Math 8 · 8.G.B.6

Proving the Pythagorean theorem by rearrangement

How can four copies of one right triangle, moved around inside a square, prove something true of every right triangle there is?

You need

The Pythagorean theoremMath 8

Where you are 4 in Proofs about similar figures

Geometry11

Geometric proof18

Proofs about similar figures4

Proving the Pythagorean theorem by rearrangement

Proving two triangles similar Proving the Pythagorean theorem with similar triangles Proving the triangle proportionality theorem

When you reach for it

The Pythagorean theorem says that in a right triangle with legs a and b and hypotenuse c,

a^2 + b^2 = c^2

Checking it on a few triangles, such as 3, 4, 5, shows that it is plausible, but a pattern is not a proof: there are infinitely many right triangles, long and flat, tall and thin, and no one can check them all. A proof has to work for a triangle whose sides are only letters.

This skill does it with area: work out one region’s area two ways, one giving c^2 and one giving a^2 + b^2, and set them equal. Four copies of the triangle and a square of side a + b are all it takes.

Reach for it when you are asked why the theorem is true rather than how to use it, when you want a proof built from nothing more than areas of squares and triangles and the angles of a triangle, and, more widely, whenever an equation about squares can be drawn as a statement about area.

The procedure

What the proof may use. Five facts, all settled before this lesson.

A square of side s has area s^2 (area of a square).
A right triangle with legs a and b has area \tfrac{1}{2}ab, half of an a by b rectangle (area of a triangle).
If a region is cut into pieces that do not overlap, its area is the sum of the pieces’ areas, and moving the pieces does not change them (area stays the same when the pieces move).
A copy of a figure, moved by sliding, turning or flipping, has the same side lengths, angles and area as the original (congruence by rigid motion).
The angles of a triangle add to 180^\circ (the triangle angle sum). In a right triangle the right angle takes 90^\circ of that, so the two acute angles add to 90^\circ. This one fact does most of the work.

Call the acute angles by what they face. In the triangle, one acute angle sits opposite the leg a and the other opposite the leg b. Wherever a copy is placed, its angle opposite a and its angle opposite b go with it, and together they make 90^\circ.

The first way to fill the square. Draw a square ABCD with side a + b. Its corners are right angles, so a copy of the triangle fits into each corner with its legs along the two sides. Place them so that, going clockwise, each side of the square gets a leg a from the corner where it starts and a leg b from the corner where it ends, as in Figure 1. Since a + b is the whole side, the two copies on a side meet at one point: P on \overline{AB}, Q on \overline{BC}, R on \overline{CD} and S on \overline{DA}.

Figure 1The first way to fill a square of side a + b. The shaded middle has four sides c.

What is left in the middle has four sides, each the hypotenuse of a copy, so each is c. It looks like a square, but that has to be proved, and the corners are the whole question. At P, three angles sit side by side along the straight side \overline{AB}: the top left copy’s angle (two arcs), the corner of the middle shape, and the top right copy’s angle (one arc). The top left copy’s angle at P faces its leg b, the side \overline{AS}. The top right copy’s angle at P faces its leg a, the side \overline{BQ}. So they are the two different acute angles of the triangle, and they add to 90^\circ. The three angles fill a straight angle of 180^\circ, which leaves 180^\circ - 90^\circ = 90^\circ for the corner. The same happens at Q, R and S, because the pattern of legs repeats around the square. Four equal sides and four right angles make the middle a square of side c, with area c^2.

The second way. Take a second square of the same side a + b and cut it with one vertical and one horizontal line, each at distance a from the top left corner, as in Figure 2. That makes a square of side a, a square of side b, and two rectangles, each a by b. Each rectangle’s diagonal splits it into two copies of the triangle: a copy set with its right angle in a corner of the rectangle and its legs along the two sides has its other two corners at the ends of the diagonal. So the same four copies fit here too, leaving a^2 and b^2.

Figure 2The second way. The same four copies, paired into rectangles, leave a^2 and b^2.

Setting the counts equal. The two big squares have the same side, so the same area. Each contains the four copies, 4 \times \tfrac{1}{2}ab in all. What remains must therefore be equal: c^2 in the first, a^2 + b^2 in the second.

What every area proof must establish. Any proof of this kind, with any arrangement, owes the reader three things.

