Pythagorean theorem — proof (by similar triangles, rearrangement), converse: Worked examples

This page is the practice half of Pythagorean Theorem — Proof & Converse. The parent note built the theorem $c^{2} = a^{2} + b^{2}$ and its converse. Here we throw every kind of problem at it — clean triangles, missing legs, decimals, near-degenerate shapes, a word problem, and an exam twist — so that no future question surprises you.

Before we start, one reminder about the symbols, because we will not re-earn them every line:

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The examples below are labelled with the cell they cover, and together they fill the whole table.

#	Cell (what makes it different)	What's asked	Example
A	Clean integer triple — find hypotenuse	$c = ?$	E1
B	Find a leg (subtract, not add)	$a = ?$	E2
C	Irrational answer — must leave a surd	$c = \cdot$	E3
D	Converse test — is it a right triangle?	yes/no	E4
E	Converse fails — acute ( $a^{2} + b^{2} > c^{2}$ ) vs obtuse ( $a^{2} + b^{2} < c^{2}$ )	classify	E5a, E5b
F	Degenerate / limiting input (a leg $\to 0$ )	behaviour	E6
G	Real-world word problem (ladder)	length	E7
H	Exam twist — theorem used twice (3-D / stacked)	length	E8

Two things to notice up front. First, side lengths are always positive — there are no negative quadrants here, so when we take a square root we always take the positive root. Second, "degenerate" means a triangle so flat it collapses to a straight line; E6 shows what the formula does at that edge.

Worked examples

E1 — Cell A: clean triple, find the hypotenuse

The figure below draws this exact triangle. Look at the two coloured legs (teal $b = 15$ along the bottom, plum $a = 8$ up the side) meeting at the small ink square — that square marks the right angle. The orange slant is the hypotenuse $c = 17$ , and you can see by eye it is the longest side.

Figure — Pythagorean theorem — proof (by similar triangles, rearrangement), converse

E2 — Cell B: a leg is missing (subtract!)

E3 — Cell C: the answer is irrational

E4 — Cell D: converse test (is it right-angled?)

E5a — Cell E (acute half): converse fails, $a^{2} + b^{2} > c^{2}$

E5b — Cell E (obtuse half): converse fails the other way, $a^{2} + b^{2} < c^{2}$

Worked example A triangle has sides

$4, 5, 8$ . Classify its largest angle.

Forecast: Now the longest side jumped to $8$ . Will $4^{2} + 5^{2}$ overshoot or fall short of $8^{2}$ ?

Step 1. Longest side is $8$ ; compare $a^{2} + b^{2} = 4^{2} + 5^{2} = 16 + 25 = 41$ against $c^{2} = 8^{2} = 64$ . Why this step? Same test, new numbers — we watch which side of the equals sign wins.

Step 2. Here $41 < 64$ , i.e. $a^{2} + b^{2} < c^{2}$ . When the legs' squares fall short, the corner opposite $c$ has to open wider than $9 0^{\circ}$ to stretch across the long side → the triangle is obtuse. Why this step? Same Law of Cosines: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ . If $a^{2} + b^{2} < c^{2}$ then $cos C < 0$ , so $C > 9 0^{\circ}$ .

Step 3. The full three-case rule (now with a worked example for each):

$a^{2} + b^{2} = c^{2}$ → right angle (E4: $9, 40, 41$ ),
$a^{2} + b^{2} > c^{2}$ → acute (E5a: $4, 5, 6$ ),
$a^{2} + b^{2} < c^{2}$ → obtuse (E5b: $4, 5, 8$ ).

Verify: $cos C = \frac{41 - 64}{2 \cdot 4 \cdot 5} = \frac{- 23}{40} = - 0.575 < 0$ , giving $C \approx 125. 1^{\circ} > 9 0^{\circ}$ ✓ — obtuse, as claimed.

E6 — Cell F: degenerate / limiting case

Worked example Keep hypotenuse

$c = 10$ fixed and one leg $b = 10$ (equal to the hypotenuse). What is $a$ , and what does the triangle look like?

Forecast: Can a leg equal the hypotenuse? What length is left for $a$ ?

Step 1. $a^{2} = c^{2} - b^{2} = 1 0^{2} - 1 0^{2} = 0$ , so $a = 0$ . Why this step? We test the boundary of the theorem, where a leg shrinks to nothing.

Step 2. Interpret: $a = 0$ means the triangle has collapsed — the two long sides lie exactly on top of each other. It is degenerate (a flat line, no area). Why this step? Every formula deserves an edge-case check so you know when a "triangle" stops being one.

