Any polygon is a pile of triangles

What to assume they know

• Half base times height, with any side available as the base — Why the area of a triangle is half base times height
• That triangles on the same base with the same height have equal areas — Triangles with the same base and height have the same area
• Area as a count of unit squares, and that the areas of the pieces total the whole — Area as a count of unit squares
• The area of a rectangle, and subtraction of areas
• What makes a polygon regular, and that a regular hexagon's six corners lie at equal distances from its centre

What they should be able to do

• State the two facts that together make every polygon computable, and say which of them the chapter proves and which it asserts
• Cut a given quadrilateral into two triangles and list exactly which measurements are then needed
• Show that a quadrilateral's area is half its diagonal times the sum of the two perpendicular distances to that diagonal, and compute it
• Cut a pentagon and a hexagon into triangles, and count how many triangles each needs
• Find the area of a shaded region by subtracting corner triangles from a rectangle
• Explain why a decomposition can be chosen for symmetry, and what one length settles about a regular hexagon
• Produce a decomposition whose answer does not depend on where a free point was drawn, and say why
• Describe a construction that halves a given quadrilateral's area
• Express the areas of the pieces of a regular hexagon as a ratio, using the small equilateral triangle as the unit

Where it usually goes wrong

• "Only nice polygons can be cut into triangles." Every polygon can. The instruction "join a diagonal" is what fails on an inward-pointing corner, not the splitting itself — and showing the failing diagonal is more instructive than asserting the rule.
• "You need every side length to find a quadrilateral's area." The Part II p.160 item gives none of the four sides. A diagonal and two perpendicular distances are enough, and that is three numbers rather than five.
• "Four sides determine a quadrilateral's area." They do not. A four-sided frame with fixed sides can be flexed, and its area changes as it flexes while every side stays the same length. This is the quadrilateral version of the point Why perimeter cannot stand in for area made about perimeter.
• "Half the product of the diagonals gives any quadrilateral's area." Only when the diagonals cross at right angles. Area of a rhombus from its diagonals is where that condition gets stated properly.
• "Decomposition means adding pieces." Subtracting corner triangles from a rectangle is a decomposition too, and on Part II p.160's second item it is by far the shorter route.
• "The answer must depend on where the free point was drawn." The fourth item on Part II p.160 is built so that it does not. Move the point and let the total refuse to change.
• "The special formulae later in the chapter are needed." They are shortcuts for shapes that turn up often. Say so before deriving them, or students will think a trapezium is a shape that cannot be handled any other way.
• "More triangles means a better cut." Fewer measurements means a better cut. The hexagon question is about exactly that, and its answer is one length.

Questions to check understanding

• Find the area of a quadrilateral from one diagonal and the two perpendicular distances to it
• State which measurements you would ask for, given an unlabelled quadrilateral or pentagon, and how many
• Find the area of a shaded region inside a rectangle by subtracting corner triangles
• Cut a given polygon into the fewest triangles, and say how many an n-sided polygon needs
• Given a figure with a free interior point, show that a stated area does not depend on where the point lies
• Find the area of one piece of a figure from the area of another, when both are built from the same square or the same small triangle
• Give the ratio of the areas into which a regular hexagon has been divided
• Describe a construction halving a given quadrilateral's area, and justify it from the median result printed on Part II p.155 (the APCQ construction) or from the quarter-triangles argument (the midpoints-of-all-four-sides parallelogram) — the open-ended item this page is built for

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.