The pieces are what they are said to be. Each triangle is a copy of the given one, and each square or rectangle has the sides claimed.
The pieces fit. There are no gaps and no overlaps, every edge that looks straight is straight, and every corner that looks square is square. This is proved from angle facts, never read from the drawing, since a diagram may show where things are but not how big. Example 6 shows a drawing that looks perfect and hides a gap.
Both counts are of the same region, or of two regions known to have the same area. Then the counts are set equal and the algebra finishes the job.

Worked examples

Example 1

Counting on a grid

In Figure 3 each small square is one unit. A copy of the right triangle with legs 3 and 4 sits in each corner of a 7 by 7 square. Without measuring, find the area of the tilted square in the middle and so the hypotenuse c. Then check the answer with the second way of filling the square.

Figure 3Legs 3 and 4 on a grid. Each small square is one unit.

Solution

The big square has side 3 + 4 = 7, so its area is 7^2 = 49. Each copy is a right triangle with legs 3 and 4, so its area is \tfrac{1}{2} \times 3 \times 4 = 6, and the four copies cover 4 \times 6 = 24. The middle is what is left:

49 - 24 = 25

The middle is a square with side c, by the corner argument in the procedure, so c^2 = 25 and c = 5.

Now the second way. The same 7 by 7 square holds a square of side 3, a square of side 4 and two 3 by 4 rectangles:

9 + 16 + 2 \times 12 = 49

Both ways fill 49, and both contain the four copies, 24. So the tilted square, 25, equals the two small squares together, 9 + 16. That is 3^2 + 4^2 = 5^2, found without a ruler.

Example 2

The whole proof, line by line

Write the rearrangement proof as a two-column proof.

Given. A right triangle with legs a and b and hypotenuse c.

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

Solution

Plan. Fill two squares of side a + b in the two ways of Figures 1 and 2. The hard step is showing that the middle of Figure 1 is a square, and it rests on the acute angles adding to 90^\circ. The rest is adding areas.

Statements	Reasons
1. ABCD is a square with side a + b	construction: a square with a given side can be drawn
2. A copy of the triangle sits in each corner of ABCD, right angle in the corner, with AP = BQ = CR = DS = a and AS = BP = CQ = DR = b	construction: a copy moved by a rigid motion keeps its sides and angles; each side a + b holds one leg a and one leg b, so two copies meet at each of P, Q, R, S: line 1
3. SP = PQ = QR = RS = c	each is the hypotenuse of a copy: line 2
4. m\angle APS + m\angle BPQ = 90^\circ	\angle APS faces the leg \overline{AS} of length b and \angle BPQ faces the leg \overline{BQ} of length a, so they are the triangle’s two acute angles, which add to 90^\circ (triangle angle sum): line 2
5. m\angle APS + m\angle SPQ + m\angle QPB = 180^\circ	angle addition postulate: P lies on \overline{AB}, so \angle APB is a straight angle
6. m\angle SPQ = 90^\circ	substitution of line 4 into line 5, then the subtraction property of equality
7. m\angle PQR = m\angle QRS = m\angle RSP = 90^\circ	lines 4 to 6 at Q, R and S: by line 2, at each of them one copy’s angle faces a leg b and the other’s faces a leg a
8. PQRS is a square with area c^2	definition of a square: lines 3, 6 and 7; area of a square
9. Area of ABCD = c^2 + 4 \times \tfrac{1}{2}ab	area addition: the four copies and PQRS fill ABCD without overlapping; area of a triangle: lines 2 and 8
10. EFGH is a square with side a + b, cut into a square of side a, a square of side b and two a by b rectangles	construction (Figure 2)
11. Each rectangle is two copies of the triangle	a copy with its right angle in a corner of the rectangle and its legs along the sides has its hypotenuse exactly on the diagonal: line 10
12. Area of EFGH = a^2 + b^2 + 4 \times \tfrac{1}{2}ab	area addition; area of a square and of a triangle: lines 10 and 11
13. Area of ABCD = area of EFGH	both are squares of side a + b: lines 1 and 10
14. c^2 + 4 \times \tfrac{1}{2}ab = a^2 + b^2 + 4 \times \tfrac{1}{2}ab	substitution: lines 9 and 12 into line 13
15. c^2 = a^2 + b^2	subtraction property of equality: line 14
16. a^2 + b^2 = c^2	symmetric property of equality: line 15

Nothing in the table used a particular length for a or b, so the proof covers every right triangle at once. The right angle was used three times: in line 2, where each copy fits a square corner; in line 4, where the acute angles add to 90^\circ only because the third angle is 90^\circ; and in line 11, where each half of a rectangle has a right angle between its legs.

