Pythagoras’ theorem: a week of ready-to-teach maths

Five lessons for Year 8 Measurement. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus every triangle drawn, Perigal’s puzzle to cut out, twelve triple cards to sort, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet. Every triangle is drawn on the worksheets.
3. Print the two cut-out sheets, one copy per pair: Perigal’s puzzle is for Day 1, the triple cards for Day 4.
4. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.
5. Teach straight from the plan. Timings, talk prompts, misconceptions and worked answers are all on the one page.

The idea in plain words

In a right-angled triangle the longest side, opposite the right angle, is the hypotenuse c, and the two shorter sides are the legs a and b. Pythagoras’ theorem says that the square on the hypotenuse has the same area as the squares on the two legs together: a² + b² = c². So the hypotenuse is c = √(a² + b²), and a shorter side is a = √(c² − b²). The theorem also works backwards: if the three sides of a triangle fit a² + b² = c², the triangle has a right angle, which is how builders check a corner with 3, 4 and 5.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

Exact and rounded answers

When a side is not a whole number, give it exactly as a square root (√65 cm) and rounded to two decimal places (8.06 cm), as the square roots unit earlier in Year 8 did. Round only at the end.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Day 1 sees the theorem as areas; Days 2 and 3 use it to find a missing side, whichever it is; Day 4 runs it backwards to test for a right angle; and Day 5 solves problems that take more than one step. It builds on square roots and irrational numbers earlier in Year 8 and on the areas of squares and triangles, and prepares trigonometry and three-dimensional problems in Year 9 and beyond.

Materials for the week (one trip)

• From the classroom: rulers, sharp pencils, calculators with a square root key, scissors, glue sticks, this pack printed.
• Cut-out sheet 1, one per pair: Perigal’s puzzle (Day 1).
• Cut-out sheet 2, one per pair: 12 triple cards (Day 4).
• Optional for Day 4: a rope with 13 knots tied at equal spacings, to make a 3-4-5 triangle.

Words for the word wall

right-angled triangle, right angle, hypotenuse, leg, square, square root, Pythagoras’ theorem, a² + b² = c², converse, Pythagorean triple, exact, rounded, diagonal, ladder, isosceles

Dear families

This week in maths, Year 8 is learning Pythagoras’ theorem: in a right-angled triangle, the square of the longest side equals the squares of the other two sides added together, a² + b² = c². It lets us find a missing side, such as the length of a ladder or a screen’s diagonal, and check whether a corner is square.

Try this at home

• Measure 30 cm along one edge of a table from a corner and 40 cm along the other. The gap between the marks should be 50 cm if the corner is square.
• Ask: a ladder reaches 4 m up a wall with its foot 3 m out. How long is it? (5 m.)
• Look up the size of a TV or phone screen. The size quoted is the diagonal, found with the theorem.
• Ask: is a triangle with sides 5, 12 and 13 right-angled? (Yes: 25 + 144 = 169 = 13².)

My right angles this week

Fill one row a day. Find a right-angled triangle or a rectangle, measure two sides, and work out the third.

Printed from the free seegongsik Pythagoras’ theorem teaching pack · seegongsik.com/au/y8/measurement/AC9M8M06/pack

The squares on the sides

The square on the hypotenuse has the same area as the squares on the two legs together. Students find the hypotenuse of any right-angled triangle, check the theorem with squares, and solve Perigal’s puzzle.

We are learning to

• find the hypotenuse: the side opposite the right angle, and the longest side,
• find the areas of the squares on the sides and see that a² + b² = c²,
• explain the theorem with a puzzle that fits two squares into a third.

Success criteria

• I can point to the hypotenuse of a right-angled triangle, whichever way it is drawn.
• I can show that 9² + 12² = 15² with squares.

You need

The worksheet and a calculator each; per pair, Perigal’s puzzle (cut-out sheet 1), scissors and a glue stick.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after the puzzle; start the second with Part A.

Watch for these ideas

• Taking the side at the top, or the sloping one, as the hypotenuse.
• Adding the sides instead of their squares: 3 + 4 is not 5.
• Using the theorem on a triangle with no right angle.

Answers

• Part A: 1. r; 2. y; 3. k.
• Part B: 25 + 144 = 169, c = 13; 64 + 225 = 289, c = 17; 36 + 64 = 100, c = 10.
• Part C: 1. 25, the longest side; 2. no: c² = a² + b² is bigger than either square, so c is longer than each leg.
• Part D: a² + a² = 50, so a² = 25 and each leg is 5 cm.

Puzzle: the square on a is 3 by 3 and the square on b is 4 by 4. Each numbered piece slides into one corner of the square on c; the square on a fills the middle.

The squares on the sides

A. Find the hypotenuse

Each triangle has a right angle, marked with a small square. Write the letter of its hypotenuse.

B. Squares on the sides

C. Think about it

1. A right-angled triangle has sides 7, 24 and 25. Which side is the hypotenuse? How do you know?
2. Can a leg of a right-angled triangle be longer than its hypotenuse? Explain with the squares.

D. Stretch: two equal legs

A right-angled triangle has two equal legs, and the square on its hypotenuse has an area of 50 cm². How long is each leg?

