angle is $\dfrac{360^\circ}{n}$ — but these per-angle formulas only apply when the polygon is regular; an irregular quadrilateral has a fixed sum but no fixed individual angle.

Figure

A quadrilateral split by one diagonal into two triangles each contributing 180 degrees to the total of 360 degrees

Derivation AInterior angle sum of a quadrilateral

The goal is to show, using only the already-known fact that a triangle's angles sum to $180^\circ$ , that any simple (non-self-intersecting) convex quadrilateral $ABCD$ has interior angles totalling exactly $360^\circ$ , and that this generalises to $(n-2)\times180^\circ$ for any convex $n$ -gon.

Step 1. Draw diagonal $AC$ in quadrilateral $ABCD$ . This single line joins two non-adjacent vertices and splits the quadrilateral cleanly into two triangles: $\triangle ABC$ and $\triangle ACD$ . Because $ABCD$ is convex, the diagonal lies entirely inside the shape, so the two triangles share only the edge $AC$ and together cover the whole quadrilateral with no gap and no overlap.

Step 2. Apply the triangle angle-sum property to each piece. In $\triangle ABC$ , the three angles $\angle BAC + \angle ABC + \angle BCA = 180^\circ$ . In $\triangle ACD$ , the three angles $\angle CAD + \angle ACD + \angle ADC = 180^\circ$ . Adding these two equations gives a combined total of $180^\circ + 180^\circ = 360^\circ$ for all six triangle-angles together.

Step 3. Regroup the six triangle-angles back into the original four quadrilateral angles. Notice $\angle BAC + \angle CAD = \angle A$ (the diagonal splits the corner at $A$ into two parts that together make the full angle of the quadrilateral there), and similarly $\angle BCA + \angle ACD = \angle C$ . The angles $\angle ABC = \angle B$ and $\angle ADC = \angle D$ are untouched by the diagonal. So the $360^\circ$ total from Step 2 is exactly $\angle A + \angle B + \angle C + \angle D$ — nothing was added or lost, only regrouped.

Step 4. Generalise to $n$ sides. From one vertex of a convex $n$ -gon, drawing every possible diagonal creates exactly $(n-2)$ non-overlapping triangles that together fill the polygon (a quadrilateral, $n=4$ , gives $(4-2)=2$ triangles, matching Steps 1–3). Each triangle contributes $180^\circ$ , so the interior angle sum is $(n-2)\times180^\circ$ , which reduces to $360^\circ$ exactly when $n=4$ .

Worked example

Problem: In quadrilateral $PQRS$ , $\angle P = 85^\circ$ , $\angle Q = 110^\circ$ , and $\angle R = 95^\circ$ . Find $\angle S$ , then check whether $PQRS$ could be a parallelogram.

Step 1. Write the angle-sum equation. Since $PQRS$ is a quadrilateral, $\angle P + \angle Q + \angle R + \angle S = 360^\circ$ .

Step 2. Substitute the three known angles: $85^\circ + 110^\circ + 95^\circ + \angle S = 360^\circ$ , which simplifies to $290^\circ + \angle S = 360^\circ$ .

Step 3. Solve for the missing angle: $\angle S = 360^\circ - 290^\circ = 70^\circ$ .

Step 4. Check the parallelogram question. In a parallelogram, opposite angles must be equal: $\angle P = \angle R$ and $\angle Q = \angle S$ . Here $\angle P = 85^\circ$ but $\angle R = 95^\circ$ , so opposite angles are not equal and $PQRS$ is not a parallelogram — just a general quadrilateral. Answer: $\angle S = 70^\circ$ , and $PQRS$ is an irregular (non-parallelogram) quadrilateral.

Competitive stretch

The four angles of a quadrilateral are in the ratio $2:3:4:6$ . Let the angles be $2x, 3x, 4x, 6x$ ; their sum must be $360^\circ$ , so $15x = 360^\circ$ , giving $x = 24^\circ$ . The angles are $48^\circ, 72^\circ, 96^\circ, 144^\circ$ , and $48+72+96+144 = 360^\circ$ checks out. Since all four differ, no pair of opposite angles matches, so this too is an irregular quadrilateral rather than a parallelogram.

Common traps

Assuming the total changes with shape. The interior sum is always $360^\circ$ for any simple quadrilateral — square, kite, or a completely lopsided four-sided figure — the individual angles vary, the total never does.
Mixing up interior and exterior sums. The interior sum of a quadrilateral is $360^\circ$ ; the exterior sum of any convex polygon (not just $n=4$ ) is also always $360^\circ$ — two separate facts that only happen to share a number when $n=4$ .
Using the regular-polygon per-angle formula on irregular shapes. $\dfrac{(n-2)\times180^\circ}{n}$ only gives each interior angle when all angles are equal. For an irregular quadrilateral, only the sum is fixed; individual angles must come from the given data (ratios, algebraic expressions, or specific measurements).
Doubting concave quadrilaterals. The $360^\circ$ rule still holds even when one interior angle is reflex (greater than $180^\circ$ ) in a concave (non-convex) quadrilateral — the diagonal-splitting proof needs convexity for the clean picture, but the sum itself does not depend on convexity.

Ask the tutor: Want to try extending the diagonal-splitting proof to a pentagon or hexagon, or work through a few more ratio-based quadrilateral angle problems?

Why does the angle sum stay 360 degrees even for a really lopsided or concave quadrilateral?

The proof only needs one diagonal to split the shape into two triangles whose angles regroup exactly into the original four corners — nothing about that argument depends on the sides being equal or the shape looking neat. For a concave quadrilateral, one interior angle is reflex (greater than $180^\circ$ ), but it still counts fully in the sum, and the total still comes out to $360^\circ$ . Try it: sketch any four-sided closed shape, split it with a diagonal, and the two $180^\circ$ triangle sums will always regroup into the four corners with nothing left over.

Is the interior angle sum the same as the exterior angle sum for a quadrilateral?

They both equal $360^\circ$ for a quadrilateral, but that is a coincidence specific to $n = 4$ , not a general rule. The interior sum depends on $n$ through $(n-2)\times180^\circ$ , so a pentagon's interior sum is $540^\circ$ and a hexagon's is $720^\circ$ . The exterior sum, however, is always $360^\circ$ for any convex polygon regardless of $n$ , because going once around the shape always turns through one full $360^\circ$ rotation. Keep the two ideas separate once $n \neq 4$ .

If I know three angles of a quadrilateral, is that always enough to find the fourth?

Yes, as long as the shape is a genuine simple (non-self-intersecting) quadrilateral: subtract the sum of the three known angles from $360^\circ$ to get the fourth, exactly as in the worked example. This works regardless of whether the quadrilateral is a square, kite, trapezium, or completely irregular shape, because the $360^\circ$ total holds for all of them. What it will not tell you is the type of quadrilateral — for that you additionally need to compare opposite angles, side lengths, or parallel sides.

Angle sum of quadrilaterals

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads