PrepShorts · Study sheet · Class 7 Mathematics · Chapter 6, Constructions and Tilings
Chapter 6 · Constructions and Tilings
Constructing a 90° angle at a chosen point on a line
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There is no second construction here. A right angle at a marked point is the perpendicular bisector run backwards.
The idea
There is no second construction here. A right angle at a chosen point O is the perpendicular-bisector construction run backwards: instead of being handed a segment and made to find its middle, you are handed the middle and you manufacture a segment around it. That reversal is the whole idea, and it pays a dividend nobody expects — because O is already known to lie on the bisector, half the arcs become unnecessary. Seeing which half, and why, is the test of whether you understood the first construction or only memorised its picture.
What you should be able to do
- State the problem: a right angle wanted at a named point of a given line
- Explain why extending the line and marking two equal distances from O turns the new problem into the old one
- Say why the perpendicular bisector of the manufactured segment is bound to pass through O
- Explain why a single crossing pair of arcs now suffices
- Carry out the construction with a compass and an unmarked ruler
- Identify what the second arc pair was doing in the earlier construction, and why it can be dropped here
- Adapt the same reversal to drop a perpendicular from a point lying off the line
- Recognise the 90° angle as the seed that later gives 45°
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| perpendicular | meeting at a right angle | printed in §6.1, Part II, pp.137, 141, 142 |
| perpendicular bisector | the line that halves a segment and meets it square on | printed in §6.1, Part II, p.137, and used throughout pp.139–142 |
| midpoint | the point splitting a segment into two equal halves | printed in §6.1, Part II, pp.137, 141 |
| arc | part of a circle drawn from a centre at a fixed radius | printed in §6.1, Part II, pp.139–141 |
| compass | the instrument that keeps one distance while it turns | printed in §6.1, Part II, pp.139–141 |
| unmarked ruler | a straight edge carrying no scale | printed in §6.1, Part II, pp.139–140 |
| line segment | the piece of a line between two named points | printed in §6.1, Part II, pp.139, 141 |
| bisection | cutting something into two identical parts | printed in bold in §6.1, Part II, p.137 |
| straight angle | the 180° angle at a point of a straight line | printed in §6.1, Part II, p.137 |
| rope | the Śulba-Sūtra stand-in for a compass, used in the follow-up question | printed in §6.1, Part II, pp.142, 144 |
| foot of the perpendicular | the point where a dropped perpendicular meets the line | an added term, not printed in this chapter — the chapter poses the outside-point construction on p.155 without naming that point |
Where people slip up
- "This is a fresh construction to be learnt separately." It is the earlier one, read from the other end. If the explanation presents it as new, the student will carry two recipes instead of one idea.
- **"You must draw arcs above and below."** Here you do not need to. The chapter asks the question explicitly on p.141 and answers no, because O is already a known point of the line being built.
- "X and Y have to be at some particular distance from O." Any opening will do; only equality matters. Students who think a special length is required will reach for a scale, which the chapter has just argued against.
- "The right angle is at A." It is at O. A is only the second point that fixes the line's direction.
- "A perpendicular through a point off the line needs a different method again." Question 4 on p.155 is the same reversal a third time: find a segment of the line that P is equally far from both ends of.
- "Right angles come from a set square or a protractor." The chapter has restricted itself to two tools from the foot of p.140 onward, and everything after that point respects the restriction.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 7 Q4
Transcript1,427 words
Here is a straight line, and here is a point sitting on it. Call it O. You are asked for a right angle at O. Not near O. At O, with one arm running along the line you already have. A set square would settle this in a second, and so would a protractor. You have neither. You have a compass, and a straight edge with nothing written on it.
That turns out to be enough, and the reason it is enough is worth more than the drawing. Because you already know how to do this. You just do not know yet that you know it. Here is what you already know how to do. Given a segment with two ends, you can find the line that cuts it in half and meets it square. Open the compass past halfway, swing from one end, swing from the other, join the crossings.
Now try to start that here. The very first instruction says open the compass past halfway. Halfway of what? There is no segment on this board. There is a point, and a line running off in both directions. So the recipe stalls on its first word, and that is not a small gap. That is the whole difficulty. The way out is not another recipe. It is reading the one you have from the other end.
What does that construction actually do? It takes a segment, and it hands you back the middle. You have the middle. You want the line. So make a segment. Manufacture the two ends yourself, one on each side of O. Then the construction you already know will hand back exactly the thing you were asked for. That reversal is the entire topic. Everything after this is detail. Extend the line a little each way first, so there is room to work.