• The nameless quadrilateral (Part II p.159, under the subheading "Area of any Polygon"). A quadrilateral ABCD drawn with A at the upper left, B at the upper right, C at the lower right and D at the lower left; it is deliberately not any special kind, and no measurement is marked on it at all. Printed: the two questions asked are how to find its area and what measurements are needed for that; the answer given is that joining BD splits it into two triangles whose areas can each be found, and hence so can the whole.
• The pentagon (Part II p.159). Printed: a plain irregular pentagon, outline only, with no vertex letters and no measurements — and the single question of how to find its area. Nothing is drawn inside it, so the cut is entirely left to the student. Not in the book: three triangles suffice, from any one corner; and in general an n-sided polygon needs n − 2.
• The general claim (Part II p.159). Printed: the chapter asks whether every polygon can be split into triangles, answers that it can, and concludes that knowing how to compute a triangle's area is therefore enough to compute any polygon's. Not in the book: the chapter offers no proof of the splitting claim, and does not need to at this class — but it is worth saying out loud that the claim is being asserted rather than shown, because a student who is told everything in the chapter has been proved will not know which parts to be careful about.
• The careless cut (not in the book; the chapter raises none of this). Draw a quadrilateral with one corner pushed inward so that it points back into the shape. One of its two diagonals lies wholly inside and cuts it into two triangles; the other lies partly outside and cuts nothing. So "join a diagonal" is not quite a recipe — for a shape with an inward-pointing corner, only one of the two diagonals works. The splitting claim survives; the naive instruction does not.
• A quadrilateral with numbers (Part II p.160, Figure it Out 1). Quadrilateral ABCD is drawn long and thin, with the diagonal AC running across it; B sits above AC and D below. The printed data: AC is 22 cm; BM is 3 cm and perpendicular to AC, with M on AC near C; DN is 3 cm and perpendicular to AC, with N on AC near A. Not in the book: ½ × 22 × 3 for the upper triangle plus ½ × 22 × 3 for the lower gives 33 + 33 = 66 cm² — and the tidier way to see it is ½ × 22 × (3 + 3), which is section 3's formula. None of the four sides, and no angle. Three numbers were enough.
• A shaded region by subtraction (Part II p.160, Figure it Out 2). ABCD is a rectangle with A at the top left, B at the top right, C at the bottom right and D at the bottom left. E lies on AB with AE 10 cm and EB 8 cm; the whole top edge is labelled and the bottom edge DC is marked 18 cm. F lies on AD with AF 6 cm and FD 4 cm, and the right side BC is marked 10 cm. The hatched region is the quadrilateral with corners D, F, E and C. Not in the book: the rectangle is 18 × 10 = 180 cm²; the corner triangle AFE is ½ × 6 × 10 = 30 cm²; the corner triangle EBC is ½ × 8 × 10 = 40 cm²; so the hatched region is 180 − 30 − 40 = 110 cm². Note that the item hands over redundant data on purpose — AE with EB, and AF with FD — so that the student has to work out which numbers the chosen decomposition needs.
• The regular hexagon question (Part II p.160, Figure it Out 3, a Math Talk). Printed: the item asks which lengths a regular hexagon's area would require you to know. Not in the book: if you may use the word regular, one sidelength is enough, because the six segments from the centre to the corners cut the hexagon into six equilateral triangles of that side, and the perpendicular distance from the centre to a side follows from it — and that is the answer the item wants. Drop the word regular and one length settles nothing at all: an irregular hexagon has no centre, so there is no perpendicular distance to measure, and the half-perimeter-times-apothem formula needs an inscribed circle and the whole perimeter, not one side. To get an irregular hexagon's area you need a triangulation, or all six sides together with the angles between them. The question is a measurement-cost question, which is why it sits under Math Talk rather than in the numbered work.
• The answer that does not depend on the drawing (Part II p.160, Figure it Out 4, a Math Talk). Inside a rectangle, two blue triangles: one stands on the whole top edge and one on the whole bottom edge, and their apexes meet at a single point inside the rectangle, drawn right of centre and slightly below the middle. The question is what fraction of the rectangle is blue. Not in the book: call the rectangle's width w and its height H, and let the meeting point sit at heights h₁ below the top and h₂ above the bottom. The two triangles come to ½·w·h₁ and ½·w·h₂, and h₁ + h₂ = H, so together they are ½·w·H — exactly half the rectangle, wherever the point was drawn. This is the best item on the page, because the free point is visibly off-centre and the answer is still a clean one half.
• Three identical squares (Part II p.158, Figure it Out 6, part Math Talk). Three equal squares ABCD, BCEF and BFGH are set in an L, with D at the bottom left, C bottom middle, E bottom right, A middle left, B middle, F middle right, H top middle and G top right. A segment runs from D up to H. The red region is bounded by D, H and C; the blue region is the small triangle cut off at the top left of square ABCD by that segment. Part (i) gives the red region as 49 sq. units and asks for the blue; part (ii) gives red and blue together as 180 sq. units and asks for the area of one square. Not in the book: with side s, the red region is a triangle on the vertical segment from H down to C, of length 2s, standing off from D by a horizontal distance s, so its area is s²; the segment from D to H crosses AB at its midpoint, so the blue triangle has base half of s and height s, giving s²/4. Hence (i) 49 gives s² = 49 and a blue region of 12.25 sq. units; and (ii) the two together are 1.25·s², so 180 ÷ 1.25 = 144 sq. units for each square. The second part coming out as a perfect square is the check that the reading of the figure is right.
• Halving a quadrilateral (Part II p.160, Figure it Out 5, a Math Talk). Printed: the item asks for a construction producing a second quadrilateral with half the area of one you are handed. Two constructions that work, both built from this chapter's own tools. Not in the book: (a) Preferred, because it is the chapter's own Part II p.155 result. Draw the diagonal AC. Let P be the midpoint of AB and Q the midpoint of CD. Then the quadrilateral APCQ has exactly half the area of ABCD: CP is a median of ∆ABC so [APC] = ½[ABC], and AQ is a median of ∆ACD so [ACQ] = ½[ACD]; adding gives ½[ABCD]. The median result — a segment from a vertex to the midpoint of the opposite side splits a triangle into two of equal area — is printed on Part II p.155, so the justification is entirely inside the chapter. (b) Join the midpoints of all four sides (the Varignon parallelogram). Each of the four corner triangles cut off is a quarter of one of the four diagonal-triangles, and ¼[ABC] + ¼[ACD] + ¼[ABD] + ¼[BCD] = ½[ABCD]. Do not offer "join the midpoints of two opposite sides" — it does not halve a general quadrilateral. Counterexample worth keeping to hand: A(0,0), B(4,0), C(5,3), D(0,2) has area 11, and the segment joining the midpoints of AB and CD splits it 5 to 6. Nor does "shrink one pair of parallel measurements by half" mean anything for a general quadrilateral, which has no pair of parallel sides to shrink.
• A regular hexagon in three pieces (Part II p.170, Figure it Out 7). Printed: a regular hexagon is shown cut into three pieces: one trapezium, one equilateral triangle and one rhombus, and the ratio of their areas is asked for. Re-measured on the printed page, the division is: the long cut is the main diagonal joining two opposite vertices — in the printed figure it runs from the lower-left vertex to the upper-right one, has length twice the side, and therefore passes through the hexagon's centre — cutting off the upper part as the trapezium; then a second cut runs from that diagonal's midpoint, which is the hexagon's own centre, down to the vertex below it, separating the equilateral triangle from the rhombus. Not a diagonal "two apart": one that skips a single vertex has length s√3, misses the centre, and cuts off one isosceles 120° triangle of a single unit area — giving no trapezium, no equilateral triangle, no rhombus and no 3 : 1 : 2. Since Figures required has the teacher redraw this from the description, getting the diagonal right is the whole of it. Not in the book: take the small equilateral triangle of the hexagon's own side as the unit. The whole hexagon is six of them; the trapezium is three, the triangle is one and the rhombus is two, so the ratio is 3 : 1 : 2. This item is printed in the trapezium exercise set and Area of a trapezium, derived two different ways notes the handover.
• What the chapter says next (Part II p.160, closing line). Printed: the chapter states that special formulae can be derived for a parallelogram, a rhombus and a trapezium — which is the honest framing. They are derivable, not necessary.

