The Pythagorean theorem through areas

• Identify the hypotenuse as the side opposite the right angle.
• Interpret each squared side length as the area of a square built on that side.
• Use the theorem to find a missing length and verify the resulting area relationship.

Build the idea

Worked example

1. Step 1

   Use six and eight for the legs because they meet at the right angle. The unknown c labels the opposite and longest side.

   $$6^2+8^2=c^2$$

2. Step 2

   The square on the six-unit leg has area thirty-six, and the square on the eight-unit leg has area sixty-four. Their combined area is one hundred.

   $$36+64=c^2=100$$

3. Step 3

   Take the square root of both sides. Although an algebraic square equation can have positive and negative roots, the triangle side length uses the positive value.

   $$c=\sqrt{100}=10$$

4. Step 4

   The result is longer than either leg and shorter than their sum, which is consistent with the hypotenuse role and the triangle inequality.

   $$8<10<6+8$$

• Using the longest-looking drawn side as a leg because the diagram is rotated or not drawn to scale.

  Locate the right-angle marker first. The two sides forming that angle are the legs, and the side directly opposite it is the hypotenuse regardless of page orientation or visual scale.

• Adding the leg lengths before squaring, so six squared plus eight squared becomes fourteen squared.

  The theorem adds two square areas, not the side lengths before squaring. Calculate each square separately: six squared plus eight squared, then take the root after adding.

A right triangle has legs of five units and twelve units. Find its hypotenuse and verify the three square areas.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Turn lengths into areas

   Begin with a right triangle. Label the perpendicular legs a and b, and label the side opposite the right angle c. Build a square on each side. Their areas are a squared, b squared, and c squared, so the familiar exponents now describe actual regions we can compare.

   • a squared plus b squared equals c squared.
2. Part 2

   Use the same pieces twice

   Place four copies of the triangle inside a large square with side a plus b. One arrangement leaves a tilted c-by-c square in the center. Rearrange the same four triangles without changing the outer boundary. The uncovered space now consists of an a-by-a square and a b-by-b square.

   • a squared plus b squared equals c squared.
3. Part 3

   Match the uncovered regions

   Both pictures have equal outer area and contain the same four triangle areas. Their leftover areas must therefore be equal. The single c-square matches the combined a-square and b-square, giving a squared plus b squared equals c squared. The equation records an area conservation argument.

   • a squared plus b squared equals c squared.
   • c equals the square root of the quantity a squared plus b squared, end quantity.
4. Part 4

   Apply and verify

   For legs six and eight, the two square areas are thirty-six and sixty-four. Their sum is one hundred, so the hypotenuse is the positive square root, ten. A ten-by-ten square also has area one hundred, providing a visual verification. Ten is longer than eight and shorter than fourteen, so the length is geometrically plausible.

   • Six squared plus eight squared equals c squared.
   • Thirty-six plus sixty-four equals c squared, which equals one hundred.
   • c equals the square root of one hundred, which equals ten.
   • Eight is less than ten, which is less than six plus eight.

Scope and limitations

• The theorem applies directly only to right triangles; non-right triangles require a relationship such as the law of cosines.
• This lesson presents one rearrangement argument and numerical examples, not a survey of the many known proofs or a treatment of three-dimensional distance.