PrepShorts · Study sheet · Class 8 Mathematics · Chapter 7, Area
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Both diagonals of a rectangle cut it into four triangles that plainly do not match — two wide and low, two narrow and tall.
The idea
The area formula can only see a base and a perpendicular distance, so anything that leaves those two numbers alone leaves the area alone. Slide a triangle's apex along a line parallel to its base and the area does not move at all — while the perimeter changes the whole time. That single observation does an unreasonable amount of work: it explains why four visibly unequal triangles cut from a rectangle by its diagonals have equal areas, why a line from a vertex to the midpoint of the opposite side always halves a triangle, and why the midpoint triangle is exactly a quarter. And because the perimeter does move, the same picture supports a genuinely different question — which apex gives the shortest perimeter — whose answer comes not from measuring but from a mirror.
What you should be able to do
- State the condition under which two triangles must have equal areas, and distinguish it from congruence
- Show that the four triangles cut from a rectangle by its two diagonals have equal areas, naming the base-height pair used for each comparison
- Prove that a line from a vertex to the midpoint of the opposite side splits a triangle into two of equal area
- Given a family of triangles on one base with the apex on a fixed parallel line, say which has the greatest and which the least area, and justify the answer
- Say which member of that family has the least perimeter, and which has the greatest, and explain why one of these questions has no answer
- Reproduce the reflection argument for the minimum-perimeter triangle, naming the congruence that makes the two path lengths equal
- Confirm that the minimising apex lies on the perpendicular bisector of the base
- Compute an area or a fraction of an area using the equal-halves result rather than by measuring anything new
- Apply the reflection argument to a route problem stated outside geometry
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| altitude | the perpendicular segment from a vertex to the opposite side, whose length is the height | printed in this chapter, Part II §7.1 (Part II pp.153, 155, 158) |
| congruent | able to be laid exactly on top of one another; a stronger condition than equal area | printed in this chapter, Part II §7.1 (Part II pp.150, 155, 161) |
| midpoint | the point that divides a segment into two equal parts | printed in this chapter, Part II §7.1 (Part II pp.155, 159, 170) |
| perpendicular bisector | the line at right angles to a segment through its midpoint | printed in this chapter, Part II §7.1 (Part II pp.156, 157) |
| reflection | the image of a point across a line, at equal distance on the other side | printed in this chapter, Part II §7.1 (Part II p.157) |
| isosceles | having two sides of equal length | printed in this chapter, Part II §7.1 (Part II pp.158, 164) |
| median | the segment from a vertex to the midpoint of the opposite side | not printed in this chapter, which states the result about that segment without giving it a name; the word does occur elsewhere in this book, but only as the statistical median in Part II printed Chapter 5 |
| equal-area family | all the triangles sharing one base with their apex on one line parallel to it | an added term; the chapter draws the family on Part II p.156 and does not name it |
| sliding the apex | moving a triangle's third vertex along a parallel to the base, which changes the perimeter and not the area | an added phrasing; not printed |
Where people slip up
- "Equal area means congruent." The chapter opens this subsection by saying the four triangles are not congruent, and then proves they are equal. Getting students to hold those two ideas apart is most of the work of the explanation.
- "The tallest-looking triangle in the family has the most area." They all have the same area. The apparent height a student sees is a slant length, not the perpendicular distance.
- "If the area does not change, nothing changes." The perimeter changes continuously, and has a minimum. Two measures, one picture.
- "A median divides a triangle into two congruent halves." Equal in area, not congruent — except in the isosceles case, where it happens to be both and therefore teaches the wrong lesson if it is met first.
- "Any line from a vertex halves the triangle." Only the one to the midpoint. Show a near-miss line and the unequal halves it leaves.
- "The minimum-perimeter apex is found by trying positions and measuring." It is found by reflecting, and the reflection converts a bent path into a straight one. That conversion is the idea worth remembering.
- "The shortest way to the river is straight down." Only if you are not going on to the tank afterwards. The second destination is what makes it a reflection problem.
- "Both parts of question (ii) have answers." The least perimeter exists; the greatest does not. A question can be well posed and still have no answer, and saying so is honest mathematics rather than a dodge.
- "The tick marks on the Part II p.157 figures are decoration." They mark the equal distances that make the perpendicular-bisector conclusion work, and they are the chapter's way of pointing at the answer to its own Math Talk.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 2 Q3, Figure it Out · 2 Q7, Figure it Out · 2 Q8, Figure it Out · 5 Q8
Transcript1,338 words
Here is a rectangle, eight across and six up, holding forty-eight square units. Draw both of its diagonals. They cut it into four triangles. Are those four equal? They certainly do not look it. Two of them are wide and low; the other two are narrow and tall. And yet each one holds twelve, and four twelves are forty-eight. Before we see why, notice the question people reach for first.
Are the four congruent - could you lift one and lay it exactly on another? No. The wide ones have sides five, five and eight. The tall ones have five, five and six. Only the opposite pairs match; the neighbours do not. So congruence is not going to settle this, and that is worth knowing early, because equal area is a WEAKER claim than congruence, and weaker claims are easier to prove.
