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Chapter 7 · Area

Area of a trapezium, derived two different ways

The special quadrilaterals11 min

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11 min.

Half the height times the sum of the parallel sides is not a fourth thing to memorise, and the half is not something to remember.

The idea

The trapezium formula says something simple once you read it as a sentence rather than as symbols: a trapezium fills the same space as a rectangle of the same height whose width is the average of its two parallel sides. Two derivations that share almost nothing arrive at that expression — slice the trapezium into a rectangle with a right triangle at each end, or paste a rotated copy against it to make a parallelogram and take half. Their agreement is the reason to trust the result, and the second one explains the half that the first only produces as algebra. And because setting the two parallel sides equal gives base times height and shrinking one of them to nothing gives half base times height, this one formula quietly contains the parallelogram's and the triangle's.

What you should be able to do

  • Cut a trapezium into a rectangle and two right triangles, and justify that the middle piece really is a rectangle
  • Write the three pieces' areas with letters for the unknown horizontal offsets, and collect the sum
  • Eliminate the two offsets using the relation between the two parallel sides
  • State the formula, and restate it as an average width
  • Complete the two sketched approaches the chapter leaves open for the sheared trapezium, and say when each one works
  • Build a parallelogram from two copies of a trapezium, and say what has to be checked before the figure may be called one
  • Rule out the six-sided alternative using an angle sum, and identify the transversal that makes the argument work
  • Derive the formula a second time from the parallelogram, and account for the half
  • Compute a trapezium's area from two parallel sides and a height, including when the figure is drawn rotated
  • Recover the parallelogram and triangle formulae as special cases

Words to know

TermDefinition in one lineFirst introduced
trapeziuma quadrilateral with one pair of parallel sidesprinted in this chapter, Part II §7.1 under the subheading "Trapezium" (Part II pp.160, 166)
parallel sidesthe pair of opposite sides of a trapezium that never meet; the two lengths the formula addsprinted in this chapter, Part II §7.1 (Part II p.167)
heightthe perpendicular distance between the two parallel sidesprinted in this chapter, Part II §7.1 (Part II pp.161, 167)
transversala line crossing two others, whose co-interior angles the argument uses twiceprinted in this chapter, Part II §7.1 (Part II pp.166, 168)
isosceles trapeziuma trapezium whose two non-parallel sides are equalprinted in this chapter, Part II §7.1 (Part II p.169)
parallelograma quadrilateral whose two pairs of opposite sides are parallel; what two trapezium copies makeprinted in this chapter, Part II §7.1 (Part II pp.161, 167, 168)
letter numbersthe chapter's own name for the letters it puts in place of unknown lengthsprinted in this chapter, Part II §7.1 (Part II p.166)
quadrilaterala four-sided figure; the shape the two rotated copies turn out to makeprinted in this chapter, Part II §7.1 (Part II pp.159, 168)
average widthhalf the sum of the two parallel sides, read as the width of an equivalent rectanglean added term, and the reading that makes the formula memorable; not printed
midlinethe segment halfway up a trapezium, whose length is that averagean added term; the chapter draws no such segment

Where people slip up

  • "The height is one of the slanted sides." It is the perpendicular distance between the two parallel sides, and in the sheared figure on Part II p.167 it is drawn entirely outside the shape.
  • "Add the parallel sides and multiply by the height." The half is missing, and this is the most common error on the topic by a wide margin. The average-width reading is the fix: half the sum is a width, and widths get multiplied by heights.
  • "The formula only works for an isosceles trapezium." The derivation on Part II pp.166–167 assumes nothing of the kind, and the chapter names the isosceles case separately on Part II p.169 precisely because it is special.
  • "It fails when the perpendicular lands outside the base." Part II p.167 exists to settle exactly this, with two approaches. The formula does not change; the picture does.
  • "Two copies make twice the area, so the trapezium must be twice the parallelogram." It is half. Students invert this reliably; show the two areas together.
  • "The two rotated copies obviously make a parallelogram." The chapter spends most of a page ruling out the six-sided alternative with an angle sum. That argument is the content of section 11, not an aside — and it is the part a student can be examined on.
  • **"a and b are the two longest sides."** They are the parallel sides, whichever they are. In the rotated figures on Part II p.169 they are not even horizontal.
  • **"The two offsets x and y have to be measured."** Only their total is needed, and the bottom side supplies it. That is the reason the formula has three inputs.
  • "A trapezium formula is a separate thing to memorise." Set the two parallel sides equal and it is the parallelogram formula; let one shrink to nothing and it is the triangle formula. Ending on that is worth more than the derivation.
Transcript1,449 words

Here is a trapezium. One pair of its sides is parallel and the other is not. The parallel ones are 6 and 27, and they are 12 apart. Notice what is not true here. Its slanted sides are 13 and 20, so it is not symmetric - and neither of them is the height. The height is not a side at all: it is the distance across, from one parallel side to the other.

We are not going to reach for a formula. We are going to cut this into things we already handle. Drop a perpendicular from each end of the short side down to the long one. A triangle, a rectangle, and another triangle. One thing has to be checked first, because the whole calculation rests on it. Is that middle piece really a rectangle? Two of its corners are right angles by construction - we drew those lines perpendicular.

The other two have to be argued. They are right angles because the top side is parallel to the bottom one: each perpendicular cuts across both parallels, and the two angles it makes on the same side of it total a straight angle. One is a right angle, so the other one is too. Four right angles. It is a rectangle: both horizontal sides 6, both uprights 12. And the two end pieces are right-angled triangles, square at the foot of each perpendicular and nowhere else.

