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.

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 $\angle A + \angle B + \angle C + \angle D = 180 + 180 = 360^{\circ}$.

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

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

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 $= (n - 2) \times 180^{\circ}$.

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

$\triangle$ 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 $360^{\circ}$.

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 $360^{\circ}$.

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?

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.

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

PROBLEM

A quadrilateral has angles $75^{\circ}, 110^{\circ}, 95^{\circ}$ and $x^{\circ}$. Find $x$.

1
Use the rule
$75 + 110 + 95 + x = 360$ (∠ sum of quad)
2
Add the known angles
$280 + x = 360$
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

PROBLEM

A quadrilateral has angles $(x + 30)^{\circ}, (2x)^{\circ}, (x + 50)^{\circ}, (2x + 40)^{\circ}$. Find $x$ and state each angle.

1
Equation
$(x + 30) + 2x + (x + 50) + (2x + 40) = 360$ (∠ sum of quad)
2
Simplify and solve
$6x + 120 = 360 \Rightarrow 6x = 240 \Rightarrow x = 40$
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

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
Use angle sum
$70 + x + 110 + x = 360$ (∠ sum of quad)
2
Simplify
$180 + 2x = 360 \Rightarrow 2x = 180$
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

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

Brain Trainer · $360^{\circ}$ Practice

4 problems

Four drills. Solve, then reveal.

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

$360 - 60 - 70 - 80$.$x = 150^{\circ}$
2 Angles $90^{\circ}, 90^{\circ}, 90^{\circ}, x^{\circ}$. Find $x$.

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

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

The interior angles of every quadrilateral add up to:

+10 XP

Three angles are $90^{\circ}, 80^{\circ}, 95^{\circ}$. Find the fourth.

+10 XP

The proof of the $360^{\circ}$ rule comes from splitting a quadrilateral into:

+10 XP

A quadrilateral has angles $x^{\circ}, 2x^{\circ}, 3x^{\circ}, 4x^{\circ}$. Find $x$.

+10 XP

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. D2 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: $36^{\circ}, 72^{\circ}, 108^{\circ}, 144^{\circ}$.

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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