What is the difference between the Pythagorean Theorem and its converse?

The theorem starts with a right triangle and concludes that a^2 + b^2 = c^2. The converse starts with three side lengths satisfying a^2 + b^2 = c^2 and concludes that the triangle is right. In short, the theorem gives you a length when you already know there is a right angle; the converse gives you a right angle when you only know lengths.

How do I know which side is the hypotenuse in a word problem?

Find the right angle first, then take the side opposite it. In practice the right angle is usually formed by something vertical meeting something horizontal — a wall and the floor, a pole and the ground, the length and width of a rectangle. The slanted object (ladder, cable, ramp, diagonal, straight-line distance) is the hypotenuse, and it is always the longest side.

Do I have to memorize the Pythagorean triples?

You are not required to, since the formula always works. But recognizing (3,4,5), (5,12,13), (8,15,17), (7,24,25) and their multiples makes homework much faster and helps you catch arithmetic errors. Just confirm the largest number is actually the hypotenuse before applying a triple.

Why does the altitude to the hypotenuse prove the theorem?

That altitude splits the right triangle into two smaller triangles, each similar to the original by AA. The resulting proportions give a^2 = cd and b^2 = ce, where d and e are the two pieces of the hypotenuse. Adding them yields a^2 + b^2 = c(d+e), and since d + e = c, the right side is c^2.

GEOM-7.1

The Pythagorean Theorem & Its Converse

Master the Pythagorean Theorem in Geometry 7.1: find missing sides, prove it with the altitude to the hypotenuse, and classify triangles as right, acute, or obtuse.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on The Pythagorean Theorem & Its Converse, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start The Pythagorean Theorem & Its Converse free

What this lesson covers

Almost every measurement problem in a right triangle starts with the same relationship: the two legs and the hypotenuse are locked together by

a^2 + b^2 = c^2

. You have used this since middle school, but Geometry asks two harder things of you. First, you must be able to justify why it is true using the similar triangles created when you drop the altitude to the hypotenuse. Second, you must run the relationship backwards — given three side lengths and no picture, decide whether the triangle is right, acute, or obtuse.

This lesson builds all three skills. Get comfortable here, because the special right triangles, the trigonometric ratios, and the distance formula in coordinate geometry all rest on this single equation.

What the Theorem Actually Claims

The Pythagorean Theorem applies to right triangles only. If a triangle has a right angle, then

a^2 + b^2 = c^2

where

a

and

b

are the legs (the two sides that form the right angle) and

c

is the hypotenuse (the side opposite the right angle, always the longest side).

The equation is a statement about areas: the square built on the hypotenuse has the same area as the two squares built on the legs combined. That is why the quantities are squared and why you must square each side length separately before adding. Writing

a + b = c

is the single most common error in this unit, and a quick check kills it: a triangle with legs 3 and 4 has hypotenuse 5, not 7.

The hypotenuse must be identified by position, not by which number looks biggest in the problem. It is the side across from the

90^\circ

angle. In a triangle labeled with the right angle at

C

, the hypotenuse is side

c

, and the theorem is

a^2 + b^2 = c^2

no matter how the triangle is rotated on the page.

One consequence worth memorizing: because

c^2 = a^2 + b^2

and

a^2, b^2 > 0

, the hypotenuse is strictly longer than either leg but shorter than their sum. If a problem hands you a "hypotenuse" that is shorter than a leg, something has been mislabeled.

Justifying the Theorem with the Altitude to the Hypotenuse

Your class needs more than the formula — you need a reason. The cleanest proof uses similarity.

Start with right triangle

ABC

with the right angle at

C

, legs

a = BC

and

b = AC

, and hypotenuse

c = AB

. Drop the altitude from

C

perpendicular to

\overline{AB}

, meeting it at

D

. This altitude splits the hypotenuse into two pieces: let

d = BD

(the piece next to leg

a

) and

e = AD

(the piece next to leg

b

), so

d + e = c

.

The altitude creates two smaller triangles,

\triangle CBD

and

\triangle ACD

. Each has a right angle at

D

, and each shares an acute angle with the original triangle (

\angle B

and

\angle A

respectively). By AA similarity,

\triangle CBD \sim \triangle ABC \sim \triangle ACD

All three triangles are similar. From

\triangle CBD \sim \triangle ABC

, matching the leg-to-hypotenuse ratios gives

\frac{d}{a} = \frac{a}{c}

, so

a^2 = cd

. From

\triangle ACD \sim \triangle ABC

\frac{e}{b} = \frac{b}{c}

, so

b^2 = ce

.

Now add the two results:

a^2 + b^2 = cd + ce = c(d + e) = c \cdot c = c^2

That final substitution

d + e = c

is where the proof closes. Each leg is the geometric mean between the whole hypotenuse and the segment of the hypotenuse adjacent to it, and adding the two geometric-mean relationships reassembles the whole hypotenuse. Students who stumble on this proof usually state the similarity but never write the proportions, or they forget to say why the small triangles are similar — always cite AA with the shared acute angle and the right angles.

Solving for a Missing Side

Two cases show up, and they are not equally easy to get right.

Given	Unknown	Setup	Example
Both legs	Hypotenuse	Add, then take a square root	$6^2 + 8^2 = c^2 \Rightarrow c = 10$
One leg and the hypotenuse	Other leg	Subtract, then take a square root	$a^2 + 5^2 = 13^2 \Rightarrow a = 12$

The subtraction case is where errors cluster. If you are missing a leg, the hypotenuse squared goes alone on one side:

a^2 = c^2 - b^2

. Students who reflexively add get an answer larger than the hypotenuse, which is geometrically impossible — that impossibility is your built-in check.

Leave answers in simplest radical form unless the problem asks for a decimal. For legs 5 and 7:

c^2 = 25 + 49 = 74

, so

c = \sqrt{74}

, which does not simplify because 74 has no perfect-square factor. For legs 6 and 6:

c^2 = 72

, and

\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}

.

Recognizing Pythagorean triples saves time:

(3,4,5)

(5,12,13)

(8,15,17)

(7,24,25)

, and every multiple of these, such as

(6,8,10)

(9,12,15)

. If you spot

9

and

15

with

15

as the hypotenuse, the missing leg is

12

instantly. But be careful — a triple only helps if the largest number really is the hypotenuse. In a triangle with legs 5 and 13, the hypotenuse is

\sqrt{194}

, not 12.

The Converse and Classifying Any Triangle

The converse reverses the logic: if the three sides of a triangle satisfy

a^2 + b^2 = c^2

(with

c

the longest side), then the triangle must be a right triangle, with the right angle opposite the longest side. This is what lets you prove a triangle is right using only a ruler — no protractor needed.

Extending the idea classifies every triangle. Always name the longest side

c

first, then compare

c^2

with

a^2 + b^2

Comparison	Triangle type	Reason
$c^2 = a^2 + b^2$	Right	Converse of the Pythagorean Theorem
$c^2 < a^2 + b^2$	Acute	Longest side is shorter than a right triangle would need, so the largest angle closes below $90^\circ$
$c^2 > a^2 + b^2$	Obtuse	Longest side is stretched past the right-triangle length, opening the largest angle past $90^\circ$

Think of the largest angle as a hinge. Hold the two shorter sides fixed and swing them apart: the opposite side grows. At exactly

90^\circ

that side equals

\sqrt{a^2+b^2}

. Longer means a wider angle, shorter means a narrower one.

Two traps to avoid. First, you must square the longest side by itself; comparing

10^2

with

8^2 + 12^2

when 12 is the longest side gives a wrong classification every time. Second, check the triangle inequality before classifying. The lengths 3, 4, and 9 cannot form a triangle at all, since

3 + 4 < 9

, so calling it "obtuse" is meaningless. Only after confirming that the two shorter sides sum to more than the longest side should you run the comparison.

Notice also that classification tells you about the largest angle only. An acute classification guarantees all three angles are under

90^\circ

, because if the biggest one is acute the others must be too.

Where Students Lose Accuracy

Four habits fix most mistakes in this topic.

Square before you add or subtract.