Each triangle stands on a stub of the bottom side, and those stubs are what we do not know. Call them x and y - here, 5 and 16. The first triangle is half of 5 times 12, which is 30. The rectangle is 6 times 12, which is 72. The last triangle is half of 16 times 12, which is 96. Thirty, seventy-two and ninety-six - together, 198, which is exactly what the four corners of the whole shape say it holds.

But an answer that needs x and y is not much of an answer, because nobody hands you those. Write the three areas with the letters left in. Half h x, plus h a, plus half h y. Every term has an h, so pull it out: what is left inside is half of x, plus a, plus half of y. And here is the whole trick. The x and the y never appear apart. Only their total does.

And their total is sitting on the bottom side: the bottom is the top plus the two stubs, so the stubs together are the bottom minus the top. Here that is 5 plus 16, which is 21, which is 27 minus 6. You never measure either one. That is why this shape needs three measurements, not five. Put the total back in: half h, times b minus a, plus 2 a. The minus a and one of the two a's cancel.

Half the height, times the sum of the two parallel sides. Twelve halved is 6. Six plus twenty-seven is thirty-three. And six times thirty-three is one hundred and ninety-eight - the answer the pieces gave. Now, the half. It is the most-dropped symbol in this subject. Leave it out and you get three hundred and ninety-six. That is not slightly wrong. It is exactly twice the shape, every time. So let us make the half impossible to forget.

Half of the sum of the two parallel sides is 16.5. Do not read that as half of an area. Read it as a length. It is a length you can point to. Go halfway up and draw a line straight across, from the middle of one slanted side to the middle of the other. That segment is 16.5 long, and parallel to both. Longer than the short side, shorter than the long one. It is the average width - the width this shape would have if it did not taper.

So the formula is not a formula at all. It is the rectangle rule: width times height. Sixteen point five by twelve. And you can build that rectangle. Swing the triangle below the halfway mark on the left outward, and it lands exactly on the gap above it. The same on the right - and those two are not the same triangle, so both swings have to be checked. Average width times height. Nobody reading it that way loses the half.

Now a shape that breaks the picture we have been drawing. The top side is pushed so far over that the perpendicular from its left end lands outside the bottom side altogether. The height has to be drawn out in space, beside the shape rather than inside it. It is still a trapezium: parallel sides 6 and 16, and 8 apart. Does the formula still hold? Half of 8 is 4, times 22, is 88. And the four corners say 88.

It holds. But the picture that produced it has fallen apart, so we owe ourselves a reason. There are two, and both are worth having. The first keeps the three pieces and changes their signs. Drop both perpendiculars anyway and take the rectangle between the two feet: 6 by 8, holding 48. At the left end the triangle between rectangle and shape lies outside the shape, so it comes off: 20. At the right end it lies inside, so it goes on: 60.

Forty-eight, less twenty, plus sixty. Eighty-eight. And the two triangles are not equal, so those signs are doing real work. On the ordinary trapezium the very same three pieces are all added, because both feet land on the bottom side. That is the whole difference: a sign flipped. Not the formula. The second draws one line: from the top right corner, parallel to the left slanted side, down to the bottom.

What it cuts off is a parallelogram - top and bottom already parallel, and the two sides parallel because we drew one that way. So it holds base times height: 6 times 8, which is 48. What is left is a triangle standing on the difference of the two parallel sides, 16 minus 6, which is 10. Half of 10 times 8 is 40. Forty-eight and forty. Eighty-eight again, from a different cut.

One warning: if the two parallel sides are equal, the shape is already a parallelogram and there is no triangle left behind. The approach is fine, but that case needs checking. There is a third way, and it explains the half rather than producing it. Take a second copy. Turn it through a straight angle about the middle of one slanted side, and slide it against the original. What you get is a parallelogram.

Its base is the two parallel sides laid end to end: 6 and 27, so 33. Its height is 12. It holds 396. And the trapezium is half of it: one hundred and ninety-eight. There is the half, and it is not algebra any more. You built two trapeziums' worth of area on purpose, and gave half of it back. But why is that a parallelogram? Why is it even four-sided?

Join two copies of a shape along a side and what you normally get is six-sided, with a corner sticking out at each end of the join. Watch a quadrilateral with no parallel sides. Turn a copy, join it, and the area doubles just the same - but the corners stay corners. Six of them. The trapezium's straighten out instead. At each end of the join two angles meet, and they are the two that sit on the same side of a line crossing the parallel pair. They total a straight angle, so the two edges become one.

That is where the parallel sides get used, and it is the only place they are used. Without it, the parallelogram is just something the picture looked like. Four to try. Two are drawn turned, which changes nothing - the parallel sides are whichever ones are parallel, and they need not be horizontal. Ten and seven, 16 apart: 136. Twenty-four and thirty-six, 14 apart: 420. Fourteen and six, 10 apart: 100. And twelve and eighteen, 8 apart: 120.

It runs backwards too. Given 420, a height of 14 and one side of 24, the other is 36. And now the last thing, worth more than the derivation. Set the two parallel sides equal - both 9 - and it gives 9 times the height. That is the parallelogram rule. Let one shrink away to nothing, and it gives half the base times the height. That is the triangle rule.

So this is not a fourth thing to memorise. It is the one that contains the other two.

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

Either side of this one

The book

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