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Lesson 10 ~25 min Unit 3 · Geometry +85 XP

Angle Sum of Quadrilaterals

Every quadrilateral — no matter the shape — has interior angles that add up to $360^{\circ}$. We'll prove this by splitting any quadrilateral into two triangles, then use the rule to find missing angles in general and special quadrilaterals.

Today's hook: Triangles add to $180^{\circ}$. Quadrilaterals add to $360^{\circ}$. Pentagons? $540^{\circ}$. There's a pattern — want to see why?
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Think First
warm-up

Draw any weird-shaped quadrilateral (not a special one — just any 4-sided shape). Measure its four angles with a protractor and add them up. What total do you get? If it's a bit off from $360^{\circ}$, can you guess why?

Record your answer in your workbook.
1
The $360^{\circ}$ Rule
+5 XP

The four interior angles of any quadrilateral always sum to $\mathbf{360^{\circ}}$. The word "any" matters — this works for squares, kites, trapeziums and weird irregular shapes alike. The reason: a single diagonal splits any quadrilateral into TWO triangles, and each triangle contributes $180^{\circ}$.

Quadrilateral $ABCD$ split by diagonal $AC$: Triangle $ABC$ has angle sum $180^{\circ}$, and triangle $ACD$ has angle sum $180^{\circ}$. Their six angles together make up the four angles of the quadrilateral, so $\angle A + \angle B + \angle C + \angle D = 180 + 180 = 360^{\circ}$.

Any quadrilateral splits into 2 triangles by 1 diagonal A B C D Triangle ABC = 180° Triangle ACD = 180° Sum = 180 + 180 = 360° for the WHOLE quadrilateral.
$\angle A + \angle B + \angle C + \angle D = 360^{\circ}$ (∠ sum of quad).
Always $360^{\circ}$
Works for squares, kites, irregular shapes — everything 4-sided.
2 triangles inside
One diagonal ⇒ two triangles ⇒ $2 \times 180^{\circ}$.
Reason name
Write "(∠ sum of quad)" in brackets when you use the rule.
2
What You'll Master
objectives

Know

  • Angle sum of a quadrilateral $= 360^{\circ}$
  • Why this follows from triangle angle sum $= 180^{\circ}$
  • How to use the rule in special quadrilaterals

Understand

  • Why splitting into triangles works for ANY quadrilateral
  • How angle-sum combines with other rules (opp. angles, co-int.)
  • How to set up an algebraic equation when angles contain $x$

Can Do

  • Find the fourth angle given the other three
  • Solve for $x$ in algebraic-angle quadrilateral problems
  • Apply the rule in kites and trapeziums to find unknowns
3
Words You Need
vocabulary
Interior angleAn angle inside the shape, at a vertex.
Angle sumThe total of all interior angles of a shape.
DiagonalA line joining two non-adjacent vertices.
Convex quadrilateralAll vertices "stick out"; no inward dents.
Concave quadrilateralHas one "inward" vertex; rule still holds.
"∠ sum of quad"The reason name written in brackets in working.
"∠ sum of $\triangle$"Triangle angle sum $= 180^{\circ}$ (used in the proof).
Pattern$n$-gon angle sum $= (n - 2) \times 180^{\circ}$.
4
Why $360^{\circ}$? (The Proof)
+5 XP

Take any quadrilateral $ABCD$ and draw the diagonal $AC$. This splits it into two triangles, $ABC$ and $ACD$. Each triangle's three angles sum to $180^{\circ}$. The six angles together cover all four of the quadrilateral's interior angles (the diagonal "shares" the angle splits at $A$ and $C$). So the total is $180 + 180 = 360^{\circ}$.

Proof outline: Diagonal $AC$ ⇒ $\triangle ABC$ with sum $180^{\circ}$ and $\triangle ACD$ with sum $180^{\circ}$. Total angles in both triangles = $360^{\circ}$. These six pieces equal the four interior angles of $ABCD$ exactly (the diagonal splits the corner at $A$ and at $C$, but both halves are inside the quadrilateral, so they all count). $\therefore \angle A + \angle B + \angle C + \angle D = 360^{\circ}$.