Figures to have open

• The unnamed quadrilateral, then the same quadrilateral with BD drawn (Part II p.159). Two states of one figure; the point is how little was needed.
• The pentagon (Part II p.159), redrawn with the fan of three triangles added. The printed figure has no internal lines and no labels at all, so everything the explanation draws on it is added here.
• The inward-pointing quadrilateral with both diagonals attempted. Not in the book; needed for section 5. Standard schematic.
• The long thin quadrilateral with AC, BM and DN (Part II p.160, item 1). Redraw accurately; the figure's extreme thinness is what makes the "no sides given" point land visually.
• The shaded region inside a rectangle (Part II p.160, item 2) with all six printed lengths, and with the two corner triangles tinted so the subtraction is visible.
• The two blue triangles with a draggable apex (Part II p.160, item 4). The apex must move. This is the topic's key visual.
• The three squares in an L with the segment from D to H (Part II p.158, item 6), with the red and blue regions coloured as printed.
• The regular hexagon in three pieces (Part II p.170, item 7), overlaid with the six small equilateral triangles so the ratio 3 : 1 : 2 can be counted rather than computed.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 7, "Area", §7.1 "Rectangle and Squares", under the unnumbered bold subheading "Area of any Polygon", Part II p.159, for the quadrilateral, the pentagon and the general claim. The subheading begins below Figure it Out item 8 on the same page.
• Part II p.160, Figure it Out items 1 to 5, and the closing line of the page which announces the three special formulae. Items 3, 4 and 5 are each marked Math Talk.
• Part II p.158, Figure it Out 6, the three identical squares, whose second part is marked Math Talk. The other items in that block belong to Why the area of a triangle is half base times height and Triangles with the same base and height have the same area.
• Part II p.170, Figure it Out 7, the regular hexagon in three pieces. That item is printed inside the trapezium exercise set; Area of a trapezium, derived two different ways notes the handover.
• Part II p.171, the chapter SUMMARY, second bullet, which states that any polygon's area is obtainable by cutting it into triangles. This is the only one of this topic's results that reaches the SUMMARY.
• Backward pointers: Why the area of a triangle is half base times height for the triangle formula this topic spends, and Triangles with the same base and height have the same area for the equal-halves result used in the halving construction.
• Forward pointers: Area of a parallelogram, by turning it into a rectangle, Area of a rhombus from its diagonals and Area of a trapezium, derived two different ways are the three special formulae the closing line of Part II p.160 announces.