PrepShorts · Teaching notes · Class 8 Mathematics · Chapter 7, AreaPrepShorts

Chapter 7 · Area

Area of a trapezium, derived two different ways

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What to assume they know

What they 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

Where it usually goes wrong

  • "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.

Questions to check understanding

  • Find a trapezium's area from its two parallel sides and its height, with the figure drawn upright and then drawn rotated
  • Given the area, the height and one parallel side, find the other parallel side
  • Given the area and both parallel sides, find the height
  • Identify which two sides of a drawn quadrilateral are the parallel pair, and which marked length is the height
  • Complete one of the two sketched approaches for the sheared trapezium
  • Show that two copies of a trapezium make a parallelogram, including the step that rules out the six-sided figure
  • Construct a trapezium of a stated area, and give a second one with the same area
  • Derive the parallelogram and triangle formulae from the trapezium formula as special cases — the reasoning item that shows the chapter has been understood as one argument rather than four

Examples worth working on the board

Items marked printed are stated or worked on the page; items marked not in the book are an added argument or arithmetic on the chapter's inputs and must not be presented as something the chapter states.

  • Three trapeziums on grid paper (Part II p.166, top). Printed: the instruction is to get each area by cutting the shape into parts the student can already handle, and three trapeziums are drawn on a square grid.
    • The first is lettered A and B along the top and D, M, C along the bottom, with A directly above D and B directly above M — so AD is a vertical side, AB and DC are the parallel pair, and BC is the single slant, with ∆BMC as the only triangle.
    • The second has S, T, U, R along the bottom and P above T, Q above U, so it splits into ∆PST, rectangle PTUQ and ∆QUR.
    • The third has Z, M, N, Y along the bottom and W above M, X above N — the same structure again, and it is the figure the chapter immediately reuses for its derivation. The grid counts are only partly transcribed. The first one is now measured on the printed page: AB = 4, DC = 6 and the height 4 grid units, with M sitting 4 units from D — so its area is 20 square units, which the decomposition confirms as rectangle ABMD = 16 plus ∆BMC = 4. Treat that as a lead for a reviewer with the printed book rather than as a settled transcription, and note that the other two were not measured; see the Notes.
  • The derivation (Part II p.166). Trapezium WXYZ with WX parallel to ZY, W and X on top, Z and Y at the bottom, and M and N the feet of the perpendiculars dropped from W and from X onto ZY. Printed: the construction of those two perpendiculars in the chapter's short notation; the question of whether WXNM is a rectangle; and the answer — the two angles at W and X are right angles, because WX is parallel to ZY and the co-interior angles at each of the two perpendiculars, taken as transversals, total two right angles. Printed: the area is then the sum of ∆WMZ, rectangle WXNM and ∆XNY, written out as half of MZ times WM, plus WX times WM, plus half of NY times XN.
  • The letters (Part II p.166). Printed: the chapter says it will assign letter numbers to the lengths it needs, and sets MZ = x, WM = XN = h, WX = a and NY = y. The printed chain then runs: ½hx + ha + ½hy, then h(½x + a + ½y), then h times (x + y + 2a) over 2, then ½h(x + y + 2a). The second figure on the page carries a along the top, h twice as the two verticals, x and y as the two base offsets and b as the whole bottom side.
  • Removing the offsets (Part II p.167). Printed: the chapter names the other parallel side b, asks whether the area can be written using only a, b and h, and observes that b = x + y + a; subtracting a gives x + y = b − a. Substituting leaves ½h(b − a + 2a), which is ½h(a + b). Printed boxed result: a trapezium's area is half its height times the sum of its parallel sides. Not in the book: the elimination is the whole trick, and it is worth naming what happened — the two offsets never had to be known separately, only their total, and the bottom side already tells you their total. That is why the formula needs three measurements rather than five.
