PrepShorts · Study sheet · Class 8 Mathematics · Chapter 2, The Baudhāyana-Pythagoras Theorem
Chapter 2 · The Baudhāyana-Pythagoras Theorem
Doubling a square: the diagonal is the construction
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Draw a square of exactly twice the area. Not roughly twice — exactly. Baudhāyana set that problem down around 800 BCE.
The idea
Doubling a square's area is not doubling its side. Stretch every side to twice its length and you have laid down four copies of the original, not two. The segment that actually does the job is the diagonal, and the reason is a counting argument, not a measurement: extend the original square's own horizontal and vertical sides across the square built on its diagonal, and both squares fall into copies of one single triangle — the original holds two of them, the new one holds four. Area is being counted in identical pieces, so "double" comes out exact, and nobody has to know how long the diagonal is.
What you should be able to do
- Explain why doubling every side of a square multiplies its area by four rather than by two, and give the general reason in terms of side × side
- Carry out Baudhāyana's construction: given a square, draw its diagonal and build a square on that diagonal
- Draw the horizontal and vertical lines that make the doubling visible, and say why those lines pass through the new square's corners
- Count the congruent triangles in each of the two squares and use the counts 2 and 4 to justify the word "double" without measuring any length
- Justify the congruence of the four triangles rather than asserting it from the drawing
- Continue the construction to produce a chain of squares in which each has twice the area of the one before, and say how many triangles each holds
- Reproduce the doubling with paper: cut two identical squares into quadrant triangles and reassemble them into one square of twice the area
- State the same result as a fact about the diagonal, ready for §2.3 to convert into a length
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| diagonal | a segment joining two corners of a square that are not next to each other | printed throughout this chapter (Part II, §2.1, p.33) |
| Śulba-Sūtra | Baudhāyana's text of geometric construction rules, dated in the chapter to about 800 BCE | printed in this chapter (Part II, §2.1, p.33) |
| Verse 1.9 | the numbered verse of the Śulba-Sūtra in which the doubling rule is given | printed in this chapter (Part II, §2.1, p.33) |
| east-west | Baudhāyana's word, as the chapter reports it, for a horizontal reference line | printed in this chapter, in single quotes (Part II, §2.1, p.34) |
| north-south | his word for the vertical line perpendicular to the east-west one | printed in this chapter, in single quotes (Part II, §2.1, p.34) |
| congruent | of exactly the same shape and size, so one can be laid on the other | printed in this chapter (Part II, §2.1, p.34) |
| area | the amount of surface a figure covers, measured here by counting equal triangles | printed throughout this chapter (Part II, §2.1, p.33) |
| sidelength | the length of one side, written as one word throughout this book | printed throughout this chapter (Part II, §2.2, p.36) |
| dotted square | the chapter's way of naming the newly constructed square in its figures, which it draws with a broken outline | printed in this chapter (Part II, §2.1, p.34) |
| tiling by congruent pieces | covering a figure exactly with copies of one shape, no gaps and no overlaps | an added phrasing; the chapter performs this and does not name it |
| scale factor | the number every length is multiplied by, whose square is what the area is multiplied by | an added term; not printed in this chapter |
Where people slip up
- "Double the area means double the side." The chapter opens on this guess precisely because it is the one students bring. It gives four times the area, and the chapter shows the four quadrants. Make the student count the quadrants before saying anything about diagonals.
- "Then double the area means side times 1.5, or thereabouts." No single familiar fraction works; the required side is a length that cannot be written as a fraction at all. That is §2.3–§2.4's business, but an explanation that lets the class hunt for a fraction now has set the later argument up properly.
- "The tilted square is smaller — it looks pinched." A square standing on its diagonal looks narrower on the page than the same square set upright. Tilting a figure changes nothing about its area. Rotate it until it is upright and the illusion dies.
- "The four triangles look the same, so they are." Looking the same in a diagram is not congruence. The chapter asks the student to explain the congruence: each triangle has two sides that are halves of the two equal diagonals of the broken tilted square built on the diagonal — not of the genuinely largest square in the figure — meeting at a right angle where those diagonals cross.
- "The extended sides pass through the corners because the drawing was made that way." They pass through because the original square's side lies along a diagonal of the new square, and a diagonal of a square runs from corner to corner. The chapter's hint puts it the other way round — via angle bisection — and either direction is fine as long as one is given.
- "Cutting and rearranging might lose a bit." Nothing is lost: the pieces are the same pieces. This is the whole reason the chapter argues by dissection rather than by arithmetic, and it is the habit the rest of the chapter needs.
- "This only works for a square." Section 9 should be honest here. The chapter's chain doubles squares; the argument used is specific to a square's diagonals being equal and perpendicular.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 2.3 Q1
Transcript1,280 words
Here is a square. Draw me another one with exactly twice its area. Not roughly twice. Exactly twice. Around eight hundred years before the common era, Baudhayana set this problem down in the Sulba-Sutra, a manual of geometric construction, and gave an answer that needs no measuring at all. The answer is a single segment you can already see in the square. But before that segment, the guess almost everyone makes.
Double the area, so double the side. It sounds inevitable. Take a square of side one, area one, and stretch every side to two. The new square does not have area two. It has area four. You can see why by cutting it: four copies of the original square fit inside it exactly. The reason is that area is a length multiplied by a length. Double the side and you have doubled that product twice over, so the area goes up by two times two.
Doubling a side never doubles an area. It quadruples it. So the side has to grow by less than double. By how much? Try one and a half. One and a half squared is two point two five, which overshoots. Try one point four, which is seven fifths. Squared that is one point nine six, which falls short. Seventeen twelfths lands inside a hundredth of two, and still misses. Ninety-nine seventieths is closer still, by a factor of three, and it misses too.
You can go on hunting. No fraction whatsoever squares to exactly two, and none of sixty denominators tried here comes close enough to be equal. Which is why the answer is not a number to multiply by. It is a line to draw. Draw the square's diagonal. Now build a new square standing on that diagonal, using it as one side. That square has exactly twice the area of the one you started with.
Not approximately. Exactly, and the proof does not use the diagonal's length anywhere. Nobody in this argument ever has to know how long the diagonal is. Instead, both squares are going to be cut into copies of one single triangle, and then counted. The whole construction is one line, drawn from one corner to the opposite one. Here is the device that makes the picture carry the argument. Take the original square's own top side and extend it straight across the new square. Baudhayana called that direction east-west.
Take its own left side and extend that too, north-south. Look where those two lines arrive. Each one passes through two corners of the new square, so between them they reach all four. They are not drawn to hit those corners. They arrive there because the original square's side lies along a diagonal of the new square, and a diagonal runs corner to corner. The two lines cross at right angles, and they cross exactly at the new square's centre.
And they have cut it into four triangles. Now go back to the original square and draw the diagonal you already have. It cuts that square into two triangles. Two triangles inside. Four triangles outside. And they are all the same triangle. So the original square is two of that piece and the new square is four of it. Four is twice two, and that is the whole proof. No lengths, no square roots, no arithmetic beyond counting to four.
Area is being measured here in identical pieces, which is why the word double comes out exact. But saying they are the same triangle is a claim, and it needs a reason. Every one of the four is built the same way. Two of its sides run from the centre out to a corner. Those are halves of the new square's two diagonals, and a square's two diagonals are equal, so all eight of those half-sides are the same length.
The corner between them, at the centre, is a right angle, because a square's diagonals meet at right angles. Two sides the same and the angle between them the same. That fixes the triangle completely. The third side, the one lying along the outside, is a side of the new square, and there are four of those, all equal. So the four are not merely similar looking. They are the same triangle, four times over.
It is worth being strict about this, because looking alike proves nothing. Here are two right triangles of exactly the same area, two square units each, and they are not the same triangle at all. One has legs four and one. The other has legs two and two. Same area, different shape. Equal area is not congruence. And there is a second trap in the picture. The four triangles' short sides are halves of the tilted square's diagonals.
Not halves of the largest square on the page. Those would be a different length entirely, and the argument would collapse. Take the sides from the right square. One more objection, and it is the one your eye keeps making. The new square is standing on a corner, and a square standing on a corner looks pinched. It looks smaller than it is. It is not. Turning a figure changes nothing whatsoever about its area.
A square drawn on a slanted step of three across and four up has area twenty-five, and so does an upright square of side five. Same area, exactly, and the tilted one is still a square by the test that matters. That test is worth stating: four equal sides is not enough, because a squashed diamond has four equal sides too. A square needs its two diagonals equal as well, and the diamond fails that.
Nothing stops you doing it again. Take the new square, draw its diagonal, and build on that. The first square has area one, the second two, the third four. Each is exactly double the one before. Count the triangles and you get two, then four, then eight. And look at the third square. Its area is four, which is where the doubled side was going to land at the very start.
So the wrong answer was not nonsense. It was the right answer to the question asked twice. Keep going and the areas run one, two, four, eight, sixteen, for as long as you have paper. You can do all of this with scissors, and there are two ways. Take two identical paper squares. Cut the second along both its diagonals into four quarter triangles. Set those four around the first square, one against each side, and the outline you get is a square of area two.
It is the square on the first square's diagonal, built out of paper. The second way is simpler. Take two identical squares and cut each along one diagonal only. Four right triangles. Put their four right angles together at a centre and their four long sides on the outside. That is a square of area two as well. Different cuts, different pieces, the same square at the end. So what is the length of that diagonal?
The new square has area two, and its side is the diagonal. So the diagonal is a length whose square is two. And we already went hunting for that number and did not find it. It is not any fraction at all. Which is exactly why the counting argument matters. It never needed the number. Two pieces in the small square, four in the big one. Four is twice two, and the area doubles.
The length is a question for another day, and it turns out to be a strange and important one. Baudhayana had the construction long before anybody had the number.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Comes up again in
- Halving a square: the doubling construction run backwardsClass 8 · Ch 2, The Baudhāyana-Pythagoras Theorem
- The isosceles right triangle: why its hypotenuse must be a√2Class 8 · Ch 2, The Baudhāyana-Pythagoras Theorem
- Combining two different squares: a² + b² = c²Class 8 · Ch 2, The Baudhāyana-Pythagoras Theorem
Either side of this one
- Tricky percentages: why a 50% margin followed by a 50% discount leaves a 25% lossClass 8 · Ch 1, Fractions in Disguise