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IGCSE · Cambridge (CIE) · 0444

Use Pythagoras and verify the right-angle condition

Identify the hypotenuse opposite the right angle · Add or subtract squared lengths in the correct roles · Apply two right triangles to an indirect distance

Section Paper 4 · Historical Extended · calculator

Mathematics (US) 0444 · historical 2025 study · C8.1 / E8.1
Exam scope and assessment notes

Core and Extended foundations. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.

Before you begin

  • Squares and roots
  • Right angles and metric lengths

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What should you establish before applying the method in this lesson?

What you will learn

  • Identify the hypotenuse opposite the right angle
  • Add or subtract squared lengths in the correct roles
  • Apply two right triangles to an indirect distance
Name sides relative to θθ12513ABCHypotenuse opposite the right angle
The right triangle has adjacent leg 12, opposite leg 5 and hypotenuse 13 relative to θ at A. The square marks the right angle at B. Switching to the acute angle at C exchanges the two leg roles, while hypotenuse AC stays 13. The side ratios are defined relative to the named angle.

The right angle is a condition

For perpendicular legs x,y and hypotenuse h, x²+y²=h². The hypotenuse is opposite the right angle and is the longest side. To find it, take √(x²+y²); to find a missing leg, subtract the other leg square from h² before taking the non-negative root.

A triangle with side lengths 5,12,13 satisfies 25+144=169. The converse establishes a right angle opposite side 13. Side lengths 6,8,9 do not satisfy the same condition, so the right-triangle formula cannot be imposed on that triangle.

More than one triangle

For a cuboid with perpendicular dimensions 3,4,12, first find its base diagonal √(3²+4²)=5. That base diagonal and height 12 form another right triangle, whose hypotenuse is the space diagonal. Draw the two triangles rather than guessing which lengths are perpendicular. Exact roots can be retained when a calculator is unavailable.

Sources for this lesson

Pause and explain

Do side lengths 6,8,9 make a right triangle?

Put the idea to work

Worked example

Find the hypotenuse when the perpendicular legs are 5 and 12 units.

Show the worked solution
  1. h²=25+144=169.
  2. Take the non-negative root.

Answer 13 units.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find the hypotenuse of a right triangle with legs 6 cm and 8 cm.

Give me a hint

Add the two leg squares.

Compare my reasoning
  1. h²=36+64=100.
  2. h=10 cm.

10 cm.

Look for these in your work

  • Use perpendicular legs
  • Take a positive length
2 · Independent

A right triangle has hypotenuse 13 m and one leg 5 m. Find the other leg.

Give me a hint

Subtract from the hypotenuse square.

Compare my reasoning
  1. Leg²=169−25=144.
  2. The leg is 12 m.

12 m.

Look for these in your work

  • Identify the hypotenuse correctly
  • Subtract before taking the root
3 · Transfer

Find the space diagonal of a cuboid with perpendicular dimensions 3 cm, 4 cm and 12 cm. Explain the two triangles used.

Give me a hint

Find the base diagonal first.

Compare my reasoning
  1. The 3 by 4 base has diagonal 5 cm.
  2. The base diagonal and height 12 are perpendicular, giving √(25+144)=13 cm.

13 cm, using a base right triangle then a vertical right triangle.

Look for these in your work

  • Locate the perpendicular quantities
  • Keep the intermediate value exact

Common mistakes

  • Use the side opposite the stated right angle as hypotenuse.
  • A sloping drawing alone does not establish perpendicularity.
Recall without your notes

What can you explain now?

Do side lengths 6,8,9 make a right triangle?

Compare with the explanation

No; 6²+8²=100 but 9²=81.

The largest side must satisfy the squared-length condition for a right angle.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Do side lengths 6,8,9 make a right triangle?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Mathematics (US)

Course scope and assessment

191 original lessons provide 573 guided, independent and application tasks. All 142 tier entries have linked teaching foundations: 55 Core and 87 Extended entries across 87 unique numbered sections in ten areas. Extended includes Core, so the tier-entry count is not a percentage of teaching time or student mastery. This is not full programme completion: qualified mathematics-teacher review, cumulative and exam-length assessment, and reliable drawn-work review remain unfinished. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.

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