  • The average-width reading (not in the book; the chapter does not offer it). ½(a + b) is a length: it is what the trapezium's width would be if it did not taper, and it is the length of the segment halfway up. So the formula is the rectangle formula applied to that width. A student who remembers "average width times height" will not lose the half, which is the single most common slip on this formula.
  • The sheared trapezium (Part II p.167, third figure). A trapezium drawn so that the shorter parallel side a on top is offset to the left of the bottom side b — equivalently b is shifted right — far enough that the perpendicular dropped from a's left end lands outside b, to its left; F is that foot, and the height h is drawn dashed outside the figure at the left. (Offset to the right would mirror the figure, put F inside b, and leave Approach 1's printed lines making no sense.) Printed: the chapter asks whether the formula still holds and says there are different ways to approach it, then sketches two and instructs the student to complete the arguments.
  • Approach 1 (Part II p.167, printed heading "Approach 1: Rectangle and Triangles"). Trapezium ABCD with A and B on top, F, D on the bottom left and C on the bottom right, E on the bottom line inside; a labels AB, b labels the bottom and h the height at the left. Printed are exactly two lines: the area of ABCD is the area of ABED plus the area of ∆BEC, and the area of ABED is the area of ABEF minus the area of ∆AFD. Not in the book completion: ABEF is a rectangle of height h; ∆AFD and ∆BEC are the two end triangles; the horizontal offsets FD and EC enter once each with opposite signs and combine with a to give b, exactly as they did on Part II p.166, so the total is again ½h(a + b).
  • Approach 2 (Part II p.167, printed heading "Approach 2: Parallelogram and Triangle"). The same trapezium with G marked on the bottom side between D and C, and the printed instruction to draw BG parallel to AD. A Math Talk then asks whether this approach works for every type of trapezium. Not in the book completion: ABGD is a parallelogram, because AB is parallel to DG by construction of the trapezium and BG is parallel to AD by construction, so its area is a × h; and ∆BGC has base GC = b − a and height h, so its area is ½h(b − a). Adding gives ah + ½hb − ½ha, which is ½h(a + b). Not in the book answer to the Math Talk: it works whenever the shorter parallel side is the top one, so that G lands between D and C. If the two parallel sides are equal the figure is already a parallelogram and there is no triangle left; if the longer side is on top, draw the parallel from the other figure instead. The approach is fine, but it needs a case check, and that is what the Math Talk is asking for.
  • Two copies (Part II p.168, under the printed subheading "Finding the Area Using Two Copies of the Trapezium"). Printed: take two copies of a trapezium in which AB is parallel to CD, rotate the second, and join them along BC. The first figure shows the two copies side by side with the angles at B and C marked x and y in the first copy and the rotated copy's corresponding angles marked on the other side. Printed question: what figure results, with the answer that it is either six-sided or four-sided, and both possibilities drawn under the printed heading "Possibilities".
  • Ruling out the six-sided case (Part II p.168). Printed: because AB is parallel to CD, the two marked angles x and y total a straight angle, being co-interior along the transversal BC; so the two joins straighten out into single lines and the result is a quadrilateral. Printed: the chapter then marks the trapezium's other two angles u and v, notes that they also total a straight angle, concludes that one further pair of sides is parallel, and — since the other pair was already parallel — that the quadrilateral is a parallelogram. Three small figures at the foot of Part II p.168 show the trapezium with a, h and b marked, and then the parallelogram with a and b laid end to end along one edge.
  • The second formula (Part II p.169, top). Printed: the trapezium is half the parallelogram, which is ½h(a + b). Not in the book: this is where the half stops being algebra. You built two trapeziums' worth of area on purpose and then gave half of it back; the parallelogram's base is a + b precisely because the two parallel sides ended up end to end. Say that, and the formula is unforgettable.
  • Four figures to measure (Part II p.169, Figure it Out 3). All four are trapeziums and two of them are drawn rotated, which is the point of the set.
