Pythagorean theorem — proof (by similar triangles, rearrangement), converse: Question bank

This bank hunts the ideas behind the Pythagorean theorem, not the arithmetic. Every item is a one-line reveal: read the prompt, answer out loud, then check. The answers give reasons, because in maths "yes" without "why" is worth nothing.

True or false — justify

The theorem $c^{2} = a^{2} + b^{2}$ only holds when the triangle has a right angle. Every claim below tests whether you know where that condition hides.

True or false:

a^{2} + b^{2} = c^{2}

holds for every triangle if you just label the longest side

c

False. It holds only for right triangles; the right angle is the earned condition, not the labelling. A 3–4–6 triangle has a longest side yet

9 + 16 \neq = 36

True or false: In

c^{2} = a^{2} + b^{2}

, the letter

c

must be the longest side.

True.

c

is the hypotenuse, which sits opposite the largest (right) angle, and the largest angle always faces the longest side — so

c

is necessarily longest.

True or false: If a triangle is right-angled, the right angle is always opposite the hypotenuse.

True. The hypotenuse is defined as the side facing the

90°

angle; that is what makes it the longest side.

True or false: The converse lets you detect a right angle using only a ruler, no protractor.

True. Measure all three sides; if

a^{2} + b^{2} = c^{2}

then the angle opposite

c

must be

90°

— that is exactly what the converse guarantees.

True or false: A triangle with sides

1, 1, 2

is right-angled.

True.

1^{2} + 1^{2} = 1 + 1 = 2 = (2)^{2}

, so by the converse the angle opposite

2

is a right angle (it is the isosceles right triangle).

True or false: If

a^{2} + b^{2} > c^{2}

(with

c

longest) the triangle can still be right-angled.

False. Equality is required for a right angle;

a^{2} + b^{2} > c^{2}

means the angle opposite

c

is acute (less than

90°

), a fact you can feel via the Law of Cosines.

True or false: If

a^{2} + b^{2} < c^{2}

the triangle is impossible.

False. It is perfectly valid but obtuse — the angle opposite

c

exceeds

90°

. Impossibility only arises if a side is longer than the other two combined.

True or false: Scaling a 3–4–5 triangle to 30–40–50 keeps it right-angled.

True. Multiplying every side by the same positive number preserves all angles (similar triangles), so the right angle survives;

3 0^{2} + 4 0^{2} = 5 0^{2}

still checks out.

True or false: The similar-triangles proof needs the perpendicular dropped from the right-angle vertex to the hypotenuse.

True. That single altitude splits the triangle into two smaller ones similar to the original, which is the whole engine of the proof; drop it from any other vertex and the similarity vanishes.

Spot the error

Each line contains a flawed argument. Name the flaw in one breath.

"Sides 5, 12, 13: check

1 2^{2} + 1 3^{2} = 313 \neq = 25

, so not right." — what went wrong?

The wrong side was called hypotenuse. The hypotenuse is the longest side,

13

, so test

5^{2} + 1 2^{2} = 169 = 1 3^{2}

✓ — it is right.

"Legs 3 and 4, so the hypotenuse is

3^{2} + 4^{2} = 25

." — what went wrong?

The square root was forgotten.

c^{2} = 25

gives

c = 25 = 5

, not

25

;

c^{2}

is an area-like quantity,

c

is the length.

3^{2} + 4^{2} \neq = 6^{2}

, so I must have made an arithmetic slip." — what went wrong?

No slip — the triangle simply isn't right-angled. The theorem is a property of right triangles, not a rule every triple must obey.

"This scalene triangle has sides 7, 8, 9, so

7^{2} + 8^{2}

should equal

9^{2}

." — what went wrong?

There is no reason to expect equality;

49 + 64 = 113 \neq = 81

, so it's obtuse-free but not right. Only right triangles give the sum-of-squares identity.

"I proved similarity, so I can write

\frac{b}{c} = \frac{c}{x}

." — what went wrong?

Corresponding sides are mismatched. In

△ A B C \sim △ A C D

the ratio is

\frac{b}{c} = \frac{x}{b}

(hypotenuse-to-hypotenuse, leg-to-leg), giving

b^{2} = c x

, not that flipped fraction.