Step 3. The general rule: for a real right triangle you need each leg strictly shorter than the hypotenuse, $0 < b < c$ . As $b \to c$ , the other leg $a \to 0^{+}$ and the triangle flattens; as $b \to 0$ , it flattens the other way.

Verify: $a^{2} + b^{2} = 0 + 100 = 100 = c^{2}$ ✓ — the algebra is consistent, but geometrically it is the limit, not a triangle.

The figure shows this flattening as a movie in one frame. Read the three dashed hypotenuses (all length $10$ ): the plum one has a tall leg $a = 6$ , the teal one a shorter $a = 3$ , and the orange one a leg $a \to 0$ that has almost lain down flat onto the base. As the vertical leg keeps shrinking, the triangle squashes toward a straight line.

E7 — Cell G: real-world word problem (ladder)

Worked example A

$5 m$ ladder leans against a wall. Its foot is $3 m$ from the wall. How high up the wall does the top reach?

Forecast: The ladder is the slant — leg or hypotenuse? Guess the height.

Step 1. Draw the setup: wall (vertical leg $= h$ ), ground (horizontal leg $= 3$ ), ladder (slant $= 5$ ). The ladder is the longest side, so it is the hypotenuse $c = 5$ ; the wall height $h$ is a leg. Why this step? Correctly naming which measurement is the hypotenuse is the whole battle in word problems.

Step 2. Missing leg → subtract: $h^{2} = c^{2} - 3^{2} = 5^{2} - 3^{2} = 25 - 9 = 16$ . Why this step? $h$ is a leg, and a leg's square is hypotenuse-square minus the other leg-square.

Step 3. $h = 16 = 4 m$ (positive root — a height cannot be negative).

Verify: $3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$ ✓. Units: all metres, answer in metres — consistent. It's the $(3, 4, 5)$ triple wearing a hat.

The figure below matches the words to the picture. The plum bar up the left is the wall (leg $h = 4$ ), the teal bar along the bottom is the ground (leg $3$ ), and the orange slant is the ladder (hypotenuse $5$ ). The little ink square where wall meets ground is the right angle that makes the theorem apply.

E8 — Cell H: exam twist — theorem used twice (space diagonal)

Worked example Find the longest straight rod that fits inside a box measuring

$3 \times 4 \times 12$ (a rectangular room's space diagonal).

Forecast: Is the answer bigger than $12$ ? By a lot or a little?

Step 1. First find the floor diagonal $d$ using the two floor edges $3$ and $4$ : $d^{2} = 3^{2} + 4^{2} = 9 + 16 = 25$ , so $d = 5$ . Why this step? The space diagonal sits in a tilted right triangle whose base is the floor diagonal — so we must build that base first.

Step 2. Now form a new right triangle: one leg is the floor diagonal $d = 5$ (flat on the floor), the other leg is the height $12$ (straight up), and the hypotenuse is the space diagonal $D$ . These two legs meet at $9 0^{\circ}$ because "up" is perpendicular to the floor. Why this step? Applying the theorem a second time in a perpendicular plane is the standard 3D Distance Formula trick.

Step 3. $D^{2} = d^{2} + 1 2^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169$ , so $D = 169 = 13$ .

Verify: Directly, $D^{2} = 3^{2} + 4^{2} + 1 2^{2} = 9 + 16 + 144 = 169 = 1 3^{2}$ ✓ — matching the 3-D distance formula $D = x^{2} + y^{2} + z^{2}$ . And $13 > 12$ , so the diagonal just beats the longest edge. This is also the modulus idea behind Vectors and Magnitude.

The figure draws the box with two diagonals highlighted. Follow the teal line across the floor — that is $d = 5$ from Step 1. Then follow the orange line lifting from the same corner up to the opposite top corner — that is the space diagonal $D = 13$ from Step 3. Notice the orange line is the hypotenuse of a right triangle standing up out of the teal floor line.

Recall Quick self-test (reveal to check)

Given legs $8$ and $15$ , the hypotenuse is ::: $17$ Given hypotenuse $13$ and leg $5$ , the other leg is ::: $12$ Sides $4, 5, 6$ : is the largest angle acute, right, or obtuse? ::: acute (since $41 > 36$ ) Sides $4, 5, 8$ : is the largest angle acute, right, or obtuse? ::: obtuse (since $41 < 64$ ) If a leg equals the hypotenuse, the other leg is ::: $0$ (a degenerate, flat triangle) The space diagonal of a $3 \times 4 \times 12$ box is ::: $13$