Math 8 · 8.G.B.6

Converse of the Pythagorean theorem

A triangle’s three sides satisfy the Pythagorean equation. Does that alone force a square corner?

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The Pythagorean theoremMath 8

Where you are 18 in Triangles

Geometry11

Triangles18

▴ 10 above Isosceles and equilateral triangles The Pythagorean theorem

Converse of the Pythagorean theorem

Pythagorean triples Using the Pythagorean theorem ▾ Show all 18 (3 more)

Why this exists

The Pythagorean theorem starts from a right angle and ends with an equation: if the corner is square, then the squares on the two short sides add up to the square on the long one. But people often meet the problem the other way round. A corner has been marked out on a floor, or drawn in a diagram, and nobody knows whether it is square. What they can do is measure lengths, and lengths are far easier to measure than angles.

So the question is whether the equation can be used backwards. Measure two sides and the gap between their ends. If the squares add up, is the corner square?

That is not a formality. A true statement turned round is often false: every square is a rectangle, but not every rectangle is a square. Turning a theorem round gives its converse, and each converse has to be proved on its own (a theorem and its converse). Here the turned-round statement turns out to be true, and it is the reason a string knotted into 12 equal spaces, pulled tight into a triangle with sides of 3, 4 and 5 spaces, makes a square corner every time.

Definition

The converse of a statement “if P, then Q” is the statement “if Q, then P” (conditionals and their converses).

The Pythagorean theorem says: if a triangle has a right angle, then the squares on the two sides that meet at it add up to the square on the side opposite it. Its converse starts from the lengths and ends at the angle.

Converse of the Pythagorean theorem. Take any triangle, and call its side lengths a, b and c. If

a^2 + b^2 = c^2

then the angle between the sides a and b, which is the angle opposite the side c, is a right angle.

Notice that the words “leg” and “hypotenuse” do not appear in the hypothesis. Those names belong to a triangle already known to be right. Here we are given only three lengths, and the conclusion is that the triangle is right, with the side c as its hypotenuse.

Intuition

Take two sticks, one of length 3 and one of length 4, and join them at one end with a hinge. Close the hinge, so the sticks lie on top of each other, then open it slowly until they lie in one straight line. The gap between the two free ends starts at 1 and grows, never shrinking, until it reaches 7 (the hinge theorem says that a wider opening always gives a longer gap).

Figure 1The same two sides, 3 and 4, opened to three angles. Only at 90^\circ does the third side square to exactly 25.

Figure 1 shows three openings. At 90^\circ the sticks are the two legs of a right triangle, and the Pythagorean theorem says the gap is \displaystyle \sqrt{3^2 + 4^2} = 5. The gap grows all the way, so it equals 5 at that opening and at no other. Any triangle with sides 3, 4 and 5 is therefore the 90^\circ triangle, because it is the only opening of the hinge that makes a gap of 5.

Nothing here depends on 3 and 4. For any two sticks, the right-angled opening gives a gap whose square is the sum of the squares of the sticks, and no other opening does. The proof below makes this exact.

Results

Converse of the Pythagorean theorem. If the side lengths a, b and c of a triangle satisfy

a^2 + b^2 = c^2

then the triangle has a right angle, opposite the side c.

Put together with the theorem itself, this gives a two-way statement. A triangle with sides a, b and c has a right angle opposite c if and only if a^2 + b^2 = c^2 (a biconditional).

The same comparison also tells you which way a triangle fails to be right. Let c be the longest side.

If c^2 < a^2 + b^2, the angle opposite c is acute, and so is every other angle: the triangle is acute.
If c^2 = a^2 + b^2, the angle opposite c is a right angle: the triangle is right.
If c^2 > a^2 + b^2, the angle opposite c is obtuse: the triangle is obtuse.

To test three lengths that do form a triangle, square the longest, add the squares of the other two, and compare.

Why it is true

The converse. Suppose a triangle ABC has sides BC = a, CA = b and AB = c, and suppose a^2 + b^2 = c^2. We do not know yet that any of its angles is a right angle. So build a second triangle where we do know, and compare.

Figure 2The triangle we are told about, and a right triangle built from the same two sides a and b.

Draw a right angle, and mark off a length a along one arm and a length b along the other, as in Figure 2. Join the two ends. The result is a right triangle DEF with legs FD = a and FE = b. Call its hypotenuse x.

The Pythagorean theorem applies to this triangle, because it is right by construction:

x^2 = a^2 + b^2

Our hypothesis says a^2 + b^2 = c^2, so x^2 = c^2. Both x and c are lengths, so neither is negative, and therefore x = c.

Now the two triangles have all three pairs of sides equal: BC = DF = a, CA = FE = b and AB = ED = c. Three equal pairs of sides is enough to make triangles congruent, by the SSS criterion (SSS, SAS, ASA, AAS):

\triangle ABC \cong \triangle EDF

with A matching E, B matching D and C matching F. Matching angles of congruent triangles are equal, so the angle at C equals the angle at F, which is 90^\circ. The triangle we were told about has a right angle at C.

The proof never assumed that ABC had a right angle. It used the theorem on a triangle that was built to have one, and then showed the two triangles are the same shape and size. This is Proposition 48 of Book I of Euclid’s Elements, the last proposition of the book, and Euclid’s argument is this one.

The sharper test. The same comparison can be made without building anything, by dropping a perpendicular. Take a triangle with sides a, b and c, where c is opposite the angle at C. Drop the perpendicular from B to the line through C and A. Call its foot D, its length h, and write x for the distance from C to D.

Figure 3A perpendicular from B to the line through C and A. Which side of C the foot D lands on depends on the angle at C.

There are three cases, according to the angle at C.

The angle at C is right. Then D is C itself, and the theorem gives c^2 = a^2 + b^2.

The angle at C is acute. Then D lies on the same side of C as A (Figure 3, left), and x > 0. The right triangle BDC gives a^2 = x^2 + h^2. The distance from D to A is b - x, or x - b if D falls beyond A, and squaring makes the two the same, so the right triangle BDA gives

\begin{aligned} c^2 &= (b - x)^2 + h^2 \\ &= b^2 - 2bx + x^2 + h^2 \\ &= a^2 + b^2 - 2bx \end{aligned}

and since b and x are positive, c^2 < a^2 + b^2. (If D happens to be A, the distance from D to A is 0 = b - x, and the same line holds.)

The angle at C is obtuse. Then D lies on the far side of C (Figure 3, right), the distance from D to A is b + x, and in the same way

c^2 = (b + x)^2 + h^2 = a^2 + b^2 + 2bx

so c^2 > a^2 + b^2.

The angle at C is right, acute or obtuse, and the three cases give three different outcomes: the squares add up to c^2, to more than c^2, or to less. So the outcome tells you the case. Equality can only come from a right angle, which proves the converse a second time. The two inequalities work the same way for the angle opposite c. When c is the longest side, that angle is the largest in the triangle (a larger angle lies opposite a longer side), so if it is acute, all three are.

Worked examples

Example 1

A right triangle in disguise

A triangle has sides of 8, 15 and 17. Is it a right triangle?

Solution

To use the converse, pick the longest side to play the part of c, and compare its square with the sum of the other two squares. The longest side is 17:

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

The two agree, so 8^2 + 15^2 = 17^2, and by the converse the triangle has a right angle. It is opposite the longest side, the one of length 17.

Check. Drawn to scale, the angle between the sides of 8 and 15 would look like the corner of a sheet of paper. A triangle with these sides cannot be drawn any other way.

Example 2

Sides given in any order

A triangle has sides of 26, 10 and 24. Is it a right triangle?

Solution

The lengths were not given smallest first, so begin by finding the longest: it is 26. That side must be c, the one left on its own. Compare:

\begin{aligned} 10^2 + 24^2 &= 100 + 576 = 676 \\ 26^2 &= 676 \end{aligned}

Equal, so the triangle is right, with the right angle opposite the side of 26.

Trying a shorter side as c would have been pointless, and would even have looked like failure: 10^2 + 26^2 = 776, which is not 24^2 = 576. If an equation of the form a^2 + b^2 = c^2 holds at all, c^2 is bigger than each of the other squares, so c is bigger than both other sides. That is why the longest side is the only one worth trying.

