Right Triangles

Try this

With the slider at the far left, four congruent right triangles sit in the four corners of a big square, leaving a tilted square in the middle. Its side is the hypotenuse $c$ of the triangles.
Drag the slider to the far right: three of the triangles slide over and pair up into rectangles, leaving two squares uncovered, with sides $a$ and $b$ .
All the way through, the big square has area $(a + b)^{2}$ and the four triangles have total area $2 ab$ ; neither changes. So the uncovered area does not change either: $c^{2} = a^{2} + b^{2}$ .
Change $a$ and $b$ and drag again. The ✓ at the bottom right never goes out.

Predict first

A right triangle has legs $5$ and $12$ . How long is the hypotenuse?

Choose an answer first; then you can see the result.

Why: the same uncovered area, counted twice

A jigsaw proof of the Pythagorean theorem

Step 1 / 4

Arrangement one
Four right triangles (legs $a$ and $b$ , hypotenuse $c$ ) sit in the corners of a square of side $a + b$ . The quadrilateral in the middle has four sides of length $c$ , and each of its angles is $18 0^{\circ}$ minus two complementary acute angles, that is, $9 0^{\circ}$ . So it is a square.

$uncovered = c^{2}$
Slide
Arrangement two
Both counts measure the same thing

The Pythagorean theorem is an equation

Once you know there is a right angle, you can write an equation in the side lengths. This is the most common way algebra enters a geometry problem.

Example: the hypotenuse is 2 longer than one leg, and the other leg is 6

Step 1 / 4

Choose an unknown
Let the longer leg be $x$ . Then the hypotenuse is $x + 2$ .

$legs 6, x; hypotenuse x + 2$
Write the Pythagorean equation
Solve it
Check

The median to the hypotenuse

Right triangles have another property that does not need the Pythagorean theorem: the median to the hypotenuse is half the hypotenuse. In other words, the right-angle vertex lies on the circle with the hypotenuse as diameter.

C

can only slide along the circle with diameter

A B

. Does

∠ A C B

ever change? What about

C M

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $a^{2} + b^{2} = c^{2}$

A right triangle. The top bar is $a^{2}$ followed by $b^{2}$ , the bottom bar is $c^{2}$ : they are the same length. Drag $A$ or $B$ (keeping the right angle) and they stay the same length.

Practice

Exercise 1Basic

A right triangle has legs $9$ and $12$ . Find the hypotenuse.

Exercise 2Basic

A triangle has sides $7$ , $24$ , $25$ . Find its area.

Exercise 3Advanced

A right triangle has legs $15$ and $20$ . Find the altitude to the hypotenuse.

Exercise 4Challenge

A rectangular sheet of paper $A B C D$ has $A B = 8$ and $B C = 4$ . The sheet is folded so that vertex $C$ lands on vertex $A$ , and the fold line meets $A B$ at $E$ . Find the length of $A E$ .

Try this

Why: the same uncovered area, counted twice

A jigsaw proof of the Pythagorean theorem

The Pythagorean theorem is an equation

Example: the hypotenuse is 2 longer than one leg, and the other leg is 6

The median to the hypotenuse

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Assuming both given sides are legs

Not taking the longest side as c in the converse