Now open the compass. To anything at all. There is no correct opening here, and nothing on the board tells you what to pick. Whatever you choose, lock it, and do not touch it again. Put the point on O and mark the line to the left. That mark is X. Put the point on O again, same opening, and mark to the right. That mark is Y. Two marks, one opening used twice, and O is now exactly halfway between X and Y.
Not because anybody measured. Because one distance was used twice, which is the only thing a compass is good at. Now look at what you have built, and ask the old question about it. Where is the line that cuts X to Y in half and meets it square? Every point that is equally far from X and from Y sits on that line. That was the whole of the last idea.
So test O. How far is O from X? One opening. How far from Y? The same opening. O is equally far from both. O is on the line you are hunting. And you have not drawn a single arc yet. That is the dividend the reversal pays. You begin already owning a point of the answer. Which turns the rest of the job into counting. A straight edge needs two points. Give it one and it can still swing anywhere it likes through that point.
Give it a second and there is exactly one line it can rule. You have one already. O. So the arcs are being asked for one more point. One. Not two. And half of the construction you memorised has just become unnecessary, for a reason you can say out loud. So here is the whole thing. Open the compass wider than the step you took, wider than O out to X.
Swing an arc from X. Swing an arc from Y with that same opening. They cross above the line, at a point. Call it A. Lay the straight edge on O and on A, and rule. That is your right angle, and you are finished. Notice what is missing. There is no crossing below the line, and there is no need for one. Draw it out of habit if you like. It will land exactly where you expect, and it will tell you nothing you did not already have.
One thing to be careful about, because it is the easiest thing to get backwards. The right angle is at O. It is not at A. The corner at A is nowhere near square, and nothing in the construction ever claimed it was. A has one job. To be a second point, so the straight edge knows which way to go. Open the compass wider and A climbs higher up.
Open it narrower and A drops. The line through O is the same line every time. A moves. The answer does not. The angle you were asked for is at the point you were handed. Put the two constructions beside each other and the missing arcs explain themselves. The first one began knowing nothing. Not one point of the answer was in hand. So it had to manufacture two of them, one crossing above and one below, and join them.
This one began holding a point already. So that is what the second pair of arcs was ever doing. Not squareness. Not accuracy. It was buying a second point, because two points are what a straight edge costs. Get one for free and you only ever needed the one pair. Now go back to the first move and loosen it, because it looks fussier than it is. Does the step from O out to X have to be some particular length?
Step out five, then cross with an opening of thirteen. The crossing lands twelve above the line. Step out twelve instead, cross with thirteen again, and the crossing drops to five above. Step out fifteen, cross with seventeen, and it lands eight above. Three different steps. Three crossings in three different places. And all three of them rule the very same line through O. So the size never mattered. What mattered is that the two steps were equal, because that is what put O in the middle.
A student reaching for a ruler here has misread the question. One more version, and it is the same move a third time. Now the point is not on the line. It is P, floating somewhere above it, and you want the perpendicular from P down to the line. Put the compass point on P and open it until the arc will reach past the line. Swing. It cuts the line in two places.
Both of those cuts are on one arc from P, so P is exactly as far from one as from the other. Which means P is equally far from two points of the line. And you know where such points live. P is on the perpendicular bisector of the piece between the cuts. So bisect that piece, the way you already can, and the line you draw runs through P and meets the line square.
Handed a point, you manufactured a segment it sits equally far from. Same reversal, third time. It is worth noticing that none of this actually needs a compass. Strip the construction down to what it uses, and it uses one thing. Hold a distance, and turn. A rope with a peg does that. Peg it at O, pull it tight, scratch the ground to the left and to the right. There are X and Y.
Take a longer rope. Peg at X and scratch an arc. Peg at Y and scratch another. Where they meet is A. Pull a line from O through A and you have a right angle in the ground. Nothing was read. Nothing was numbered. A rope with a knot in it is a compass that happens to be made of rope. So why build a right angle this way, when a set square would have done it?
Because of what happens next. A right angle is a hundred and eighty halved. The next thing anybody wants is half of that again. Forty five degrees. And you get it by cutting this angle in two. The same word as before, bisecting, doing the same work one dimension up. But you cannot cut an angle in half until you have one to cut. That is what this construction really is. Not a trick for right angles. The seed.
And it was built from a point you were given, a distance you chose at random, and one crossing.
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
- The perpendicular bisector, and the equidistance property that justifies itClass 7 · Ch 6, Constructions and Tilings
Comes up again in
- A stretched rope as compass and straightedge: the Śulba-Sūtra constructionsClass 7 · Ch 6, Constructions and Tilings
- Bisecting an angle, and halving 90° to get 45°Class 7 · Ch 6, Constructions and Tilings