Example 3

When the squares do not add up

Decide whether each triangle is right, acute or obtuse: one with sides 7, 10 and 12, and one with sides 4, 5 and 7.

Solution

For the first triangle the longest side is 12:

\begin{aligned} 7^2 + 10^2 &= 49 + 100 = 149 \\ 12^2 &= 144 \end{aligned}

The squares on the two shorter sides add to more than the square on the longest, 149 > 144, so c^2 < a^2 + b^2. The angle opposite the side of 12 is acute, and since it is the largest angle, all three are. The triangle is acute.

For the second triangle the longest side is 7:

\begin{aligned} 4^2 + 5^2 &= 16 + 25 = 41 \\ 7^2 &= 49 \end{aligned}

Now the two shorter squares add to less than the longest square, 41 < 49, so c^2 > a^2 + b^2. The angle opposite the side of 7 is obtuse. The triangle is obtuse.

Both triangles do exist: in each, the two shorter sides together are longer than the longest, as any triangle needs (the triangle inequality). The first is only just short of right, since 149 is close to 144.

Example 4

Sides that are square roots

A triangle has sides \displaystyle \sqrt{5}, \displaystyle \sqrt{11} and 4. Is it a right triangle?

Solution

The longest side is 4, because \displaystyle 4 = \sqrt{16} and 16 is more than 11 and more than 5. The square of a square root is the number inside it, so each square is easy:

\begin{aligned} \left(\sqrt{5}\right)^2 + \left(\sqrt{11}\right)^2 &= 5 + 11 = 16 \\ 4^2 &= 16 \end{aligned}

They agree, so the triangle is right, with the right angle opposite the side of 4. No decimal was needed, and none would have been exact.

Example 5

Which side is the longest?

A triangle has sides of 6 and 8 and a third side x. For which values of x is it a right triangle?

Solution

The converse needs the longest side in the place of c, and here the longest side might be x or might be 8, so both cases have to be tried.

If x is the longest side, then

\begin{aligned} 6^2 + 8^2 &= x^2 \\ x^2 &= 100 \\ x &= 10 \end{aligned}

If 8 is the longest side, then

\begin{aligned} 6^2 + x^2 &= 8^2 \\ x^2 &= 64 - 36 = 28 \\ x &= \sqrt{28} = 2\sqrt{7} \end{aligned}

which is about 5.29, shorter than 8 as that case requires.

If 6 were the longest side, we would need x^2 + 8^2 = 6^2, so x^2 = -28, which no length can satisfy.

So the triangle is right when x = 10 and when \displaystyle x = 2\sqrt{7}. Both give a genuine triangle: in each case the two shorter sides add to more than the longest. In the first the right angle is between the sides 6 and 8; in the second it is between 6 and \displaystyle 2\sqrt{7}.

Example 6

Checking a corner with a tape measure

A builder marks a corner on a concrete slab. She measures 1.2 m along one edge and 1.6 m along the other, and then measures the straight gap between the two marks. It is exactly 2.0 m. Is the corner square? The next day a second corner gives a gap of 2.05 m for the same two distances. What does that say?

Solution

The two marked distances and the gap are the three sides of a triangle, and the corner is the angle between the two marked distances. For the first corner, the gap is the longest side:

\begin{aligned} 1.2^2 + 1.6^2 &= 1.44 + 2.56 = 4.00 \\ 2.0^2 &= 4.00 \end{aligned}

The squares add up exactly, so by the converse the corner is square.

For the second, 1.2^2 + 1.6^2 = 4.00 as before, but the gap squared is 2.05^2 = 4.2025, which is more than 4.00. So c^2 > a^2 + b^2 and the angle is obtuse: the two edges open out wider than a square corner. A gap that is too long means the corner is too wide, and a gap that is too short means it is too narrow.

The lengths 1.2, 1.6 and 2.0 are 0.4 times 3, 4 and 5, which is why the first corner came out square.