    • (i) drawn tilted; the two parallel sides are marked 10 ft and 7 ft, and the perpendicular distance between them is marked 16 ft as a dashed segment with a right angle at its upper end. Not in the book: 136 ft².
    • (ii) upright; parallel sides 24 m on top and 36 m at the bottom, height 14 m dashed inside near the right. Not in the book: 420 m².
    • (iii) drawn rotated a quarter turn, so that the parallel sides are the left side, marked 14 in, and the right side, marked 6 in, with the perpendicular distance between them marked 10 in as a dashed horizontal segment. Not in the book: 100 in².
    • (iv) upright; parallel sides 12 ft on top and 18 ft at the bottom, height 8 ft dashed inside near the left. Not in the book: 120 ft².
  • Trapezium into rectangle (Part II p.169, Figure it Out 4 and 5). Printed: item 4, a Śulba-Sūtras problem, asks for a method converting an isosceles trapezium into a rectangle by dissection. Printed: item 5, a Math Talk, exhibits one route from a trapezium ABCD to an equal-area rectangle EFGH. Measured on the printed page: H lies on line AB extended beyond A, and E on line AB extended beyond B — both outside the trapezium, not inside A and B. So the rectangle's top side HE is longer than AB and shorter than DC, its length being exactly (AB + DC)/2; on the printed page AB = 713 px, HE = 940 px and DC = 1180 px, and (713 + 1180)/2 = 946, which matches HE to within the line width. That has to be so — the brief's own completion makes the rectangle's width the average width, and an average of a and b exceeds a. I and J are drawn as dots at the marked midpoints of the slant sides AD and BC, each carrying tick marks on both halves; that is the fact the printed hint (∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ) rests on. HG and EF are the dashed verticals through I and J, with right angles marked at H, E, G and F; D and G sit at the bottom left, F and C at the bottom right. Not in the book: the construction swings ∆DGI — which is inside the trapezium and outside the rectangle — outward onto ∆AHI, and ∆CFJ outward onto ∆BEJ. It does not fold the slants inward. The rectangle's height is the trapezium's height and its width is the average width of section 6, which is why the construction exists at all.
  • A trapezium of a given area (Part II p.170, Figure it Out 6, a Math Talk). Printed: the student is told to use the rectangle conversion in both directions and to build a trapezium of area 144 cm². Not in the book: any a, b and h with h(a + b) = 288 works — for instance h = 12 cm with parallel sides 10 cm and 14 cm — and the fact that infinitely many answers exist is the interesting part of the item.
  • Two items printed here that belong elsewhere. Part II p.170 item 7, the regular hexagon in three pieces, is carried by Any polygon is a pile of triangles; Part II p.170 item 8, the trapezium shown equal in area to ∆ZWB, is carried by Triangles with the same base and height have the same area. Both are printed inside this exercise set, so anyone working from the page will meet them here.

Figures to have open

  • The three grid trapeziums (Part II p.166, top) with their vertex letters, redrawn on a visible grid. The grid counts must be read off a printed copy.
  • **WXYZ with the two perpendiculars, and the same figure lettered with x, a, y, b and the two hs** (Part II p.166). Both states are needed; the second is the one the algebra refers to.
  • The midline drawn on a trapezium with the equivalent rectangle behind it. Not in the chapter; this is the topic's key visual and the whole of section 6.
  • The sheared trapezium with the external height (Part II p.167), then the two approaches drawn on it (Part II p.167, both labelled figures). The chapter's own auxiliary points F, E and G must be kept, since the printed argument lines name them.
  • The two-copies construction (Part II p.168), including both drawn possibilities. The six-sided figure has to be shown before it is ruled out, or section 11 is arguing against nothing.
  • The angle-marking figures (Part II p.168), with x, y, u and v in place. The angles are marked inside the artwork on the printed page and are easy to lose in a redraw.
  • The four exercise trapeziums (Part II p.169, item 3) in their printed orientations. Items (i) and (iii) must stay rotated.
  • The trapezium-to-rectangle construction EFGH (Part II p.169, item 5) with all eight labelled points.
  • No photograph is needed.

Where this sits in the book

The book

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