"In the rearrangement proof the outer square has side

c

." — what went wrong?

The outer square has side

a + b

(leg plus leg along each edge); the inner tilted square has side

c

. Confusing the two collapses the whole area argument.

"The converse is obvious — it's just the theorem read backwards, no proof needed." — what went wrong?

A statement and its converse are logically independent; the converse needs its own argument (construct a reference right triangle, then use SSS congruence).

Why questions

Answer the reason, not just the fact.

Why does only the right angle give the clean

a^{2} + b^{2} = c^{2}

(and no other angle)?

Because the general relation is

c^{2} = a^{2} + b^{2} - 2 ab cos C

; the correction term

2 ab cos C

vanishes exactly when

C = 90°

since

cos 90° = 0

. See Law of Cosines.

Why is the theorem often described as being about areas, not lengths?

Because

a^{2}

b^{2}

c^{2}

are literally the areas of squares built on the sides, and the theorem says the two smaller square-areas exactly fill the big one — a conservation of area.

Why must the perpendicular in Proof 1 create three similar triangles?

Each small triangle shares one acute angle with the original and has its own right angle, so by AA (two matching angles) all three are similar — that shared-angle structure is what proportions the sides.

Why does adding equations

b^{2} = c x

and

a^{2} = cy

finish the proof so neatly?

Because the two hypotenuse segments satisfy

x + y = c

, so

a^{2} + b^{2} = c (x + y) = c \cdot c = c^{2}

; the geometry that

x

and

y

tile the hypotenuse is doing the real work.

Why does the theorem underlie the Distance Formula in Coordinate Geometry?

The horizontal and vertical gaps between two points are the two legs of a right triangle, and the straight-line distance is its hypotenuse, so distance

= (Δ x)^{2} + (Δ y)^{2}

is Pythagoras in disguise.

Why can a carpenter trust the "3-4-5 method" to square a corner?

By the converse: since

3^{2} + 4^{2} = 5^{2}

, any triangle with those side ratios is forced to have a right angle opposite the 5-side — measurement alone certifies the corner.

Why does the theorem generalise to

x^{2} + y^{2} + z^{2}

in three dimensions?

Applying Pythagoras twice — once in a base plane, once for the vertical rise — stacks two right triangles, giving the 3D Distance Formula; each new perpendicular axis adds one more squared term.

Edge cases

The theorem must survive degenerate and boundary inputs. These probe the limits.

What happens to the triangle as one leg shrinks toward

0

It flattens into a straight segment; in the limit

b \to 0

we get

c^{2} = a^{2}

, i.e.

c = a

, the sensible statement that a "triangle" with no height is just its own base.

Can

a

b

, or

c

equal zero and still satisfy

c^{2} = a^{2} + b^{2}

Only degenerately: if

a = 0

then

c = b

and there is no triangle (two vertices coincide). A genuine triangle needs all three sides strictly positive.

If a triangle has sides

2, 3, 4

, what does

a^{2} + b^{2}

c^{2}

tell you?

4 + 9 = 13 < 16 = 4^{2}

, so the angle opposite the

4

-side is obtuse (exceeds

90°

); Pythagoras' inequality classifies the largest angle.

For which triangle does

a = b

and the theorem hold with the smallest whole-number-free values?

The isosceles right triangle

1, 1, 2

: equal legs force a

45°

–

45°

–

90°

split and an irrational hypotenuse, the classic non-integer Pythagorean case.

Does the theorem work for a triangle drawn on a sphere (a globe's surface)?

No. Pythagoras is a fact of flat (Euclidean) geometry; on a curved surface the angle sum and side relations change, so

a^{2} + b^{2} = c^{2}

fails — the right angle no longer locks the sides that way.

Why is the diagonal of a unit square,

2

, guaranteed to be irrational yet still a legitimate hypotenuse?

Length being irrational is no obstacle — the theorem constrains the square of the length (

c^{2} = 2

), and geometry happily realises

2

as an exact segment even though no fraction names it.

Recall One-line summary of the traps

The theorem lives or dies on the right angle; the hypotenuse is always the longest side; squaring and un-squaring are different steps; and the converse — the part that certifies right angles — needs its own proof.