Example 7

A right angle inside a figure

In Figure 4, D lies on the side \overline{BC} of triangle ABC. The lengths are AB = 13, AC = 15, BD = 5, DC = 9 and AD = 12. Show that \overline{AD} \perp \overline{BC}, and find the area of triangle ABC.

Figure 4Is \overline{AD} perpendicular to \overline{BC}? The lengths will tell.

Solution

The figure shows no right angle at D, and none is given. Look at the small triangle ABD, whose sides are 5, 12 and 13. The longest is 13:

\begin{aligned} 5^2 + 12^2 &= 25 + 144 = 169 \\ 13^2 &= 169 \end{aligned}

By the converse, triangle ABD has a right angle opposite its longest side \overline{AB}, which puts it at D. So \angle ADB = 90^\circ, and \overline{AD} \perp \overline{BC}.

That makes \overline{AD} the height of triangle ABC from A. The base is BC = BD + DC = 5 + 9 = 14, so the area is

\tfrac{1}{2} \times 14 \times 12 = 84

Check. The other small triangle should be right at D as well. Its sides are 12, 9 and 15, and 12^2 + 9^2 = 144 + 81 = 225 = 15^2. It is. That had to happen, because the two angles at D lie along one straight line, so if one is 90^\circ the other is too.

Boundaries

The equation has to hold exactly. The converse concludes a right angle only from exact equality. Squares that almost add up give an angle close to 90^\circ, but not equal to it. Lengths that came from a tape measure or a ruler are only as exact as the measuring, and the test can then only say “close to square”. The sharper test is useful here: it says which way a corner is out, even if not by how many degrees.

c must be the longest side. The equation a^2 + b^2 = c^2 can only hold when c is the longest, so that is the only side worth trying. If the longest side fails, no other choice can succeed, and the triangle is not right. When it is not known which side is longest, as in Example 5, each candidate has to be tried.

Positive lengths that satisfy the equation always make a triangle. If a, b and c are positive and a^2 + b^2 = c^2, then (a + b)^2 = c^2 + 2ab > c^2, so a + b > c, and the two shorter sides are together longer than the longest. You never need to check that the triangle exists separately. The same is true whenever c is the longest side and c^2 < a^2 + b^2, since then (a + b)^2 > a^2 + b^2 > c^2. But when c^2 > a^2 + b^2, the lengths may fail to make a triangle at all: 1, 1 and 5 give 5^2 > 1^2 + 1^2, yet no triangle has those sides. Before calling a triple obtuse, check that the two shorter sides add to more than the longest.

It is about the angle opposite c, and about flat figures. The converse says nothing about the other two angles except what follows from the angle sum: they are acute, since a triangle cannot have two angles of 90^\circ or more. And it is a fact about triangles drawn on a flat surface. On a curved surface such as a globe, the equation is not the test for a right angle.

Squares, and only squares. The test compares a^2 + b^2 with c^2. Adding the lengths themselves, a + b against c, is the triangle inequality and says nothing about right angles.

Where this goes next

The most useful use of the converse is checking for square corners: squaring up a foundation, a frame, a garden bed, or a line on a sports field, with nothing but a tape. Whole numbers that make the equation work, such as 3, 4, 5 and 5, 12, 13, are what make that easy, and they have their own lesson: Pythagorean triples. Turning a real situation into a right triangle and then using the theorem is the skill using the Pythagorean theorem.

On a coordinate grid the converse becomes a way to prove that a plotted triangle has a right angle by comparing the squares of its three side lengths, found with the distance formula; that is showing a plotted triangle is right. A second way to do the same job compares slopes, and the two methods can be used to check each other.

The pair of inequalities, less than and greater than, is a first look at something exact: the difference between c^2 and a^2 + b^2 is fixed by the two sides a and b and the angle between them, its sign by the angle alone, and the law of cosines measures it. The Pythagorean theorem is the case where that difference vanishes, which is the subject of why the law of cosines contains the Pythagorean theorem.

Exercises

Right, acute or obtuse?

Decide whether a triangle with the given side lengths is right, acute or obtuse. Show the comparison you made.

