The Pythagorean Theorem and Converse

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Learning goals

Identify the hypotenuse before finding an exact missing side
Derive the Pythagorean theorem from the hypotenuse altitude relations
Prove and apply the converse using a constructed right triangle
Classify possible triangles by comparing the squares of their sides
Recognize common Pythagorean triples and their positive multiples

Choose the side across from the right angle

The hypotenuse is the side opposite the right angle in a right triangle. The two sides meeting at that angle are its legs. The hypotenuse is the longest side because it faces the largest angle, as proved in Sides, Angles, and Shortest Distances.

For example, a right triangle with legs $8$ and $15$ units has a hypotenuse of $17$ units. The squares fit a useful pattern:

\begin{aligned} 8^2+15^2&=64+225,\\ 8^2+15^2&=289,\\ 17^2&=289. \end{aligned}

The right-angle mark identifies the two legs and the side across from them.

This relationship is about the side opposite the right angle, wherever the triangle points. A letter such as $c$ names a length; it does not make that side a hypotenuse.

Why similarity gives the sum of squares

Take a nondegenerate triangle $ABC$ with a right angle at $C$ . Use the chapter’s notation: $a$ names $BC$ , $b$ names $CA$ , and $c$ names $AB$ . Let $H$ be the perpendicular foot from $C$ to the open segment $AB$ . Write $p$ for $AH$ , $q$ for $HB$ , and $h$ for $CH$ .

The altitude splits the hypotenuse into two interior pieces. These are the same labels used in Lesson 7.1.

In Similarity in Right Triangles, AA gave the correctly ordered similarities

\triangle ABC\sim\triangle ACH\sim\triangle CBH.

The corresponding side ratios gave these prior results:

\begin{aligned} b^2&=cp,\\ a^2&=cq. \end{aligned}

Each leg’s square is a product of the whole hypotenuse and the piece next to that leg. Adding those products uses both pieces of the same hypotenuse.

The Pythagorean theorem from similarity#

First use a right triangle whose legs are $9$ and $12$ units, with hypotenuse $15$ units. Its altitude divides the hypotenuse into pieces $5.4$ and $9.6$ units, adjacent to the $9$ -unit and $12$ -unit legs, respectively. The prior similarity relations give

\begin{aligned} 9^2&=15(5.4),\\ 12^2&=15(9.6). \end{aligned}

Add these relations and factor out the common whole hypotenuse:

\begin{aligned} 9^2+12^2&=15(5.4+9.6),\\ 9^2+12^2&=15(15),\\ 9^2+12^2&=15^2. \end{aligned}

For the general triangle above, the same addition gives

\begin{aligned} a^2+b^2&=cq+cp,\\ a^2+b^2&=c(q+p). \end{aligned}

Since $H$ lies between $A$ and $B$ , segment addition gives

p+q=c.

Substitute that whole length for the two pieces:

a^2+b^2=c^2.

Nothing in the argument depends on the numerical example. The two similar-triangle relations and the interior foot are what make it work for every right triangle.

The Pythagorean theorem states that in a nondegenerate right triangle, the sum of the squares of the legs equals the square of the hypotenuse. All three lengths must use the same unit.

Check your understanding

In the altitude construction, which fact turns $c(p+q)$ into $c^2$ ?

A The altitude bisects every hypotenuse B

p+q=c

because H is between A and B C

p=q

in every right triangle D

h=c

Find a missing side

For a missing hypotenuse, add the leg squares. For a missing leg, subtract the known leg square from the hypotenuse square. Take the positive root because a side length is positive. Keep a radical for an exact answer; use an approximation sign for a rounded decimal.

Worked example 1 Two known legs

Problem. A right triangle has legs $7$ cm and $10$ cm. Find its hypotenuse exactly and to the nearest tenth of a centimeter.

The unknown side lies across from the marked right angle.

Solution. The unknown side $t$ is opposite the marked right angle, so

\begin{aligned} t^2&=7^2+10^2,\\ t^2&=149,\\ t&=\sqrt{149}. \end{aligned}

There is no perfect-square factor greater than $1$ in $149$ . The exact length is $\sqrt{149}$ cm. Rounding only at the end gives

t\approx12.2\text{ cm}.

This is longer than either leg and shorter than their sum, as expected.

Worked example 2 A rotated triangle

Problem. Triangle $QRT$ has a right angle at $R$ . Side $QT$ is $14$ m and side $RQ$ is $9$ m. Find $RT$ exactly.

The orientation changes; the side across from R is still QT.

Solution. The hypotenuse is $QT$ , across from $R$ . The unknown $RT$ is a leg. Therefore,

\begin{aligned} RT^2+9^2&=14^2,\\ RT^2&=196-81,\\ RT^2&=115,\\ RT&=\sqrt{115}. \end{aligned}

The exact answer is $\sqrt{115}$ m. Its square is less than $196$ , so this leg is shorter than the hypotenuse.

Check your understanding

A right triangle has a hypotenuse of $11$ units and one leg of $6$ units. What is the other leg?

\sqrt{157}

units B

5

units C

\sqrt{85}

units D

85

units

Turn the theorem around

The theorem starts with a right angle and concludes an equality of squares. Its converse starts with that equality and concludes a right angle. That direction needs its own proof.

For example, lengths $12$ , $16$ , and $20$ units satisfy

12^2+16^2=20^2.

Construct a right triangle with legs $12$ and $16$ . The theorem forces its hypotenuse to be $20$ . SSS then matches any triangle with these three side lengths to the constructed right triangle.

The first triangle has the given lengths. The dashed triangle is constructed with a right angle; its third side is found before SSS is applied.

