PrepShorts · Study sheet · Class 7 Mathematics · Chapter 1, Geometric Twins
Chapter 1 · Geometric Twins
RHS: the right-angled special case
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RHS is not an exception bolted onto the SSA warning. It is the case where the SSA ambiguity turns harmless.
The idea
RHS is not an exception bolted onto the SSA warning; it is the case where the SSA ambiguity turns harmless. Both times the chapter draws a base, raises a ray at the given angle and swings the given radius across it, and both times the arc cuts twice. What differs is where the second cut lands. Make the angle a right angle and the two cuts sit on opposite sides of the base, so the second triangle is the first one flipped over — and a flip was always allowed. Leave the angle at the 30° of the p.11 counterexample and both cuts land on the same side, where no flip can rescue you. So the right angle is not a magic ingredient; it is the thing that puts the two answers back into mirror positions.
What you should be able to do
- State the RHS condition, naming which three parts have to match
- Identify the hypotenuse in a right-angled triangle
- Carry out the four-step construction from a base, a perpendicular and an arc
- Explain why the arc necessarily meets the perpendicular on both sides of the base
- Argue that the two resulting triangles are congruent, using the base as a fold line
- Say precisely how this case differs from the SSA counterexample
- Recite the chapter's list of five sufficient conditions, and say what is not on it
- Apply RHS to a labelled pair of right-angled triangles
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| right-angled triangle | a triangle one of whose angles is 90° | printed in §1.2, Part II, pp.16–17, and in the SUMMARY, p.22 |
| hypotenuse | the side of a right-angled triangle that faces its right angle | printed in §1.2, Part II, p.17, and in bold in the SUMMARY, p.22 |
| RHS condition | a right angle, a hypotenuse and one other side, which forces congruence | printed in bold in §1.2, Part II, p.17 |
| perpendicular | at right angles to a given line | printed in §1.2, Part II, p.16 |
| arc | part of a circle, swung from a centre at a fixed radius | printed in §1.2, Part II, pp.16–17 |
| sufficient | enough on its own to force the conclusion | printed in §1.2, Part II, p.17, in the bold heading over the list of five |
| guarantee | what a condition does when it forces congruence rather than merely permitting it | printed in §1.2, Part II, pp.12, 16–17 |
| special case | a narrower situation in which a generally unreliable set of givens does work | printed in §1.2, Part II, p.16 |
| extension | the continuation of a drawn line beyond the point it stopped at | printed in §1.2, Part II, p.17 |
| leg | the common name for a side of a right-angled triangle other than the hypotenuse | an added term, not printed in this chapter — the book calls it simply a side |
| Pythagoras' theorem | the relation among the three sides of a right-angled triangle | the explanation's reference, not printed in this chapter — do not invoke it in the script |
| fold line | the explanation's name for the line the second triangle is flipped about | an added phrase, not printed in this chapter |
Where people slip up
- "RHS proves that SSA works after all." It does not. SSA still fails, and the chapter keeps that failure on the books. What RHS says is that this particular SSA-shaped situation is safe, for a reason you can see in the picture.
- "Any two sides of a right-angled triangle will do." One of the two has to be the hypotenuse. Give a right angle and the two shorter sides instead and you are in the SAS case, which also works — but for a different reason, and the chapter is careful to name only the hypotenuse version RHS.
- "The hypotenuse is the bottom side" or "the slanted one." It is whichever side faces the right angle, wherever the triangle happens to be sitting on the page. The p.16 sketch has it slanting and the p.20 exercise has it named only by its letters — use both.
- "The arc might miss the line below the base." It cannot. The base is perpendicular to l at Q, so Q is the closest point of l to R; whatever distance the arc reaches beyond that, it reaches equally far up and down. Say this in words, without a formula.
- "A second solution always means the givens were not enough." This is the belief the whole chapter is training against, and it has now been wrong twice and right once. What matters is whether the second solution is a flip of the first.
- "There are five conditions, so there are five things to memorise." AAS reduces to ASA by the angle sum, and RHS is a repaired SSA. Two of the five are consequences, not axioms.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 1.3 Q2
Transcript1,360 words
Two sides, and an angle that is not between them. We watched that fail. The arc cut the ray twice, and the two triangles were genuinely different. So that set of measurements does not settle a triangle. That still stands. But it does not fail every single time. There is a case where it works. And it is worth seeing, because the reason is not that some new rule arrives.
The reason is that the same failure happens, and then turns out to be harmless. One ingredient changes. Everything else about the construction stays exactly as it was. The ingredient is a right angle. Here is the set. A right angle, and two of the three sides. The angle at B is ninety degrees. The side B C is four centimetres. And the side A C is five centimetres. Look carefully at where those two sides are, because this is the whole condition.
B C runs from the right angle along to C. It is one of the two shorter sides. A C does not touch B at all. It is the side facing the right angle. So the angle sits at one end of the four, and the five runs from the other end. That is exactly the shape that failed last time. Same shape. Different angle. So build it, in four steps, the same way as before.