\sqrt{a^2 + b^2}

is not

a + b

. If your work ever shows

\sqrt{9 + 16} = 3 + 4

, stop; the correct value is

\sqrt{25} = 5

, while

3 + 4 = 7

— the two are not equal, and for positive legs they never are.

Identify the hypotenuse from the diagram, not the numbers. In word problems the hypotenuse is usually the slanted thing: a ladder leaning on a wall, a guy wire from the ground to a pole, the diagonal of a rectangle, the straight-line distance between two points. The vertical height and horizontal ground distance are the legs.

Keep exact values until the last step. If a two-part problem asks for a missing leg and then a perimeter, carry

\sqrt{74}

through the work rather than rounding to 8.6 early, then round once at the end. Rounding twice can shift a final answer enough to disagree with the answer key.

Answer the question that was asked. Classification problems want a word — right, acute, or obtuse — plus the supporting comparison. Writing only "

c^2 > a^2 + b^2

" is incomplete work; writing only "obtuse" with no arithmetic shown is also incomplete. Show both.

Finally, sanity-check with size. In any right triangle the hypotenuse is the longest side, so a computed hypotenuse smaller than a given leg, or a computed leg larger than the given hypotenuse, signals that you added when you should have subtracted. This one check catches the majority of arithmetic slips on homework and quizzes.

Key terms

Hypotenuse.: The side of a right triangle opposite the right angle; always the longest side of that triangle.
Leg.: Either of the two sides of a right triangle that form the right angle.
Pythagorean Theorem.: In a right triangle with legs $a$ and $b$ and hypotenuse $c$ , $a^2 + b^2 = c^2$ .
Converse of the Pythagorean Theorem.: If three side lengths satisfy $a^2 + b^2 = c^2$ with $c$ the longest, the triangle is a right triangle with the right angle opposite $c$ .
Pythagorean triple.: Three whole numbers that satisfy $a^2 + b^2 = c^2$ , such as $(3,4,5)$ , $(5,12,13)$ , and $(8,15,17)$ , along with all their multiples.
Altitude to the hypotenuse.: The perpendicular segment from the right-angle vertex to the hypotenuse; it divides the right triangle into two triangles similar to it and to each other.
Geometric mean relationship.: Each leg is the geometric mean of the hypotenuse and the adjacent hypotenuse segment: $a^2 = cd$ and $b^2 = ce$ .
Triangle inequality.: Three lengths form a triangle only if the sum of the two shorter lengths is greater than the longest length.

Worked example

A support cable runs from the top of a vertical antenna to a point on level ground 10 meters from the base of the antenna. The cable is 26 meters long. (a) How tall is the antenna? (b) A second triangular brace has sides of 7 m, 9 m, and 12 m. Classify it as right, acute, or obtuse.

Part (a). The antenna is vertical and the ground is horizontal, so those two sides meet at a right angle and are the legs. The slanted cable is the hypotenuse, so

c = 26

and one leg is

b = 10

. Let the height be

a

.

Because the hypotenuse is known, subtract:

a^2 + 10^2 = 26^2

a^2 + 100 = 676

a^2 = 576

a = 24

The antenna is 24 meters tall. Check the size: 24 is less than the 26-meter hypotenuse, as it must be. You could also have recognized

(5,12,13)

doubled to

(10,24,26)

.

Part (b). First confirm a triangle exists:

7 + 9 = 16 > 12

, so it does. The longest side is 12, so set

c = 12

a = 7

b = 9

.

Compare the two quantities separately:

c^2 = 12^2 = 144

a^2 + b^2 = 49 + 81 = 130

Since

144 > 130

, we have

c^2 > a^2 + b^2

, so the brace is an obtuse triangle, with the obtuse angle opposite the 12-meter side.

Notice how close 144 and 130 are — the triangle is only slightly obtuse. That is exactly why you must compute rather than eyeball a sketch.

Practice questions

A triangle has side lengths 11, 13, and 17. Which classification is correct?

