068i - Deriving Pythagoras' Theorem

Lesson 68 — Explicit Instruction: Deriving and Stating Pythagoras’ Theorem

Strand: Measurement | Descriptor: AC9M8M06 | Duration: 45 minutes

Learning Intentions

• To derive Pythagoras’ theorem by comparing the areas of squares built on the sides of a right-angled triangle.
• To state Pythagoras’ theorem correctly and use it to test whether a triangle is right-angled.

Success Criteria

I can:

1. Explain, using an area diagram, why the sum of the areas of the two smaller squares equals the area of the largest square in a right-angled triangle.
2. State Pythagoras’ theorem as , correctly identifying the hypotenuse .
3. Use the theorem to check whether a triangle with three given side lengths is right-angled.
4. Recall common Pythagorean triples and generate new ones by scaling.

Warmup

(5 minutes — mini whiteboards, rapid-fire)

Recall from Year 7: two squares have areas and .

1. Find the side length of each square.
2. Find the side length of a square whose area equals the sum of these two areas.
3. Write the three side lengths you found as a set of numbers. Do you recognise them?
4. In each diagram below, circle the side that is opposite the right angle (the hypotenuse).

Answers: 1. cm and cm; 2. cm; 3. — this is the same Pythagorean triple previewed in the Year 7 squares-and-roots unit; 4. The hypotenuse is always the side opposite the right angle — the longest side — regardless of how the triangle is rotated on the page.

Activities

Activity 1 — Explicit Instruction: the Area Proof (14 min)

I do: Arrange four identical right-angled triangles, with legs and and hypotenuse , inside a large square of side , so that a smaller square of side is left uncovered in the middle.

The large square’s area can be found two ways:

Since both expressions describe the same total area, they must be equal:

This is Pythagoras’ theorem: in a right-angled triangle with legs and hypotenuse , .

We do: Verify the theorem numerically for the –– triangle:

You do: Verify the theorem for: (a) (b) (c) .

Activity 2 — Explicit Instruction: Identifying the Hypotenuse and Testing Right Angles (12 min)

I do: The hypotenuse is always the side opposite the right angle — not “the side that looks longest” or “the bottom side.” Model identifying it in three differently-rotated triangle diagrams.

To test whether a triangle with sides is right-angled, check whether using the two shortest sides as and :

Since , the triangle is not right-angled.

We do: Test .

You do: Test (a) (b) (c) .

Activity 3 — Inquiry: Spotting the Pattern in Pythagorean Triples (8 min)

Pairs. Given the table:

Triple

1. What do you notice about the second and third columns compared to the first?
2. Predict the next triple in the family, and check it.
3. Can you find a Pythagorean triple that is not a multiple of or ?

Teacher prompt if stuck: “Try scaling by . Does the pattern still work?”

Checks for Understanding

(6 minutes — exit ticket, collected)

1. State Pythagoras’ theorem, defining each letter used.
2. In a right-angled triangle, which side is always the hypotenuse?
3. Verify whether a triangle with sides is right-angled.
4. A triple is scaled by a factor of . State the new triple, and verify it still satisfies Pythagoras’ theorem.
5. Reasoning. Explain, using the area diagram from Activity 1, why the theorem is really a statement about areas, not just numbers.

Answers: 1. , where and are the two legs and is the hypotenuse (the side opposite the right angle); 2. The side opposite the right angle; 3. ; ; since , not right-angled; 4. ; ✓; 5. The squares built on each side represent literal areas; the theorem states that the two smaller square areas together exactly fill the largest square’s area — the "" terms are areas, not just squared numbers.

Common Misconceptions

Misconception	How to pre-empt it
The hypotenuse is “the side on the bottom” or “the longest-looking side on the page.”	Rotate triangle diagrams randomly; always locate the right angle first, then the side opposite it.
applies to any triangle.	State explicitly, every time: this only works for right-angled triangles.
Adding the side lengths instead of squaring them.	Model the area-square diagram alongside every numeric example.
Believing any three numbers can be “plugged in” in any order as , , .	Insist (the largest value) is always isolated on one side of the equation before substituting.
Assuming all triples are multiples of .	Introduce and explicitly as separate “primitive” families.

Enrichment — Competition-Style Problems

E1 (AMC Junior style). A right-angled triangle has legs and . What is its hypotenuse?

Answer

(This is the triple scaled by .)

E2 (Kangaroo style — alternative proof). US President James Garfield’s proof of Pythagoras’ theorem uses a trapezium made from two copies of a right triangle (legs , hypotenuse ) plus the original, arranged so the trapezium’s area can be computed two ways. Research or sketch this arrangement — what area equation results?

Answer

The trapezium has parallel sides and and height , giving area . This is also the sum of three triangle areas: two copies of plus one right-angled triangle of legs giving . Setting them equal and simplifying gives , the same result as the square-in-a-square proof.

E3 (Challenge). How many Pythagorean triples exist with hypotenuse less than (counting only triples in “lowest terms,” i.e. not multiples of a smaller triple)?

Answer

Three: ; ; — all others under hypotenuse (such as ) are scaled multiples of these.

Homework

1. State Pythagoras’ theorem in words and in symbols.
2. Verify whether each triangle is right-angled: (a) (b) (c) .
3. A Pythagorean triple begins . Find the missing value using the theorem, then check your answer is a whole number.
4. Scale the triple by a factor of . State the new triple.
5. Reasoning. Explain why swapping which side is called the hypotenuse in the equation would give a wrong answer, using a labelled diagram to support your explanation.
6. Challenge. Two right-angled triangles share the same hypotenuse of . One has legs and . Find a different pair of whole-number legs also satisfying .

Answers: 1. In a right-angled triangle, the sum of the squares of the two legs equals the square of the hypotenuse: ; 2. (a) , not right-angled (b) , right-angled (c) , not right-angled; 3. , so the missing value is ; 4. ; 5. Swapping for a leg would place the largest value’s square on the wrong side, e.g. using with as the true hypotenuse would produce a negative or impossible result, since must always be the largest value squared alone; 6. ( scaled by ), since .