Academy — Sanjib Sikder

The square on the hypotenuse equals the sum of the squares on the other two sides. Always. For every right triangle that ever existed.

Why does this exist?

You already know the hypotenuse is the longest side. You know the right angle forces it. But how much longer is it? Is there a precise relationship between the three sides — not just an inequality, but an exact equation?

There is. And it was known to the Babylonians, the Indians, the Chinese, and the Greeks — independently, across centuries. Pythagoras did not discover it first. But his name stuck, and the theorem is this:

In any right triangle, the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides.

Not approximately. Not usually. Exactly. Always.

See it

A right triangle. A square built outward on each side. Drag any vertex.

Canvas 400 520
Scale 52
Grid 0
Origin at 220 300

Point A at -2, -1.5
Point B at  2, -1.5
Point C at -2,  1.5

Triangle T vertices A B C
Fill T 0.15 blue
Color T blue

# Square on leg AB (area = 16)
Rotate A around B by  90 as AB1
Rotate B around A by -90 as AB2
Quadrilateral sqAB vertices A B AB1 AB2
Fill sqAB 0.20 orange
Color sqAB orange

# Square on leg CA (area = 9)
Rotate C around A by  90 as CA1
Rotate A around C by -90 as CA2
Quadrilateral sqCA vertices C A CA1 CA2
Fill sqCA 0.20 green
Color sqCA green

# Square on hypotenuse BC (area = 25)
Rotate B around C by  90 as BC1
Rotate C around B by -90 as BC2
Quadrilateral sqBC vertices B C BC1 BC2
Fill sqBC 0.20 red
Color sqBC red

Mark right-angle at A

Label A "A"
Label B "B"
Label C "C"

Drag A
Drag B
Drag C

The orange square and the green square together fill exactly the same area as the red square. Every single drag. Every shape of right triangle.

The right angle at A is the anchor. Try moving A — the square mark follows. The colours always balance.

Measure it

Now with live areas. Watch the three numbers in the panel.

Canvas 400 520
Scale 52
Grid 0
Origin at 220 300

Point A at -2, -1.5
Point B at  2, -1.5
Point C at -2,  1.5

Triangle T vertices A B C
Fill T 0.10 blue
Color T blue

# Square on leg AB
Rotate A around B by  90 as AB1
Rotate B around A by -90 as AB2
Quadrilateral sqAB vertices A B AB1 AB2
Fill sqAB 0.15 orange
Color sqAB orange

# Square on leg CA
Rotate C around A by  90 as CA1
Rotate A around C by -90 as CA2
Quadrilateral sqCA vertices C A CA1 CA2
Fill sqCA 0.15 green
Color sqCA green

# Square on hypotenuse BC
Rotate B around C by  90 as BC1
Rotate C around B by -90 as BC2
Quadrilateral sqBC vertices B C BC1 BC2
Fill sqBC 0.15 red
Color sqBC red

Mark right-angle at A

Show area sqAB
Show area sqCA
Show area sqBC

Label A "A"
Label B "B"
Label C "C"

Drag A
Drag B
Drag C

Add the first two numbers. They equal the third.

Drag to a tall thin triangle. Add again. Drag to a nearly flat triangle. Add again. The equation holds every time — not because you were told it does, but because you can see it.

State it

The relationship you just measured has a name.

Call the two legs a and b. Call the hypotenuse c. Then:

$a^2 + b^2 = c^2$

That is the Pythagorean theorem. The areas of the squares are a², b², and c². So "the sum of the squares on the legs equals the square on the hypotenuse" is exactly the same statement as a² + b² = c².

Canvas 400 360
Scale 52
Grid 1
Origin at 200 180

Point A at -2, -1.5
Point B at  2, -1.5
Point C at -2,  1.5

Triangle T vertices A B C
Fill T 0.12 blue
Color T blue

Segment leg_a from A to B
Segment leg_b from A to C
Segment hyp_c from B to C

Mark right-angle at A

Show length leg_a
Show length leg_b
Show length hyp_c

Text at -3.8, -3.0 gray "a² + b² = c²"

Label A "A"
Label B "B"
Label C "C"

Drag B
Drag C

The panel shows a, b, c live. Square them yourself and check. Any right triangle, anywhere, satisfies the equation.

Why is it true?

There are over 370 known proofs of the Pythagorean theorem — more than any other theorem in mathematics. Here is the cleanest one to picture.

Take the red square — the one on the hypotenuse. Tile it with four copies of the original triangle plus a small square in the middle. The small square has side (a − b). The four triangles each have legs a and b.

Area of red square = 4 × (½ab) + (a − b)² = 2ab + a² − 2ab + b² = a² + b²

That is exactly the sum of the orange and green squares. The algebra closes.

This is one proof. There are hundreds more — from similar triangles, from coordinates, from folding paper. All of them arrive at the same place: a² + b² = c².

The seed planted here

The Pythagorean theorem does not just connect three sides of a triangle. It connects distance to coordinates.

If a point has coordinates (x, y), its distance from the origin is √(x² + y²). That formula is Pythagoras — the horizontal leg is x, the vertical leg is y, the hypotenuse is the distance.

Every distance formula in mathematics, every length calculation in physics, every signal strength in engineering — all of them are this theorem, applied to a right triangle that may not even be drawn.