Proof: split into triangles A B C D x + y + z = 180° p + q + r = 180° Sum of all 6 angles = 360° = angle sum of the quadrilateral.
$\triangle$ sum $+ \triangle$ sum $= 180 + 180 = 360^{\circ}$.
One diagonal
Pick either $AC$ or $BD$ — both work.
Pattern extends
$n$-gon angle sum $= (n - 2) \times 180^{\circ}$. Pentagon $= 540^{\circ}$.
Works for concave too
Even an "arrowhead" quadrilateral still sums to $360^{\circ}$.
Book notes · Card 4
  • Proof: a diagonal splits any quadrilateral into 2 triangles ⇒ $180 \times 2 = 360^{\circ}$.
  • Reason name: "(∠ sum of quad)".
  • Pattern: $n$-gon angle sum $= (n - 2) \times 180^{\circ}$.
Quick check+1 coin

True or False: Every quadrilateral — including concave (dented) ones — has interior angles that add to $360^{\circ}$.

5
Finding a Missing Angle
+5 XP

When three angles of a quadrilateral are known, find the fourth by subtracting their total from $360^{\circ}$. Layout:

Steps:
1. State the rule: $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$ (∠ sum of quad)
2. Substitute the three known angles
3. Subtract: missing angle $=$ $360^{\circ} - \text{sum of three known}$
4. Always check that all four add to $360^{\circ}$.

Example: three angles given 80° 100° 130° x° x = 360 - 80 - 100 - 130 x = 50°
Missing angle $= 360^{\circ} - $ (sum of other three).
Show the rule
Mention the rule before subtracting — markers want to see it.
Then subtract
$360 - $ (the three given angles) gets the answer in one step.
Sanity check
Verify the four angles really add to $360^{\circ}$.
Book notes · Card 5
  • 4th angle = $360^{\circ} - $ (sum of 3 known).
  • Always cite "(∠ sum of quad)".
  • Verify the four angles total $360^{\circ}$.
Quick check+1 coin

Three angles of a quadrilateral are $90^{\circ}, 100^{\circ}, 50^{\circ}$. What is the fourth?

6
Algebraic Angles
+5 XP

When the angles contain $x$, the same rule applies. Add all four expressions, set equal to $360^{\circ}$, and solve. Always finish by stating the actual angle values.

For example, if the angles are $(x + 20)^{\circ}, (2x)^{\circ}, (x - 10)^{\circ}, (3x + 30)^{\circ}$:
1. $(x + 20) + 2x + (x - 10) + (3x + 30) = 360$  (∠ sum of quad)
2. Simplify: $7x + 40 = 360$
3. $7x = 320 \Rightarrow x = \dfrac{320}{7}$... here you'd recompute with friendlier numbers. With $(x + 10), (2x), (x - 10), (3x + 20)$ the simpler form is $7x + 20 = 360 \Rightarrow x = \dfrac{340}{7}$... carefully chosen numbers in classroom problems give whole numbers.

Algebra: angles contain x (x+10)° (2x)° (x+20)° (2x-30)° (x+10)+(2x)+(x+20)+(2x-30) = 360 6x = 360 ⇒ x = 60
Set the sum of all four angle expressions equal to $360^{\circ}$, then solve for $x$.
Add expressions
Combine all $x$-terms and constants carefully.
Set to $360^{\circ}$
Write a clean equation, then isolate $x$.
Don't stop at $x$
Sub $x$ back to find each actual angle. State them.
Book notes · Card 6
  • Sum all four angle-expressions, set $= 360^{\circ}$.
  • Solve for $x$, then substitute back to find each angle.
  • Verify by adding the four final angles.
Fill the blank+1 coin

