Pythagoras finds any side. For the hypotenuse, add the squares: c² = a² + b². For a shorter side, subtract from the hypotenuse's square: b² = c² − a². Then take the square root, and check the hypotenuse is still the longest.
A 5 m ladder leans against a wall with its foot 3 m out. How high up the wall does it reach? Screens, ramps, roofs and ladders all hide a right-angled triangle.
Pythagoras' theorem links all three sides, so it can find any one of them:
The hypotenuse: add the squares. c2=a2+b2.
A shorter side: the hypotenuse's square minus the other side's square. b2=c2−a2.
Then take the square root. The answer often lies between two whole numbers: give it to two decimal places, and check that the hypotenuse is still the longest side.
Predict: Read the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
Try: Try it on the board and compare the result with your prediction.
Investigate: Work out why it happens, one step at a time.
Modify: Change the problem on purpose and see what happens.
Practise: A few on your own, to check you have it.
Make: Make your own problem using what you have learned. Expect a few tries; that is normal.
Make your prediction first. Help opens once you've locked it in.
PredictRead the problem and decide what you think before you try anything. A wrong guess is fine: it shows you what to look for.
The ladder
Which side of the triangle is the ladder?
TryTry it on the board and compare the result with your prediction.
The height that fits
Watch the biggest square as b changes: you need it to be 25.
Try it yourselfYour edits are saved in this browser.
The ladder is the hypotenuse, 5 m, and the foot is a = 3 m out. Change the height b until c = 5. Was your prediction right?
Use the board. This ticks when you get there.
InvestigateWork out why it happens, one step at a time.
Add or subtract?
Here are both kinds.
Worked exampleBefore each step, say what you think comes next.
c2=62+82=100,c=10Missing the hypotenuse: add the squares.
b2=102−62=100−36=64,b=8Missing a shorter side: subtract from the hypotenuse's square.
b2=72−42=33,b=33≈5.74Not a perfect square: between 5 and 6 (25 and 36), about 5.74.
5.74<7Check: the hypotenuse must still be the longest.
Now try one: hypotenuse 13 cm, one side 5 cm.
Read Ken's working. Can a shorter side be longer than the hypotenuse?
Your working can be short notes. Then choose the reason for the key step.
A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. How long is the other side, in cm?
Ken finds a shorter side. Click the line where his working goes wrong.
A rectangular screen is 48 cm wide and 36 cm high. How long is its diagonal? Write your steps.
Your working
A model solution
The diagonal is the hypotenuse of a right-angled triangle with sides 48 and 36 the corners of a rectangle are right angles
482+362=2304+1296=3600for the hypotenuse, add the squares
3600=60 cm take the square root
Check your own working
ModifyChange the problem on purpose and see what happens.
One change at a time
Each task changes one thing.
Need a hint?
The big square must be 100. Take away 36.
Try it yourselfYour edits are saved in this browser.
a = 6. Change b until the hypotenuse is 10.
Use the board. This ticks when you get there.
A ramp is 2.6 m long (the sloping side) and goes 2.4 m across the ground. How high does it rise, in m?
Hypotenuse 7 cm, one side 4 cm. The other side, in cm, to two decimal places?
Walking round a 30 m by 40 m park is 70 m. How many metres shorter is the path straight across, corner to corner?
PractiseA few on your own, to check you have it.
Any side
Hypotenuse 25 cm, one side 7 cm. The other side, in cm?
A square has sides of 5 cm. How long is its diagonal, in cm, to two decimal places?
A 10 m ladder reaches 8 m up a wall. How far is its foot from the wall, in m?
Review: the sides at the right angle are 9 cm and 12 cm. Hypotenuse, in cm?
MakeMake your own problem using what you have learned. Expect a few tries; that is normal.
Design a ramp
Then: real ramps should rise no more than 1 m for every 14 m along. Is this ramp too steep?
Need a hint?
The sloping length is the hypotenuse.
Another hint
1.3² − 0.5² = 1.69 − 0.25.
A wheelchair ramp must rise 0.5 m and its sloping surface is 1.3 m long. How far out from the step does it reach along the ground, in m? Write your steps.
Your working
A model solution
The sloping surface is the hypotenuse: c=1.3, a=0.5it is opposite the right angle at the step
b2=1.32−0.52=1.69−0.25=1.44for a shorter side, subtract the squares
b=1.44=1.2 m take the square root
Check your own working
For teachers
Students use Pythagoras to find the hypotenuse or a shorter side (subtracting the squares), with non-whole answers to two decimal places and a sense check, in contexts such as ladders, screens and ramps (AC9M8M06, node 2). About 45 minutes.
I can find a shorter side of a right-angled triangle.
I can decide whether to add or subtract the squares.
I can solve problems with ladders, screens and ramps, and check my answer makes sense.
Before class
Do the Run board yourself first. Students need the hypotenuse method (previous lesson) and square roots between whole numbers (Between the squares).
The one idea: add the squares to find the hypotenuse, subtract them to find a shorter side. The common mistake is adding for a shorter side (Ken's mistake); the check "the hypotenuse is the longest" catches it.
Running the lesson
Predict: students who chose 5.83 added; ask whether the wall can be taller than the ladder.
Run: ask what the big square has to be.
Investigate: for the screen, share sketches; label the hypotenuse first.
Modify: for the shortcut, ask what is being compared.
Make: compare 0.5 m in 1.2 m with the 1 in 14 rule.
The hard part (on purpose)
Deciding whether to add or subtract. Ask students to point to the hypotenuse and say whether it is the side they know or the side they want, before any calculation.
When students get stuck
A student adds the squares for a shorter side (Ken's mistake). Ask: Is your answer shorter than the hypotenuse? Fix: Subtract: hypotenuse squared minus the known side squared.
A student subtracts the lengths, not the squares. Ask: What does the theorem say about the squares? Fix: Square both, subtract, then root.
A student cannot find the triangle in a word problem. Ask: Where is the right angle? Which length is opposite it? Fix: Sketch it, mark the right angle, label c first.
Stretch and support
Stretch: A 2 m by 1.5 m door: can a 2.6 m wide board be carried through it, tilted along the diagonal?
Support: Use a decision card: 'Know c? Subtract. Want c? Add.' and keep squares to 20² visible.
Where it leads
Next: Right, acute or obtuse? (node 3): using the theorem backwards to test a triangle.
Where it leads: distance between two points, trigonometry (Year 10), and 3D diagonals of boxes.
What each step is for
the ladder: Committing to a height for the 5 m ladder with its foot 3 m out shows whether they subtract the squares or add them as for a hypotenuse.
the height that fits: Changing b until the hypotenuse is 5 shows that the missing square is the big square minus the known one.
add or subtract?: For a shorter side subtract the squares; Ken adds, and gets a side longer than the hypotenuse. A real problem starts by spotting the right angle and the hypotenuse; students own the working and pick the key reason.
one change at a time: One change at a time: a side on the board, a ramp with decimals, a side that is not a whole number, a shortcut saving, then the edge case of a shorter side longer than the hypotenuse.
any side: A shorter side, a diagonal, a ladder and a review, mixed; adding for a shorter side is the planted slip.
design a ramp: Designing a ramp from its rise and sloping length is the subtract case in a real setting, with decimals and a sense check.
Now you can
Find a shorter side of a right-angled triangle.
Decide whether to add or subtract the squares.
Solve problems with ladders, screens and ramps, and check my answer makes sense.