7, 24, 25
12, 16, 20
8, 9, 12
15, 20, 26
0.6, 0.8, 1
\displaystyle \sqrt{7}, 3, 4
\displaystyle 2\sqrt{2}, \displaystyle 2\sqrt{3}, 5
28, 45, 53

Finding a side that makes it right

A triangle has two sides of lengths 5 and 12. Find every possible length of the third side that makes it a right triangle.
A triangle has sides 10, x and 26, and 26 is the longest. Find x so that it is a right triangle.
A triangle has sides x, x + 1 and x + 2, where x is positive. Find x so that it is a right triangle.
Show that for every positive number t, a triangle with sides 3t, 4t and 5t is a right triangle.
Let n be a whole number greater than 1. Show that a triangle with sides n^2 - 1, 2n and n^2 + 1 is a right triangle. Then write out the three sides when n = 5.

Using the test

A builder marks 0.9 m along one edge of a slab and 1.2 m along the other, and measures the gap between the marks as 1.5 m. Is the corner square?
A brace runs from a point 50 cm from the corner along one side of a frame to a point 120 cm from the corner along the other side. It measures 131 cm. Is the corner square? If not, is it wider or narrower than a square corner?
In triangle ABC, the point D lies on \overline{BC}, with AB = 20, BD = 16, AD = 12, DC = 5 and AC = 13. Show that \overline{AD} \perp \overline{BC}, and find the area of triangle ABC.
In the triangle of Exercise 16, what kind of angle is \angle BAC: right, acute or obtuse?
A triangle has vertices A(1, 1), B(4, 5) and C(8, 2). Find the squares of its three side lengths, and show that the triangle has a right angle. At which vertex is it?

Reasoning

Let a, b and c be positive numbers with a^2 + b^2 = c^2. Explain why c is greater than a and greater than b.
To test a triangle with sides 6, 8 and 9, it is enough to compare 6^2 + 8^2 with 9^2. Explain why there is no need to compare 6^2 + 9^2 with 8^2 or 8^2 + 9^2 with 6^2.
One of these statements is the Pythagorean theorem and the other is its converse. Say which is which. (i) If a triangle has a right angle, then the square of the longest side equals the sum of the squares of the other two. (ii) If the square of one side of a triangle equals the sum of the squares of the other two, then the triangle has a right angle.
Let a, b and c be positive numbers with a^2 + b^2 = c^2. Show that a + b > c, so that lengths a, b and c always form a triangle.
In the proof that the converse is true, a right triangle is built with legs a and b. Explain why its hypotenuse is exactly c, and say where the hypothesis a^2 + b^2 = c^2 is used.
A triangle has longest side c and c^2 > a^2 + b^2. Using the three cases in the proof of the sharper test, explain why the angle opposite c must be obtuse.