A quadrilateral has angles $x, x, 2x, 2x$. Then $6x = 360^{\circ}$, giving $x = $ $^{\circ}$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Find the fourth angle
+15 XP per step
Q1
PROBLEM
A quadrilateral has angles $75^{\circ}, 110^{\circ}, 95^{\circ}$ and $x^{\circ}$. Find $x$.
  1. 1
    Use the rule
    $75 + 110 + 95 + x = 360$ (∠ sum of quad)
  2. 2
    Add the known angles
    $280 + x = 360$
  3. 3
    Solve
    $x = 360 - 280 = 80^{\circ}$
    Check $75 + 110 + 95 + 80 = 360^{\circ}$ ✓
Answer$x = 80^{\circ}$.
Watch Me Solve It · Algebraic angles
+15 XP per step
Q2
PROBLEM
A quadrilateral has angles $(x + 30)^{\circ}, (2x)^{\circ}, (x + 50)^{\circ}, (2x + 40)^{\circ}$. Find $x$ and state each angle.
  1. 1
    Equation
    $(x + 30) + 2x + (x + 50) + (2x + 40) = 360$ (∠ sum of quad)
  2. 2
    Simplify and solve
    $6x + 120 = 360 \Rightarrow 6x = 240 \Rightarrow x = 40$
  3. 3
    State the angles
    $70^{\circ}, 80^{\circ}, 90^{\circ}, 120^{\circ}$.
    Check $70 + 80 + 90 + 120 = 360^{\circ}$ ✓
Answer$x = 40$; angles $70^{\circ}, 80^{\circ}, 90^{\circ}, 120^{\circ}$.
Watch Me Solve It · Kite with one unknown
+15 XP per step
Q3
PROBLEM
Kite $ABCD$ has $\angle B = \angle D$. The angles are $\angle A = 70^{\circ}, \angle B = x^{\circ}, \angle C = 110^{\circ}, \angle D = x^{\circ}$. Find $x$.
  1. 1
    Use angle sum
    $70 + x + 110 + x = 360$ (∠ sum of quad)
  2. 2
    Simplify
    $180 + 2x = 360 \Rightarrow 2x = 180$
  3. 3
    Solve
    $x = 90^{\circ}$, so $\angle B = \angle D = 90^{\circ}$.
    Check $70 + 90 + 110 + 90 = 360^{\circ}$ ✓
Answer$x = 90^{\circ}$.

Common Pitfalls

8
Common Pitfalls
heads-up
Using $180^{\circ}$ instead of $360^{\circ}$
Triangle = $180^{\circ}$. Quadrilateral = $360^{\circ}$. Mixing them up is the most common error.
Fix: Count the sides — 4 sides ⇒ $360^{\circ}$.
Forgetting to add expressions correctly
Combining $(2x + 10) + (3x - 20)$ as $5x - 10$ instead of $5x - 10$ — tiny sign or coefficient errors snowball.
Fix: Line up like terms, add coefficients, add constants separately.
Stopping at $x$
If the question asks for the angles, just writing "$x = 40$" loses marks.
Fix: Substitute $x$ back to find each angle, and state all four.
Copy Into Your Books

The rule

  • $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$
  • Reason: (∠ sum of quad)
  • Works for ALL quadrilaterals

Why $360^{\circ}$

  • 1 diagonal ⇒ 2 triangles
  • Each $\triangle = 180^{\circ}$
  • Total $= 2 \times 180 = 360^{\circ}$

Find 4th angle

  • 4th $= 360 - $ (sum of 3)
  • State the rule
  • Check final sum

Algebraic angles

  • Sum of expressions $= 360$
  • Solve for $x$
  • Substitute back, state angles

How are you completing this lesson?

Brain Trainer · 4 problems

D
Brain Trainer · $360^{\circ}$ Practice
4 problems

Four drills. Solve, then reveal.

  1. 1 Angles $60^{\circ}, 70^{\circ}, 80^{\circ}, x^{\circ}$. Find $x$.

    $360 - 60 - 70 - 80$.$x = 150^{\circ}$

  2. 2 Angles $90^{\circ}, 90^{\circ}, 90^{\circ}, x^{\circ}$. Find $x$.

    $360 - 270$.$x = 90^{\circ}$ (rectangle)

  3. 3 Angles $x, x, x, x$. Find $x$.

    $4x = 360 \Rightarrow x = 90$.$x = 90^{\circ}$ (square/rectangle)

  4. 4 Angles $2x, 3x, 4x, 3x$. Find $x$.

    $12x = 360 \Rightarrow x = 30$.$x = 30^{\circ}$ (angles $60, 90, 120, 90$)
Complete in your workbook.

Quick Check · 5 questions

1
The interior angles of every quadrilateral add up to:
+10 XP
2
Three angles are $90^{\circ}, 80^{\circ}, 95^{\circ}$. Find the fourth.
+10 XP
3
The proof of the $360^{\circ}$ rule comes from splitting a quadrilateral into:
+10 XP
4
A quadrilateral has angles $x^{\circ}, 2x^{\circ}, 3x^{\circ}, 4x^{\circ}$. Find $x$.
+10 XP
5
Using the pattern $(n - 2) \times 180^{\circ}$, the angle sum of a pentagon (5 sides) is:
+10 XP

Show Your Working · 3 questions

Show Your Working
9 marks total
Apply Easy 3 MARKS

Q6. A quadrilateral has angles $\angle A = 105^{\circ}, \angle B = 75^{\circ}, \angle C = 110^{\circ}$.
(a) State the angle-sum rule with reason.
(b) Compute $\angle D$.
(c) Verify your four angles add to $360^{\circ}$.