Example 3

One square, counted twice

Prove the theorem using only Figure 1, with algebra in place of the second square.

Solution

The area of ABCD can be found directly, since it is a square of side a + b, and it can be found from its pieces, as in line 9 of Example 2. Set the two equal:

(a + b)^2 = c^2 + 4 \times \tfrac{1}{2}ab

On the right, 4 \times \tfrac{1}{2}ab = 2ab. On the left, multiply out the brackets with the distributive property:

\begin{aligned} (a + b)^2 &= (a + b)(a + b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + 2ab + b^2 \end{aligned}

\begin{aligned} a^2 + 2ab + b^2 &= c^2 + 2ab \\ a^2 + b^2 &= c^2 \end{aligned}

by taking 2ab from both sides.

The second square of Figure 2 is a picture of the expansion itself: a square of side a + b made of a^2, b^2 and two rectangles ab. That identity is squaring a binomial, so the rearrangement proof and the algebra proof are the same proof told two ways.

Check. With a = 5 and b = 12: (5 + 12)^2 = 289, and c^2 + 2ab = 169 + 120 = 289, since c = 13.

Example 4

Four copies inside the square on the hypotenuse

Let the legs be a and b with a < b. Place four copies of the triangle inside a square of side c, each with its hypotenuse along one side of the square, arranged as in Figure 4. Prove that they leave a square of side b - a in the middle, and use it to prove the theorem.

Figure 4Four copies inside the square on the hypotenuse leave a small square of side b - a.

Solution

The pieces fit. At each corner of the big square, two copies meet. Place them so that one puts its angle facing b into the corner and the next puts its angle facing a, turning the same way at every corner. Those two angles add to 90^\circ, which is exactly the corner. So the second copy’s leg starts along the very ray where the first copy’s leg lies: the short leg a of one copy runs along the long leg b of the next. Because a < b, the short leg stops partway, and the long leg carries on for b - a further.

The middle is a square. Each side of the middle shape is one of those leftover pieces, so each has length b - a. At each corner of the middle shape, one side continues a copy’s short leg straight on, and the other side runs along the same copy’s long leg. The angle between them is what is left of a straight angle after the copy’s right angle: 180^\circ - 90^\circ = 90^\circ. Four equal sides and four right angles make a square, with area (b - a)^2.

Counting. The big square has area c^2, and it is filled exactly by four copies and the small square:

\begin{aligned} c^2 &= 4 \times \tfrac{1}{2}ab + (b - a)^2 \\ &= 2ab + b^2 - 2ab + a^2 \\ &= a^2 + b^2 \end{aligned}

The square (b - a)^2 was multiplied out as (b - a)(b - a) = b^2 - 2ab + a^2.

Check. With a = 3 and b = 4 the middle square has side 1, and 4 \times 6 + 1 = 25 = 5^2. The same count, four triangles around a small square, is described in Zhao Shuang’s commentary on the Chinese classic Zhoubi suanjing, written around the third century.

Example 5

Half a square: a trapezoid

In Figure 5, two copies of the triangle stand on one straight base \overline{UW}: the first has its right angle at U, its leg \overline{UV} upright with UV = a and its leg \overline{UM} along the base with UM = b; the second has its right angle at W, its leg \overline{WM} along the base with WM = a and its leg \overline{WX} upright with WX = b. Prove the theorem by finding the area of UVXW two ways.

Figure 5Two copies and a shaded half square make a trapezoid.

Solution

The middle triangle has a right angle at M. Three angles meet at M along the straight base: \angle UMV, \angle VMX and \angle XMW. In the first copy, \angle UMV faces the leg \overline{UV} of length a. In the second copy, \angle XMW faces the leg \overline{WX} of length b. So they are the two acute angles of the triangle, adding to 90^\circ, and m\angle VMX = 180^\circ - 90^\circ = 90^\circ. Its sides \overline{VM} and \overline{MX} are both hypotenuses, so \triangle VMX is a right triangle with both legs c and area \tfrac{1}{2}c^2.

The whole shape is a trapezoid. \overline{UV} and \overline{WX} are both perpendicular to the base, so they are parallel, and UVXW is a trapezoid with parallel sides a and b and height UW = a + b. By the area of a trapezoid, its area is

\tfrac{1}{2}(a + b)(a + b) = \tfrac{1}{2}(a + b)^2

Counting. The same area is two copies and the middle triangle:

\begin{aligned} \tfrac{1}{2}(a + b)^2 &= \tfrac{1}{2}ab + \tfrac{1}{2}ab + \tfrac{1}{2}c^2 \\ (a + b)^2 &= 2ab + c^2 \\ a^2 + 2ab + b^2 &= 2ab + c^2 \\ a^2 + b^2 &= c^2 \end{aligned}

The second line doubles both sides, the third multiplies out (a + b)^2, and the last takes 2ab from both sides. This proof was published in 1876 by James Garfield, then a member of the United States Congress and later President of the United States. Exercise 14 shows it is Example 3 cut in half.

Example 6

When a rearrangement lies

An 8 by 8 square is cut into four pieces, as on the left of Figure 6: a 3 by 8 strip along the top, cut by its diagonal into two right triangles with legs 3 and 8, and a 5 by 8 block, cut into two trapezoids with parallel sides 3 and 5 and height 5. The pieces are laid out as on the right, which looks like a 5 by 13 rectangle. The square had area 64 and the rectangle seems to have area 65. Where is the mistake?

Figure 6Area 64 on the left, apparently 65 on the right. The pieces do not meet along a straight line.

Solution

Moving pieces cannot change their total area, so the pieces still cover exactly 64. The mistake must be in the claim that they make a rectangle, and the place to test is the long edge where the pieces are supposed to meet in a straight diagonal.

Measure how steeply each edge climbs. The triangle’s hypotenuse rises 3 for every 8 across, a steepness of \tfrac{3}{8} = 0.375. The trapezoid’s slanted side goes from a height of 3 to a height of 5 over 5 across, rising 2 for 5, a steepness of \tfrac{2}{5} = 0.4. A true diagonal of a 5 by 13 rectangle would rise \tfrac{5}{13} \approx 0.385 all the way.

The three numbers differ, so the “diagonal” is not one straight line. It bends slightly at the two places where a triangle meets a trapezoid, and the lower pieces and the upper pieces leave a long, thin gap between them, shaped like a parallelogram. Its area is 65 - 64 = 1, spread along a diagonal more than 13 long, so on average it is less than a tenth of a unit wide. Figure 6 is drawn to scale, and there the gap shows only as a doubled line along the diagonal, easy to miss.

In Example 2 every corner of PQRS was proved to be 90^\circ (line 6). Here nothing proved that the edges line up, and they do not.

Boundaries

The triangle must have a right angle. Example 2 uses it three times. With any other angle between the two sides, the copies do not fit a square at all (Exercise 15), and the equation is false: a triangle with sides 2, 2 and 2 has 2^2 + 2^2 = 8 \ne 2^2.

A drawing is not a proof. Every straight edge and square corner that the count relies on must be proved. Example 6 looks just as convincing as Figure 1 and is wrong.

The plane must be flat. The proof leans on the angles of a triangle adding to 180^\circ and on the area of a square being its side squared. Both hold on a flat plane. On a curved surface such as a sphere, the angles of a triangle add to more than 180^\circ and a^2 + b^2 = c^2 fails.

It proves one direction. The proof starts from a right angle and ends at a^2 + b^2 = c^2. That a triangle whose sides satisfy the equation must have a right angle is a different statement, the converse of the Pythagorean theorem, with its own proof.

Where this goes next

The theorem proved here drives the rest of the right triangle lessons, from Pythagorean triples to the distance formula, and the converse turns it around to test whether an angle is right.

A second, quite different proof uses no area at all. The height from the right angle to the hypotenuse cuts the triangle into two smaller triangles similar to the whole, and their ratios combine into the same equation: proving the theorem with similar triangles.

When the angle between a and b is not right, the equation needs a correction term that depends on the angle. That corrected equation is the law of cosines, and when the angle is 90^\circ it reduces to the Pythagorean theorem. Reading algebra as area comes back in the area model for multiplying binomials and in the perfect square pattern.

Exercises

Counting the areas

Four copies of a right triangle sit in the corners of a square, as in Figure 1. For the legs given, find the area of the big square, the area of the four copies together, the area of the middle square, and the hypotenuse c (exactly, and to the nearest hundredth when it is not a whole number).

Legs 5 and 12
Legs 6 and 8
Legs 1 and 2
Legs 2 and 5
For legs 5 and 12, find the areas of the pieces in the second way of filling the square (Figure 2), and check that they add to the area of the big square.

Other arrangements

Four copies of the triangle with legs 5 and 12 are placed inside a square of side c, as in Figure 4. Find the side and area of the middle square, then find c^2 and c from the count.
Do the same for legs 2 and 3.
Do the same for legs 4 and 4. What happens to the middle square?
Two copies of the triangle with legs 3 and 4 make a trapezoid with a half square, as in Figure 5. Find the area of the trapezoid, the area of the two copies, and the area of the middle triangle, and use them to find c.

Working backwards

Four copies of one right triangle are placed in the corners of a square of side 17, as in Figure 1, and the tilted square in the middle has area 169. Find the legs of the triangle.
Four copies of one right triangle are placed in the corners of a square of side 10, as in Figure 1, and the tilted square in the middle has area 58. Find the legs.
In Figure 1, one acute angle of the triangle measures 37^\circ. Find the other acute angle, and the three angles that sit side by side at P.

Checking a rearrangement

A 13 by 13 square is cut into four pieces: a 5 by 13 strip cut by its diagonal into two right triangles with legs 5 and 13, and an 8 by 13 block cut into two trapezoids with parallel sides 5 and 8 and height 8. The pieces are laid out, as in Example 6, to look like an 8 by 21 rectangle. Compare the two areas, and compare the steepness of the triangle’s hypotenuse with that of the trapezoid’s slanted side. What has gone wrong?

Reasoning

Two copies of the trapezoid in Figure 5 fit together into the square of Figure 1. Explain how, and explain why the trapezoid proof of Example 5 is the proof of Example 3 cut in half.
Try the first arrangement with a triangle whose sides a and b meet at 100^\circ instead of 90^\circ. Explain what goes wrong.
Someone draws Figure 1 carefully, measures the corners of the middle shape with a protractor, finds 90^\circ each time, and writes “the middle is a square because it looks like one”. What is missing from the proof, and which fact supplies it?
The proof in Example 2 was written with letters, and Figure 1 was drawn with one particular pair of legs. Explain why the proof still holds for every right triangle.
Example 4 was drawn with a < b. Explain what changes if a > b, and why the proof still works when a = b.