Step one. Draw the base Q R, four centimetres long. Step two. At Q, raise a line at right angles to that base. Step three. The corner is five centimetres from R, so swing an arc of five, centred on R. Step four. Where the arc meets the line, that is the corner. Call it P, and join it to R. There is the triangle. A right angle at Q, four along the bottom, five on the slope.
The remaining side comes out at three centimetres, which is only how we draw it to scale. Nothing about that number was handed to us, and nothing in the argument needs it. Now, a question about step two that is very easy to walk straight past. We raised a line at Q. Not a ray going upward. A line. And a line goes both ways. So continue it downward, below the base.
Now swing that arc again, and let it go all the way round. It crosses up here, where we already knew. And it crosses again, down there. Two crossings. Which is exactly the trouble we had last time. So there is a second triangle. Q, R, and this lower corner. And it has the same three measurements. A right angle at Q, a four, and a five. So are those two triangles different? Look at where the crossings actually sit.
The upper one is three centimetres above the base. The lower one is three below. Not roughly three. Exactly three, both of them. So the base runs exactly halfway between the two corners. Now fold the lower triangle upward, hinging along the base. Q is on the fold line, so Q does not move. R is on it too, so R stays where it is. And the lower corner swings up and lands on the upper corner. Exactly on it.
Two crossings again. But this time the second triangle is the first one turned over. Why did that work? Because of what the fold does to the line we drew. Fold along the base, and every point above swaps with the point directly below it. Now think about that line standing at Q. Where does the fold send it? It is at right angles to the base, so it points straight away from the fold.
Folding sends it straight back onto itself. The line does not move at all. And the arc does not move either, because its centre R is sitting on the fold line. So the whole construction comes back to itself, and the two crossings have to swap. The right angle is not a magic ingredient. It is what makes the fold leave everything alone. Now set that beside the case that failed, and the difference is visible.
There the base was six, the angle was thirty degrees, and the swung side was four. The arc cut at about two and a half, and then again at about seven point eight. Both of those crossings were in front, on the same ray, above the base. Fold along the base and that ray does not come back to itself. It swings away below. So the fold carries the first triangle somewhere new. It does not carry it onto the second.
And you can see that in the numbers. Those two triangles had third sides of two and a half, and seven point eight. Different lengths. Nothing was ever going to match them up. It is worth being exact about what the good case actually needs. It is tempting to say the two crossings just have to be on opposite sides of the base. That sounds right, and it is not enough.
Extend the line far enough in the failing case and you can get crossings on both sides as well. But they sit at different distances out, so folding still does not match them. What is needed is stronger than that. The second crossing has to be the mirror image of the first. The same distance out, straight across the base, directly opposite. And the only angle in the world that arranges that is the right angle.
So the condition works. Now it needs its names. First, that longest side. The one facing the right angle. It has a word of its own. It is called the hypotenuse. And notice that it is defined by where it sits, not by how the triangle is drawn. It is not the bottom side. And it is not always the slanted one. Turn the whole triangle upside down and the hypotenuse is still the very same side.
It is whichever one faces the right angle. That is the only test there is. It is also always the longest of the three, which is going to matter shortly. And now the condition itself gets its name. A right angle. The hypotenuse. And one other side. R H S. Right angle, hypotenuse, side. And that middle letter is doing real work. It is not enough to say a right angle and any two sides you like.
One of the two has to be the one facing the right angle. Give the right angle and the two shorter sides instead, and that is a different set. There the angle sits between the two sides. Here it does not, and this is the one with the name. One more thing, and it is the part that makes this condition comfortable. Back in the failing case, the arc sometimes missed the ray completely.
Sometimes it reached and cut twice, and sometimes it never got there at all. Here it can never miss. And the reason takes one line. The base runs from Q, and the line stands at right angles to it there. So Q is the closest point of that line to R. Nothing on the line is nearer. And the hypotenuse is longer than the base. Always. So the arc always gets past Q.
I checked every whole-number pair of shorter sides up to twelve. It reached every single time. So here is the finished list. Five sets of measurements that settle a triangle. Three sides. Two sides with the angle between them. Two angles with the side between them. Two angles with a side somewhere else. And a right angle with the hypotenuse and one more. Five of them. And now look hard at what is not on that list.
Two sides with an angle beside them is not there. It never was, and it still fails. Three angles is not there either. Those hand you the shape and never the size. And of the five that are there, two are not really separate facts at all. One of them is arithmetic done first. One is the failing case, with its ambiguity folded away.
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
- SAS, and why the angle has to be the included oneClass 7 · Ch 1, Geometric Twins
- SSS: three sidelengths fix a triangle completelyClass 7 · Ch 1, Geometric Twins
Comes up again in
- Angles opposite equal sides are equalClass 7 · Ch 1, Geometric Twins
Either side of this one
- ASA and AAS: two angles are enough to fix the thirdClass 7 · Ch 1, Geometric Twins