Take two neighbouring triangles and use the halves of one diagonal as their bases. A rectangle's diagonals cut each other in half, so those two bases are the same length. Now drop a perpendicular from the corner they share, down to that diagonal. This is the step nobody thinks of: that single perpendicular is the height of BOTH triangles at once. One line, doing two jobs. Same base, same height, so the same area. Run the argument round the figure and all four fall into line.
Strip that down to what actually did the work and you get a rule about any triangle at all. Join a corner to the midpoint of the opposite side. The two pieces have equal bases, because it is a midpoint, and they share the height, because they share the apex. So the segment halves the triangle. Twenty-four becomes twelve and twelve. And look what is NOT true. The two halves are not congruent - their third sides are different lengths.
You could not lay one on the other. They are still equal in area, and that is the whole point: the area formula cannot see anything except a base and a perpendicular distance. One warning, and then a payoff. Sometimes the two halves ARE congruent. Take an isosceles triangle and drop the perpendicular from its apex: it lands exactly on the midpoint, so if one half holds twenty-four, the whole holds forty-eight.
That is a special case, not the reason. The reason is still base and height. Now the payoff. Take any triangle and mark the midpoints of two of its sides. Join one of those midpoints to the far corner. That halves the triangle: twenty-four goes to twelve. Now inside that half, the other midpoint halves it again: twelve goes to six. So the little triangle in the corner, sitting on two half-sides, is a quarter of the original. Nothing was measured. The rule was applied twice.
Here is the same idea wearing a different hat. A trapezium: two parallel sides, six and ten, four apart, holding thirty-two. Mark the midpoint of one of the slanted sides, and draw a line from the opposite corner through it, carrying on until it meets the base line. That swings a triangle off one side and drops an identical one on the other, because the midpoint gives equal bases and the parallel gives equal heights.
What is left is a triangle. Its base is sixteen - which is six and ten laid end to end - and it holds thirty-two, exactly what the trapezium held. Now let us push the idea as far as it goes. Draw a base, eight long. Draw a line parallel to it, three above. Now put the apex anywhere at all on that parallel, and keep putting it somewhere else.
Every one of these triangles has the same base, and every one has the same height, because two parallel lines stay the same distance apart forever. So every one of them holds twelve. Twelve, twelve, twelve, twelve, twelve. The shapes really do differ - and where two of them match, it is only because one apex is the mirror of the other. So: which of them has the greatest area, and which the least?
It is tempting to say the question has no answer. That is not right, and the difference matters. Every one of them is a greatest. Every one is a least. The question has an answer; what it does not have is a unique one. That is a completely different situation from a question where nothing wins - and we are about to meet one of those, so it is worth keeping the two apart.
Because now ask the other question about the very same picture. Which one has the shortest way round? Suddenly the family is not all alike. The base never changes, so only the two slanted sides matter, and those clearly do change as the apex slides. Intuition says the winner is the apex directly above the middle. But intuition is a guess, and guessing is not why we are here. And in the other direction there really is nothing. Slide the apex far enough and the two sides grow past any length you care to name. There is no longest.
So one half of that question has an answer and the other half has none at all. Here is the trick, and it is a beautiful one. Treat the parallel line as a mirror, and reflect the right-hand end of the base up through it. It was three below the line, so its image is three above. Now take any apex on the line, and look at the two distances: from the apex down to that end, and from the apex up to its image.
They are the same. The mirror line cuts the join between them at right angles and in half, so the two triangles formed are congruent, and the two distances have to match. So the journey from the left end, up to the apex, and back down to the right end is exactly as long as the journey from the left end, up to the apex, and on to the IMAGE. Wherever the apex sits.
And that second journey is one we can settle immediately. It starts at a fixed point and ends at a fixed point, and it is allowed to bend anywhere on the line. The shortest route between two points is the straight one. So stop bending it: draw the segment from the left end of the base straight to the image, and let the apex be wherever that segment crosses the line.
That segment is ten long. Add the base, eight, and the shortest way round is eighteen. And where did the winning apex land? Directly above the middle of the base - so the guess was right, and now it is not a guess. Its two sides come out five and five. The winner is the isosceles one, and every other member of the family goes further round. Now watch the triangle disappear.
Gopal is at his house. He has to fetch water from the river, and then carry it to a tank. Both the house and the tank are on the same side of the river. Where should he meet the water? There is no triangle here, no base, no apex, no area. But the shape of the problem is identical: two fixed points, one line to touch, and a bent path to make short.
So do the same thing. Reflect the tank across the river. Draw the straight line from the house to that image. Where it crosses the bank is where Gopal should go - five and five eighths of the way along. Any other point on the bank makes the walk longer, and we have not measured a single route to know that. That is what the whole video has been about. Sliding the apex leaves the area alone, because the formula cannot see anything else. Moving it changes the way round, because the way round can see everything.
One picture, two questions, and completely different answers.
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
- Why the area of a triangle is half base times heightClass 8 · Ch 7, Area
Comes up again in
- Any polygon is a pile of trianglesClass 8 · Ch 7, Area