The converse of the Pythagorean theorem#

In the numerical construction, the right triangle’s hypotenuse $d$ satisfies

\begin{aligned} d^2&=12^2+16^2,\\ d^2&=400,\\ d&=20. \end{aligned}

Its three sides match the given triangle’s $12$ , $16$ , and $20$ units. SSS congruence transfers the right angle to the angle opposite the $20$ -unit side. The particular numbers served only to match the third sides.

Now let a nondegenerate triangle have positive side lengths $a$ , $b$ , and $c$ , with

a^2+b^2=c^2.

Construct perpendicular rays and copy lengths $a$ and $b$ onto them. Join their endpoints to make a right triangle, and call its hypotenuse $d$ . Apply the already proved theorem to this constructed triangle:

d^2=a^2+b^2.

The assumed equality gives

d^2=c^2.

Since both lengths are positive, they are equal. SSS makes the constructed triangle congruent to the given triangle. The angle opposite $c$ therefore matches the constructed right angle.

Worked example 3 Three supplied sides

Problem. A triangle has side lengths $20$ , $21$ , and $29$ cm. Determine whether it has a right angle and identify the side opposite that angle.

Solution. The lengths form a triangle because the two shorter lengths add to more than the longest:

20+21>29.

Compare the squares with $29$ as the longest length:

\begin{aligned} 20^2+21^2&=841,\\ 29^2&=841. \end{aligned}

By the converse, the angle opposite the $29$ cm side is right. That side is the hypotenuse.

The right-angle mark is now justified by the converse.

Check your understanding

In the converse proof, why can we use SSS rather than assume both triangles are right?

A Equal perimeters give congruence B Every triangle has a right angle C The given triangle is right by definition D The theorem finds the constructed triangle's third side, so all three sides match

Compare with a right angle

Before classifying lengths, check that they are positive and can make a triangle. When the longest is $z$ and the others are $x$ and $y$ , it is enough to check

x+y>z.

For fixed sides $6$ and $8$ , a right triangle has third side $10$ . A third side of $9$ closes the included angle below $90^\circ$ ; a third side of $12$ opens it above $90^\circ$ .

The same two sides can enclose a smaller or larger angle. Every displayed length set passes the triangle inequality.

Here is why this comparison works without a new formula. Construct a right triangle with legs $x$ and $y$ , and call its hypotenuse $r$ . The theorem gives

r^2=x^2+y^2.

The original triangle and this reference triangle have two matching side pairs. By the converse hinge theorem, a third side shorter than $r$ faces a smaller included angle; a third side longer than $r$ faces a larger one. Equal third sides give a right angle by the converse proved above. Positive lengths have the same order as their squares.

Since $z$ is longest, its opposite angle is largest. If that angle is acute, all three angles are acute. Thus, for an existing nondegenerate triangle:

Longest-side comparison	Triangle type
$z^2<x^2+y^2$	Acute
$z^2=x^2+y^2$	Right
$z^2>x^2+y^2$	Obtuse

Worked example 4 Three length sets

Problem. Decide whether each set makes a nondegenerate triangle. Classify each triangle that exists.

Case A: $3$ , $7$ , and $10$ units.

Case B: $7$ , $9$ , and $11$ units.

Case C: $7$ , $9$ , and $13$ units.

Solution. Case A is flat because

3+7=10.

It is not a nondegenerate triangle, so it receives no acute, right, or obtuse classification.

Case B exists because $7+9>11$ . Compare the squares:

\begin{aligned} 7^2+9^2&=130,\\ 11^2&=121. \end{aligned}

The longest-side square is smaller, so Case B is acute.

Case C exists because $7+9>13$ . Its longest-side square is

13^2=169.

This exceeds $130$ , so Case C is obtuse. Failing the right-triangle equality alone would not distinguish these last two cases.

Check your understanding

Which description fits positive lengths $2$ , $6$ , and $9$ units?

A They do not form a triangle B They form an acute triangle C They form a right triangle D They form an obtuse triangle

Recognize triples and their multiples

A Pythagorean triple is a set of three positive integers whose two smaller squares add to the largest square. Common examples are $(3,4,5)$ , $(5,12,13)$ , $(7,24,25)$ , and $(8,15,17)$ . Verify a triple by squaring, rather than relying on a partly remembered list.

Multiplying all three lengths by the same positive factor preserves the right angle by similarity. Algebra gives the same reason: for example, multiplying $(3,4,5)$ by $2$ multiplies both sides of the squared equality by $4$ .

\begin{aligned} 6^2+8^2&=4(3^2+4^2),\\ 6^2+8^2&=4(5^2),\\ 6^2+8^2&=10^2. \end{aligned}

In general, a positive factor $k$ gives

(ka)^2+(kb)^2=(kc)^2.

An integer multiple is another integer triple. A noninteger positive factor still gives a right triangle, though its lengths need not be integers. Adding the same number to all three lengths does not have this property.

Worked example 5 A scaled set

Problem. A triangle has sides $15$ , $36$ , and $39$ cm. Identify a common triple and its scale factor, then decide whether the triangle is right.

Solution. Divide every length by $3$ to obtain $(5,12,13)$ . The scale factor from that triangle to the given triangle is $3$ . Check the original integer triple:

5^2+12^2=13^2.

The given sides are all multiplied by the same positive factor, so their squares are all multiplied by $9$ . The given triangle is right, with the $39$ cm hypotenuse.

Every length is three times the corresponding length in the 5, 12, 13 triangle.

Check your understanding

Which side set is obtained by multiplying every side of a $7,24,25$ right triangle by $2$ ?

9,26,27

14,24,25

14,48,50

49,576,625

The Pythagorean Theorem and Converse

Learning goals

Choose the side across from the right angle

Why similarity gives the sum of squares

Find a missing side

Turn the theorem around

Compare with a right angle

Recognize triples and their multiples

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress