PrepShorts · Study sheet · Class 8 Mathematics · Chapter 2, The Baudhāyana-Pythagoras Theorem
Chapter 2 · The Baudhāyana-Pythagoras Theorem
Halving a square: the doubling construction run backwards
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Draw a square of exactly half the area — and the thing you draw has to be a square. That last condition is the whole difficulty.
The idea
Halving a square needs no new idea — it is the doubling construction read from the other end. In §2.1 the given square was the inner one and the new square was built outward on its diagonal; here the given square is the outer one, so the new square is the tilted one whose corners sit at the midpoints of its sides. The same pair of perpendicular midlines that proved the outward case proves this one: they cut the tilted square into four copies of a triangle and the outer square into eight, so the ratio is fixed at one half. And the halving is where the side-versus-area confusion becomes impossible to dodge: a square with half the sidelength has a quarter of the area, because four of them fit inside — the chapter makes the student find those four.
What you should be able to do
- Construct, inside a given square, a square of exactly half its area, by joining the midpoints of the four sides
- Explain why that inner figure is a square and not merely a rhombus, using the angles its diagonals create
- Prove the halving by counting congruent triangles: four inside the tilted square, eight in the outer one
- Answer the chapter's own question about halving the sidelength, and state the count of small squares that fill the original
- Perform the paper fold in which four corner flaps turn inward on creases through the side midpoints, and use the flaps themselves as the area argument
- Identify the halving construction as the doubling construction with the roles of given and constructed square exchanged
- Extend the halving into a chain — half, quarter, eighth — and state the sidelength ratio at each step in words the chapter has supplied
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| midpoints of the sides | the four points halfway along a square's sides, where the inner square's corners go | printed in this chapter (Part II, §2.2, p.36) |
| crease lines | the fold lines made in the paper activity, running through those midpoints | printed in this chapter (Part II, §2.2, p.36) |
| PQRS | the chapter's labelling of the inner half-area square in the folding figure | printed in this chapter (Part II, §2.2, p.36) |
| east-west | Baudhāyana's word, as the chapter reports it, for a horizontal reference line | printed earlier in this chapter (Part II, §2.1, p.34) |
| north-south | his word for the vertical line perpendicular to it | printed earlier in this chapter (Part II, §2.1, p.34) |
| congruence | the relation between figures of identical shape and size, used here on triangles | printed in this chapter (Part II, §2.2, p.37), with the adjective congruent earlier at Part II p.34 |
| sidelength | the length of one side, written as one word throughout this book | printed in this chapter (Part II, §2.2, p.36) |
| rhombus | a four-sided figure with four equal sides, which need not have right angles | printed later in this chapter (Part II, p.53, Figure it Out no.3) |
| corner flap | one of the four triangles folded inward in the paper activity | an added term; the chapter shows the flaps and gives them no name |
| quarter-size square | a square with half the sidelength of a given one | an added phrasing; not printed in this chapter |
Where people slip up
- "Half the side, half the area." The dominant error, and the chapter attacks it head-on with its own question. Draw the four quarter-squares inside the original and count them before saying anything else.
- "A square of half the area is half a square — cut it down the middle." Cutting a square in half gives a rectangle. The task is a square, and the only way to get one is to tilt it.
- "The inner figure is a diamond, not a square." It is a square, tilted. The chapter deliberately asks for the proof, so an explanation that draws it and moves on has skipped the content. Show the equal, perpendicular, mutually bisecting diagonals.
- "The creases go corner to corner." Folding on the diagonals gives a smaller triangle, not a square. The creases run midpoint to midpoint — four of them, each joining the midpoints of two adjacent sides. Two creases could not turn four corners in, and creasing along a midline would fold half the sheet over and never produce PQRS at all.
- "The flaps overlap in the middle, so the fold proves nothing." They meet at the centre and do not overlap. This is worth showing slowly in the figure, because if a student believes the flaps overlap then the whole area argument collapses for them.
- "Halving and doubling are two constructions to memorise." They are one picture with the labels swapped. If a student can say which square is given, they can produce either result from the same figure.
- "You can keep halving and reach zero." You can keep halving forever and never reach zero; each step is another factor of one half. Worth one sentence, since section 10 walks the chain.
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Worked answers to this chapter’s exercises
Transcript1,337 words
Last time the question asked for more area. Now turn it round and ask for less. Here is a square. Draw me another one with exactly half its area. Not roughly half. Exactly half, and the thing you draw has to be a square. That last condition is the whole difficulty, and it is very easy to read straight past. There are several honest ways to cut a square into two equal pieces, and almost none of them leaves you holding a square.
So before the construction, look at the two cuts that everybody tries first. The first is to cut straight down the middle. That does halve the area. Start with a square of area four and each piece has area two, exactly as asked. But look at the piece you are holding. Its two sides are two and one. It is a rectangle. The second attempt is to fold corner to corner, along a diagonal.
That halves the area as well. Each piece is area two again. And each piece is a triangle. Two honest halvings, and neither one answers the question. The half you want has to be tilted. Here is the construction, and it is four points. Mark the middle of each of the square's four sides. Join them up, in order, and the figure you get is the answer. Its corners are those four midpoints, and it stands on one of them, tilted inside the square you began with.
That square had area four. This one has area two. Half, exactly, and again nothing has been measured. But two things still have to be shown. That this figure really is a square, and why its area is half. Take the first one, because a picture is not a proof. The four sides are certainly equal. Each one runs from the middle of one side across to the middle of the next.
But four equal sides is not enough. A squashed diamond has four equal sides and is nothing like a square. The thing that settles it is the pair of diagonals. Draw them in. They are equal in length. They cross at right angles. And each one cuts the other exactly in half. A four-sided figure whose diagonals do all three of those things is a square. This one does all three. So it is one.
It is worth seeing that all three conditions are working, because dropping any one of them lets something else in. A long rectangle has diagonals that are equal and that cut each other in half. They do not meet at right angles, and it is not a square. A diamond has diagonals that meet at right angles and cut each other in half. They are not equal, and it is not a square.
A kite has diagonals that meet at right angles, and they are neither equal nor cutting each other in half. You can even build a figure whose diagonals are equal and perpendicular but miss each other's middles entirely. Also not a square. Three conditions, and every one of them is doing real work. Which is why the diamond objection is a genuine objection and not a quibble. Now for the area, and here the old picture walks back in.
Take the outer square's own horizontal and vertical lines through its centre. The line halfway up, and the line halfway across. Look at where they land. Each one runs from one midpoint straight to the opposite midpoint. Which means each one of them is a diagonal of the tilted square. That is the same coincidence as before, seen from the other side. What used to be the extensions of the inner square's sides are now the inner square's diagonals.
It is one figure. Which square you were handed is the only thing that changes. And those two lines are what do the counting. The two lines cut the tilted square into four triangles. They are all the same triangle. Two sides of each run from the centre out to a corner, those are halves of two equal diagonals, and the corner between them is a right angle. Now look outside the tilted square. Four corner pieces are left over, one at each corner of the outer square.
Every single one of those is that same triangle again. So count. The tilted square is four pieces. The outer square is those four, plus the four corners. Eight. Four out of eight. One half, exactly, and it is a count rather than a measurement. The same picture read outward gave two against four. Read inward it gives four against eight, and the ratio never moves. There is a far more tempting answer than the tilted square, and it is wrong.
Halve the sidelength. Take a square of side two and draw one of side one. That certainly looks like it ought to give you half. It gives a quarter. Area one, where the square you started from had four. The reason is last time's reason running downhill instead of uphill. An area is a side multiplied by a side, so halving the side halves the answer twice over. A half times a half is a quarter.
One square, two different answers, and the difference is which word you halved. You can see the quarter without doing any arithmetic at all. Draw the small square in one corner of the big one. Now draw three more just like it. They fit. Two across and two up, four of them, filling the original exactly, nothing left over and nothing overlapping. Four of them make the whole, so each one is a quarter of it.
That is the doubling statement written backwards. Double a side and the area goes up four times; halve a side and it goes down four times. The half-area square is the tilted one. The quarter-area square is the little upright one. Two answers from one square. All of this can be done with a sheet of paper and nothing else. Take a square sheet and find the middle of each of its edges.
Now make four creases, each one joining the middles of two neighbouring edges. Those four creases together are the four sides of the tilted square. Fold each corner inward along its crease. Four corners, four flaps, all turning in towards the centre. Watch where the corners go. Every one of them lands exactly on the centre of the sheet. Crease corner to corner instead and each flap folds straight onto its neighbour, covering ground you already had.
The fold is not merely a way of drawing the square. It is the proof that the area is half. Look at the four folded flaps. They lie inside the tilted square. Nothing hangs over an edge. They do not overlap each other either. They meet along the two diagonals and stop there. And they leave nothing uncovered. So the four flaps together cover the tilted square exactly once, which means they have the same area as it does.
The flaps plus the tilted square make the whole sheet. And the flaps are equal to the tilted square. So the tilted square is half the sheet. Nothing anywhere in that argument measured anything. Nothing stops you doing it again. Halve the tilted square the same way and you have a quarter of what you started with. Halve that and you have an eighth. The areas run one, a half, a quarter, an eighth, and they never arrive at nothing.
The sidelengths are the interesting part. Every second square in the chain has a side you can write down. The first has side one. The third has side one half. The fifth would be one quarter. The ones in between have sides whose squares are a half and an eighth, and those are lengths you can draw but cannot write as a fraction. It is the same awkward length that turned up when we were doubling, and it is still a question for another day.
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.
Builds on
- Doubling a square: the diagonal is the constructionClass 8 · Ch 2, The Baudhāyana-Pythagoras Theorem
Comes up again in
- The isosceles right triangle: why its hypotenuse must be a√2Class 8 · Ch 2, The Baudhāyana-Pythagoras Theorem