Answer in your workbook.
Apply Medium 3 MARKS

Q7. A quadrilateral has angles $(2x)^{\circ}, (3x + 10)^{\circ}, (x + 20)^{\circ}, (4x + 10)^{\circ}$.
(a) Set up an equation.
(b) Solve for $x$.
(c) State each angle.

Answer in your workbook.
Reason Hard 3 MARKS

Q8. Explain (using triangles) why a quadrilateral's angle sum equals $360^{\circ}$. Include a diagram with one diagonal, and label which triangle contributes which $180^{\circ}$.

Answer in your workbook.
Comprehensive Answers

Quick Check

1. C — $360^{\circ}$.

2. A — $360 - 90 - 80 - 95 = 95^{\circ}$.

3. D — 2 triangles (one diagonal).

4. B — $10x = 360 \Rightarrow x = 36$.

5. A — $(5 - 2) \times 180 = 540^{\circ}$.

Show Your Working Model Answers

Q6 (3 marks): (a) $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$ (∠ sum of quad) [1]. (b) $\angle D = 360 - 105 - 75 - 110 = 70^{\circ}$ [1]. (c) $105 + 75 + 110 + 70 = 360^{\circ}$ ✓ [1].

Q7 (3 marks): (a) $(2x) + (3x + 10) + (x + 20) + (4x + 10) = 360$ (∠ sum of quad) [1]. (b) $10x + 40 = 360 \Rightarrow 10x = 320 \Rightarrow x = 32$ [1]. (c) Angles: $64^{\circ}, 106^{\circ}, 52^{\circ}, 138^{\circ}$ (sum $= 360$) [1].

Q8 (3 marks): Draw quadrilateral $ABCD$ with diagonal $AC$. Triangle $ABC$ has angle sum $180^{\circ}$ (∠ sum of $\triangle$). Triangle $ACD$ has angle sum $180^{\circ}$ (∠ sum of $\triangle$). [1 for diagram, 1 for two triangles] The six angles of the two triangles together form the four interior angles of $ABCD$, so $\angle A + \angle B + \angle C + \angle D = 180 + 180 = 360^{\circ}$. [1 for conclusion]

Stretch Challenge · +25 XP, +10 coins

Patterns in $n$-gons

(a) The formula $(n - 2) \times 180^{\circ}$ gives the interior angle sum of any $n$-sided polygon. Use it to find the angle sum of a hexagon (6 sides), a heptagon (7) and a decagon (10). (b) A regular polygon has all interior angles equal. If a regular polygon has an interior angle of $144^{\circ}$, how many sides does it have? (c) Quadrilateral $ABCD$ has $\angle A : \angle B : \angle C : \angle D = 1 : 2 : 3 : 4$. Find all four angles.

Reveal solution

(a) Hexagon: $(6 - 2) \times 180 = 720^{\circ}$. Heptagon: $(7 - 2) \times 180 = 900^{\circ}$. Decagon: $(10 - 2) \times 180 = 1440^{\circ}$. (b) For a regular $n$-gon, each interior angle $= \dfrac{(n - 2) \times 180}{n}$. Setting $= 144$: $144n = 180n - 360 \Rightarrow 36n = 360 \Rightarrow n = 10$. So it's a decagon. (c) Let the angles be $k, 2k, 3k, 4k$. Sum $= 10k = 360 \Rightarrow k = 36$. Angles: $36^{\circ}, 72^{\circ}, 108^{\circ}, 144^{\circ}$.

R
Quick Review

The rule

$\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.

Proof

One diagonal ⇒ two triangles ⇒ $2 \times 180^{\circ}$.

Reason name

(∠ sum of quad)

4th angle

$= 360 - $ (sum of 3 known).

Algebra

Sum expressions $= 360$, solve for $x$.

$n$-gons

$(n - 2) \times 180^{\circ}$ for any polygon.

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