229. The Pythagorean Balance · squares on legs equal square on hypotenuse

Try to break it

Try to break it: drag A and C into any right triangle you like. The right angle at B is locked, so AC² always stays exactly AB² + BC² — Pythagoras cannot be broken.

How you build it

Construct a right triangle with squares on its sides.

• Point tool: mark point B — the right-angle corner.
• Point tool: mark point C — BC is one leg.
• Segment tool: join B to C.
• Perp tool: click B (on BC), then click upward. This builds the exact right angle; leg BA lies on this perpendicular.
• Arc tool: centre B, drag out to your chosen leg length — the arc crosses the perpendicular at A, so BA is the second leg.
• Point tool: mark A where the arc meets the perpendicular. Angle B is now exactly 90 degrees.
• Segment tool: join C to A — the hypotenuse, facing the right angle.
• Square tool: click A then B — an exact square springs up on the outside of leg AB (its area is AB²).
• Square tool: click B then C — the exact square on leg BC (area BC²).
• Square tool: click C then A — the square on the hypotenuse (area CA²). Its area equals the other two squares added together.

The proof, step by step

Prove that AC² = AB² + BC² in a right-angled triangle.

• Draw squares ACDE, ABFG, and BCHI on sides AC, AB, and BC respectively.
• Draw BMN perpendicular to AC at M and DE at N. Join GC and BE.
• Prove △ABE ≅ △GBC using SAS congruence (AB=GB, BE=BC, ∠ABE=∠GBC).
• Area(△ABE) = ½ Area(ABFG) and Area(△GBC) = ½ Area(BMNC).
• Hence Area(ABFG) = Area(BMNC). Similarly, Area(BCHI) = Area(MNED).
• Adding both: Area(ABFG) + Area(BCHI) = Area(ACDE).
• Therefore, AC² = AB² + BC². Hence Proved.

Worked example

In a right-angled triangle ABC, right-angled at B, if AB = 6 cm and BC = 8 cm, find the length of AC.

By Pythagoras theorem, AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100. Thus, AC = √100 = 10 cm.

• 10 cm — correct
• 12 cm
• 14 cm
• 9 cm