Finding the hypotenuse

Square the legs, add, and take the square root: c = √(a² + b²). Students find hypotenuses exactly and rounded, then use them for a ladder, a screen, a short cut across a park and the distance between two points.

We are learning to

• find the hypotenuse with c = √(a² + b²),
• give it exactly as a square root and rounded to two decimal places,
• spot the right-angled triangle hidden in a real problem.

Success criteria

• I can find the hypotenuse of a triangle with legs 5 cm and 9 cm, exactly and rounded.
• I can draw the right-angled triangle in a ladder or a diagonal problem.

You need

The worksheet, a ruler and a calculator with a square root key each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with Part B.

Watch for these ideas

• Stopping at c² and giving 100 instead of 10.
• Doubling instead of squaring: 6² is 36, not 12.
• Rounding before the square root is taken.

Answers

• Part A: 9 and 12: c = 15 cm; 20 and 21: c = 29 mm; 4 and 7: c = √65 ≈ 8.06 cm; 1.5 and 2: c = √6.25 ≈ 2.50 m.
• Part B: ladder √18.25 ≈ 4.27 m; screen √18956.25 ≈ 137.68 cm (screens are sold by the diagonal).
• Part C: the diagonal is 50 m, so the short cut saves 70 − 50 = 20 m.
• Part D: across 6, up 8, so PQ = √(6² + 8²) = 10 units.

Finding the hypotenuse

c² = a² + b², so c = √(a² + b²). Give each answer exactly, and to two decimal places when it is not whole.

A. Hypotenuses

B. A ladder and a screen

C. A short cut

A rectangular park is 40 m by 30 m. How much shorter is it to walk straight across, corner to corner, than along two edges?

D. Stretch: two points

Finding a shorter side

When the hypotenuse is known, take away instead of adding: a = √(c² − b²). Students find shorter sides, decide which kind of side a problem asks for, and find the height of an isosceles triangle.

We are learning to

• find a shorter side with a = √(c² − b²),
• decide whether a problem asks for the hypotenuse or a shorter side,
• check an answer: a shorter side must be less than the hypotenuse.

Success criteria

• I can find the missing leg of a right-angled triangle with hypotenuse 17 and one leg 8.
• I can say whether to add or subtract the squares, and why.

You need

The worksheet, a ruler and a calculator with a square root key each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the ladder.

Watch for these ideas

• Adding the squares when the hypotenuse is given, so the leg comes out longer than the hypotenuse.
• Subtracting the sides before squaring: 13 − 5 = 8 is not the answer.
• Taking the slanted side as the hypotenuse when the right angle is somewhere else.

Answers

• Part A: 13 and 5: 12 cm; 25 and 7: 24 m; 10 and 6: 8 cm; 9 and 4: √65 ≈ 8.06 mm.
• Part B: √(5² − 1.4²) = √23.04 ≈ 4.80 m up the wall.
• Part C: shorter side 40 m; hypotenuse 4.10 m.
• Part D: the height cuts the base in half: √(10² − 6²) = 8, so the area is ½ × 12 × 8 = 48.

Finding a shorter side

a² = c² − b², so a = √(c² − b²). Give each answer exactly, and to two decimal places when it is not whole.

A. Shorter sides

B. A ladder

A 5 m ladder leans against a wall with its foot 1.4 m out from the wall. How far up the wall does it reach?

C. Which side?

Sketch each one, mark the hypotenuse, then find the side asked for.

D. Stretch: an isosceles triangle

A triangle has two sides of 10 cm and a base of 12 cm. Draw its height from the top to the middle of the base. Find the height and the area of the triangle.

Is it a right angle?

The theorem works backwards: if a² + b² = c² for the longest side c, the triangle has a right angle. Students test triangles, sort triple cards, check a builder’s corner and make new Pythagorean triples.

We are learning to

• test whether three sides make a right angle, using the longest side as c,
• make new triples by multiplying a known triple,
• use the converse to check that a corner is square.

Success criteria

• I can show whether a triangle with sides 9, 40, 41 is right-angled.
• I can explain how a builder uses 3, 4 and 5 to check a corner.

You need

The worksheet and a calculator each; the triple cards from cut-out sheet 2, one set per pair, cut out.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after the cards; start the second with Part B.

Watch for these ideas

• Using the wrong side as c: it must be the longest.
• Adding all three squares instead of comparing a² + b² with c².
• Thinking close is good enough: 145 and 144 are not equal.

Answers

• Part A: right-angled 1, 3, 4, 7, 9, 10, 12; not 2, 5, 6, 8, 11.
• Part B: 6-8-10, 9-12-15, 30-40-50; 10-24-26, 15-36-39, 50-120-130.
• Part C: 3² + 4² = 25, so a square corner has a diagonal of 5 m. At 5.1 m the angle is a little more than a right angle, so the corner is not square.
• Part D: m = 2, n = 1: 3, 4, 5; m = 3, n = 2: 5, 12, 13; m = 4, n = 1: 15, 8, 17; each fits a² + b² = c².

Is it a right angle?