Right, because $11^2 + 13^2 = 17^2$
Acute, because $11^2 + 13^2 > 17^2$
Obtuse, because $11^2 + 13^2 < 17^2$
Not a triangle, because $11 + 13 < 17$

Answer: Acute, because $11^2 + 13^2 > 17^2$

First check existence:

11 + 13 = 24 > 17

, so the triangle is valid, ruling out the last choice. The longest side is 17, so compare

17^2 = 289

with

11^2 + 13^2 = 121 + 169 = 290

. Since

290 > 289

, the sum of the squares of the shorter sides exceeds the square of the longest side, which places the largest angle just under

90^\circ

. The triangle is acute — barely. This is a good reminder that a one-unit difference in the squares changes the answer, so the arithmetic must be exact.

In right triangle

ABC

, the right angle is at

C

and the altitude from

C

meets hypotenuse

\overline{AB}

D

. If

AD = 4

and

DB = 9

, find the length of leg

\overline{AC}

and the length of the hypotenuse, and explain which similarity relationship you used.

Answer: $AB = 13$ and $AC = \sqrt{52} = 2\sqrt{13}$ .

The hypotenuse is the whole segment:

AB = AD + DB = 4 + 9 = 13

. The altitude to the hypotenuse makes

\triangle ACD \sim \triangle ABC

by AA — both contain a right angle (at

D

and at

C

) and both share

\angle A

. Matching the shorter leg to the hypotenuse in each triangle gives

\frac{AD}{AC} = \frac{AC}{AB}

, so

AC^2 = AD \cdot AB = 4 \cdot 13 = 52

and

AC = \sqrt{52} = 2\sqrt{13}

. This is the geometric mean relationship used in the proof of the theorem. As a check, the other leg satisfies

BC^2 = 9 \cdot 13 = 117

, and

52 + 117 = 169 = 13^2

, exactly as

a^2 + b^2 = c^2

requires.

A rectangular gate is 6 feet wide and 8 feet tall. A carpenter wants to add a diagonal brace from one corner to the opposite corner. How long must the brace be, and how could the carpenter use the converse of the Pythagorean Theorem to confirm the gate's corners are square before cutting?

Answer: The brace is 10 feet long; measuring the diagonal and checking that $6^2 + 8^2 = 10^2$ confirms a right angle.

The width and height are perpendicular, so they act as legs:

6^2 + 8^2 = 36 + 64 = 100

, giving a diagonal of

\sqrt{100} = 10

feet. This is the

(3,4,5)

triple doubled. For the second part, the converse runs the logic backwards: if the carpenter measures the two sides as 6 and 8 and the diagonal comes out to exactly 10, then the side lengths satisfy the equation and the corner must be a right angle. If the measured diagonal is longer than 10, the corner has been pushed open past

90^\circ

; if shorter, the corner is pinched below

90^\circ

. This is the standard 3-4-5 method used on real construction sites.

FAQ

What is the difference between the Pythagorean Theorem and its converse?: The theorem starts with a right triangle and concludes that $a^2 + b^2 = c^2$ . The converse starts with three side lengths satisfying $a^2 + b^2 = c^2$ and concludes that the triangle is right. In short, the theorem gives you a length when you already know there is a right angle; the converse gives you a right angle when you only know lengths.
How do I know which side is the hypotenuse in a word problem?: Find the right angle first, then take the side opposite it. In practice the right angle is usually formed by something vertical meeting something horizontal — a wall and the floor, a pole and the ground, the length and width of a rectangle. The slanted object (ladder, cable, ramp, diagonal, straight-line distance) is the hypotenuse, and it is always the longest side.
Do I have to memorize the Pythagorean triples?: You are not required to, since the formula always works. But recognizing $(3,4,5)$ , $(5,12,13)$ , $(8,15,17)$ , $(7,24,25)$ and their multiples makes homework much faster and helps you catch arithmetic errors. Just confirm the largest number is actually the hypotenuse before applying a triple.
Why does the altitude to the hypotenuse prove the theorem?: That altitude splits the right triangle into two smaller triangles, each similar to the original by AA. The resulting proportions give $a^2 = cd$ and $b^2 = ce$ , where $d$ and $e$ are the two pieces of the hypotenuse. Adding them yields $a^2 + b^2 = c(d+e)$ , and since $d + e = c$ , the right side is $c^2$ .

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