Answers

Right. 7^2 + 24^2 = 49 + 576 = 625 and 25^2 = 625.
Right. 12^2 + 16^2 = 144 + 256 = 400 and 20^2 = 400.
Acute. 8^2 + 9^2 = 64 + 81 = 145, and 12^2 = 144, which is less, so c^2 < a^2 + b^2.
Obtuse. 15^2 + 20^2 = 225 + 400 = 625, and 26^2 = 676, which is more, so c^2 > a^2 + b^2. The triangle exists, since 15 + 20 > 26.
Right. 0.6^2 + 0.8^2 = 0.36 + 0.64 = 1 and 1^2 = 1.
Right. The longest side is 4, and \displaystyle \left(\sqrt{7}\right)^2 + 3^2 = 7 + 9 = 16 = 4^2.
Obtuse. The longest side is 5, since \displaystyle 2\sqrt{3} is about 3.46. Then \displaystyle \left(2\sqrt{2}\right)^2 + \left(2\sqrt{3}\right)^2 = 8 + 12 = 20, and 5^2 = 25, which is more, so c^2 > a^2 + b^2. The triangle exists, since \displaystyle 2\sqrt{2} + 2\sqrt{3} is about 6.29, more than 5.
Right. 28^2 + 45^2 = 784 + 2025 = 2809 and 53^2 = 2809.
13 or \displaystyle \sqrt{119}. If the third side is longest, 5^2 + 12^2 = 169, so it is 13. If 12 is longest, 5^2 + x^2 = 12^2 gives x^2 = 119, so \displaystyle x = \sqrt{119}, about 10.91, which is shorter than 12 as required. Both make a triangle, since the two shorter sides add to more than the longest.
x = 24. The longest side is 26, so 10^2 + x^2 = 26^2, which gives x^2 = 676 - 100 = 576.
x = 3. The longest side is x + 2, so x^2 + (x + 1)^2 = (x + 2)^2, which simplifies to x^2 - 2x - 3 = 0, and (x - 3)(x + 1) = 0 gives x = 3 or x = -1. A length cannot be negative, so x = 3, the triangle with sides 3, 4, 5.
(3t)^2 + (4t)^2 = 9t^2 + 16t^2 = 25t^2 = (5t)^2. The equation holds, and 5t is the longest side, so by the converse the triangle is right.
(n^2 - 1)^2 + (2n)^2 = n^4 - 2n^2 + 1 + 4n^2, which is n^4 + 2n^2 + 1 = (n^2 + 1)^2. Since n > 1, all three sides are positive and n^2 + 1 is the longest, so the triangle is right. For n = 5 the sides are 24, 10 and 26.
Yes. 0.9^2 + 1.2^2 = 0.81 + 1.44 = 2.25 and 1.5^2 = 2.25, so the corner is square.
No, it is wider. 50^2 + 120^2 = 2500 + 14400 = 16900, and 131^2 = 17161, which is more, so the angle is obtuse. A brace of exactly 130 cm would have made a square corner.
Triangle ABD has sides 12, 16 and 20, and 12^2 + 16^2 = 144 + 256 = 400 = 20^2, so it is right at D and \overline{AD} \perp \overline{BC}. The base is BC = 16 + 5 = 21 and the height is AD = 12, so the area is \tfrac{1}{2} \times 21 \times 12 = 126.
Acute. The sides of triangle ABC are AB = 20, AC = 13 and BC = 21, and the angle \angle BAC is opposite the longest side, 21. Then 13^2 + 20^2 = 169 + 400 = 569, and 21^2 = 441, which is less, so c^2 < a^2 + b^2 and the angle is acute.
The squares are AB^2 = 3^2 + 4^2 = 25, BC^2 = 4^2 + (-3)^2 = 25 and AC^2 = 7^2 + 1^2 = 50. Then AB^2 + BC^2 = 25 + 25 = 50 = AC^2, so the triangle is right, at B, which is opposite the longest side AC.
c^2 = a^2 + b^2, and b^2 is positive, so c^2 > a^2. Two positive numbers are in the same order as their squares, so c > a. The same argument with a^2 > 0 gives c > b.
If 6^2 + 9^2 = 8^2 were true, then by the reasoning in Exercise 19 the side 8 would be longer than the side 9, which it is not. In the same way 8^2 + 9^2 = 6^2 would make 6 longer than 9. Neither can hold, so the only comparison that can succeed is the one with 9 alone.
(i) is the theorem: it starts from the right angle and ends with the equation. (ii) is the converse: it starts from the equation and ends with the right angle.
(a + b)^2 = a^2 + 2ab + b^2 = c^2 + 2ab, and 2ab > 0, so (a + b)^2 > c^2. Both a + b and c are positive, so a + b > c. The other two inequalities, a + c > b and b + c > a, hold because c is longer than both a and b (Exercise 19).
The Pythagorean theorem applies to the built triangle because it is right by construction, so its hypotenuse x has x^2 = a^2 + b^2. The hypothesis says a^2 + b^2 = c^2, so x^2 = c^2, and since both are positive lengths, x = c. The hypothesis is used in exactly that step: the equality of the two squares.
The angle opposite c is right, acute or obtuse. If it were right, the proof gives c^2 = a^2 + b^2. If it were acute, the proof gives c^2 = a^2 + b^2 - 2bx < a^2 + b^2. Both disagree with c^2 > a^2 + b^2, so the angle can only be obtuse, which gives c^2 = a^2 + b^2 + 2bx.