Take c as the longest side. If a² + b² = c², the triangle has a right angle opposite c. If not, it has no right angle.

A. Triple cards

Show your test for one card from each pile: ______________________________________________

B. New triples

Multiply each side of a right-angled triangle by the same number. Is the new triangle right-angled too?

C. A builder’s corner

A builder marks 3 m along one edge of a concrete slab from a corner and 4 m along the other. The distance between the marks is 5.1 m. Is the corner square? Explain.

D. Stretch: a triple machine

For whole numbers m > n, the sides m² − n², 2mn and m² + n² make a right-angled triangle. Try m = 2, n = 1; m = 3, n = 2; m = 4, n = 1. Check one of them.

Problems with Pythagoras

Real problems hide their right-angled triangles, and some need the theorem twice. Students sketch first, solve a walk, a roof and a pole in a box, find a mistake, and play the self-check quiz as a class game.

We are learning to

• sketch a problem and mark the right angle and the hypotenuse before calculating,
• use the theorem twice when one triangle gives a side of the next,
• check that an answer is shorter than the hypotenuse, or longer than each leg.

Success criteria

• I can find a rafter’s length from a roof’s span and rise.
• I can explain why 3 + 4 = 7 is not the hypotenuse of a 3-4 triangle.

You need

The worksheet and a calculator each; the self-check quiz at the bottom of the unit page for the game.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with Part C and the quiz.

Watch for these ideas

• Using the whole span of a roof instead of half of it.
• Adding lengths instead of squares, as in 3 + 4 = 7.
• Using the floor’s length instead of its diagonal for the pole.

Answers

• Part A: 1. straight across √(600² + 800²) = 1000 m, so 400 m shorter than the streets; 2. rafter √(4² + 3²) = 5 m.
• Part B: the floor diagonal is √(40² + 30²) = 50 cm; then √(50² + 120²) = 130 cm.
• Part C: Jo added the sides. The squares add: 9 + 16 = 25, so c = 5.
• Part D: any triangle with a² + b² = 100 works, for example 6 and 8, or √50 and √50.

Problems with Pythagoras

Sketch each problem first and mark the right angle and the hypotenuse.

A. A walk and a roof

B. A pole in a box

A box is 40 cm long, 30 cm wide and 120 cm tall. What is the longest straight pole that fits inside, from a bottom corner to the opposite top corner? (Find the diagonal of the floor first.)

C. Find the mistake

Jo says the hypotenuse of a right-angled triangle with legs 3 cm and 4 cm is 3 + 4 = 7 cm. What went wrong? Find the right length.

D. Stretch: many triangles, one hypotenuse

Find two different right-angled triangles with a hypotenuse of exactly 10 cm. Can the legs be whole numbers? Can they be equal?

Perigal’s puzzle

For Day 1, one per pair

Cut out the square on a and the four numbered pieces of the square on b, along the dashed lines. Do not cut the shaded square on c. Fit all five pieces on the square on c without turning any of them over or round. When they fit exactly, glue them down.

Teacher note: each numbered piece slides, without turning, into one corner of the square on c, and the square on a fills the gap in the middle: 9 + 16 = 25. Henry Perigal, a nineteenth-century amateur mathematician, made this cut famous; it works for any right-angled triangle.

Triple cards

For Day 4, one set per pair: twelve triangles

Each card gives the three sides of a triangle. Sort the cards into right-angled and not right-angled.

Teacher note: right-angled: 1, 3, 4, 7, 9, 10, 12; not right-angled: 2, 5, 6, 8, 11.

What we know: Pythagoras’ theorem

Work on your own with a calculator. Give answers exactly and to two decimal places where they are not whole.

1. Which side is the hypotenuse of this triangle? ______________________
2. Find the hypotenuse of a right-angled triangle with legs 15 cm and 20 cm. ______________________
3. Find the hypotenuse of a right-angled triangle with legs 5 cm and 7 cm. ______________________
4. A right-angled triangle has a hypotenuse of 26 m and one leg of 10 m. Find the other leg. ______________________
5. A 6 m ladder stands with its foot 2 m from a wall. How far up the wall does it reach? ______________________
6. Is a triangle with sides 9, 40, 41 right-angled? Show your test. ______________________
7. Is a triangle with sides 8, 9, 12 right-angled? Show your test. ______________________
8. Write Pythagoras’ theorem in words, for a triangle with legs a and b and hypotenuse c. ______________________

Answers and marking guide

Answers with working

1. h, the side opposite the right angle.
2. √(225 + 400) = 25 cm.
3. √(25 + 49) = √74 ≈ 8.60 cm.
4. √(676 − 100) = 24 m.
5. √(36 − 4) = √32 ≈ 5.66 m.
6. 81 + 1600 = 1681 = 41²: right-angled.
7. 64 + 81 = 145, but 12² = 144: not right-angled (close is not equal).
8. The square on the hypotenuse equals the sum of the squares on the other two sides: a² + b² = c².

A quick three-level guide

Eight questions, four ideas. A student at standard answers most questions with working, exact where needed and rounded at the end.

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: the hypotenuse and the squares, finding c, finding a leg, the converse, and problems.