Answers

Big square 17^2 = 289; copies 4 \times \tfrac{1}{2} \times 5 \times 12 = 120; middle 289 - 120 = 169; c = 13.
Big square 14^2 = 196; copies 4 \times \tfrac{1}{2} \times 6 \times 8 = 96; middle 196 - 96 = 100; c = 10.
Big square 3^2 = 9; copies 4 \times \tfrac{1}{2} \times 1 \times 2 = 4; middle 9 - 4 = 5; \displaystyle c = \sqrt{5} \approx 2.24.
Big square 7^2 = 49; copies 4 \times \tfrac{1}{2} \times 2 \times 5 = 20; middle 49 - 20 = 29; \displaystyle c = \sqrt{29} \approx 5.39.
The squares are 5^2 = 25 and 12^2 = 144, and the two rectangles are 2 \times 5 \times 12 = 120. Together 25 + 144 + 120 = 289 = 17^2, the big square.
The middle square has side 12 - 5 = 7 and area 49. Then c^2 = 4 \times \tfrac{1}{2} \times 5 \times 12 + 49 = 120 + 49 = 169, so c = 13.
The middle square has side 3 - 2 = 1 and area 1. Then c^2 = 4 \times \tfrac{1}{2} \times 2 \times 3 + 1 = 12 + 1 = 13, so \displaystyle c = \sqrt{13} \approx 3.61.
The middle square has side 4 - 4 = 0: it shrinks to a point, and the four copies fill the big square on their own. c^2 = 4 \times \tfrac{1}{2} \times 4 \times 4 = 32, so \displaystyle c = \sqrt{32} \approx 5.66. Check: 4^2 + 4^2 = 32.
The trapezoid has parallel sides 3 and 4 and height 7, so area \tfrac{1}{2}(3 + 4) \times 7 = 24.5. The two copies are 2 \times 6 = 12, so the middle triangle is 24.5 - 12 = 12.5. That is \tfrac{1}{2}c^2, so c^2 = 25 and c = 5.
5 and 12. The side gives a + b = 17. The four copies cover 289 - 169 = 120, so 2ab = 120 and ab = 60. Two numbers with sum 17 and product 60: from a(17 - a) = 60, a^2 - 17a + 60 = 0, so (a - 5)(a - 12) = 0. Check: 5^2 + 12^2 = 169.
3 and 7. Here a + b = 10 and the copies cover 100 - 58 = 42, so 2ab = 42 and ab = 21. From a(10 - a) = 21, a^2 - 10a + 21 = 0, so (a - 3)(a - 7) = 0. Check: 3^2 + 7^2 = 9 + 49 = 58.
The other acute angle is 90^\circ - 37^\circ = 53^\circ. At P the two copies put one of each, 37^\circ and 53^\circ, and the corner of the middle shape is 180^\circ - 37^\circ - 53^\circ = 90^\circ.
The square has area 13^2 = 169 and the “rectangle” 8 \times 21 = 168. The triangle’s hypotenuse rises 5 over 13, a steepness of \tfrac{5}{13} \approx 0.385; the trapezoid’s slanted side goes from 5 to 8 over 8 across, rising 3 over 8, a steepness of \tfrac{3}{8} = 0.375. They differ, so the long edge is not straight. This time the pieces would have to overlap along the diagonal, by 169 - 168 = 1 unit of area, to fit inside the rectangle.
Give a copy of the trapezoid a half turn about the midpoint of \overline{VX}. It fits against the first along \overline{VX}, with V and X trading places, and the two together form a square of side a + b: the bottom and top are the two bases \overline{UW}, each upright side is an upright side a of one trapezoid on top of an upright side b of the other, and each corner is one of the four right angles at U, W and their images. A copy of the triangle sits in each corner, as in Figure 1. The two shaded half squares, legs c and c, meet along their shared long side \overline{VX} to make the middle square of side c. Every area in Example 5 is half the matching area in Example 3: \tfrac{1}{2}(a + b)^2 against (a + b)^2, two copies against four, and \tfrac{1}{2}c^2 against c^2. Doubling Example 5’s equation gives Example 3’s.
A 100^\circ angle does not fit into the 90^\circ corner of a square, so the copies cannot be placed. Changing the outer shape does not rescue it: a four-sided outline with a 100^\circ angle at every corner would have angles adding to 400^\circ, but the angles of a quadrilateral add to 360^\circ. The middle fails the same way: the other two angles of the triangle add to 180^\circ - 100^\circ = 80^\circ, which would leave 100^\circ at every corner of the middle shape. The arrangement cannot be built, and for such a triangle a^2 + b^2 = c^2 is false.
A reason for the right angles. A protractor measures one drawing, to within a degree, and a proof must hold for every triangle exactly. The fact that supplies it: the two copies at each corner contribute the triangle’s two acute angles, which add to 90^\circ because the angles of a triangle add to 180^\circ; with the straight side that leaves exactly 90^\circ for the corner.
No step in Example 2 used the size of a or b, or read a length or an angle off the figure. Every length comes from the Given or the construction, and every angle from the triangle angle sum. The figure only shows where the pieces go, and that arrangement can be built for any positive a and b.
If a > b, the long leg is a, so swap the roles of the letters: the middle square has side a - b and area (a - b)^2 = a^2 - 2ab + b^2, the same as (b - a)^2. The count c^2 = 2ab + (a - b)^2 = a^2 + b^2 is unchanged. If a = b, the middle square has side 0, the four copies fill the square of side c exactly, and c^2 = 4 \times \tfrac{1}{2}a^2 = 2a^2 = a^2 + b^2.