Every Shape Secretly Wants to Be a Rectangle Area

Two rectangles, one 7 cm by 4 cm and the other 8 cm by 3 cm, look about as likely as each other to need more rangoli powder to fill. Counting unit squares settles it exactly: the first packs in 7 x 4 = 28 unit squares, the second only 8 x 3 = 24, so the first needs more colouring even though its two sides look, at a glance, less extreme than the second's. This is the definition worth holding onto for the rest of the chapter: area is the number of unit squares (including fractional ones) a region contains, and for a rectangle this number is always its length times its width. It is tempting to reach for perimeter as a shortcut for comparing areas, since both feel like they measure 'how big' a shape is, but they measure genuinely different things. It is entirely possible to find two regions where one has the larger perimeter and yet the smaller area, and this book asks exactly that: find two rectangles like this yourself, then find two irregular regions where the one with the larger perimeter visibly has the smaller area. A diagonal cut through a rectangle also settles a smaller, useful fact early: it always produces two congruent triangles, so each is exactly half the rectangle's area, and that single observation becomes the seed of everything that follows.

Class 6 showed only that a triangle formed by a rectangle's own diagonal is exactly half that rectangle. This chapter finishes the job properly, for every triangle, including ones where no obvious enclosing rectangle exists at all. Start with a triangle ABC where the foot of the altitude from A lands neatly inside BC, at some point X. Then BXAE (built by completing the rectangle around the triangle using the altitude AX as one side) is a rectangle, and by the same diagonal argument, Area(ABC) = 1/2 x base x height, using BC as the base and AX as the height. But what if the altitude's foot D lands OUTSIDE segment BC entirely, so no single enclosing rectangle sits neatly around the triangle? Split the difference instead: Area(ABC) is exactly the difference of Area(ADC) and Area(ADB), and each of those two triangles DOES have its altitude landing inside its own base, so each is a clean half-rectangle case already handled. Area(ABC) = 1/2 x h x DC - 1/2 x h x DB = 1/2 x h x (DC - DB) = 1/2 x h x BC, exactly the same formula, base BC and height h, holding true regardless of where the altitude's foot actually falls. Two further facts follow almost for free from this one formula. First, the 4 triangles formed by both diagonals of a rectangle are all equal in area: comparing any two adjacent ones, say sharing base OD and OB respectively (equal, since a rectangle's diagonals bisect each other) with the same shared altitude, gives equal areas, and the same argument chains around all four. Second, a median (the segment from a vertex to the midpoint of the opposite side) always splits a triangle into two equal-area halves, since the two smaller triangles share the exact same height from that vertex, and equal bases by definition of a midpoint.

Take a fixed base BC and a line l parallel to it, and imagine every possible triangle with B and C fixed but the third vertex free to sit anywhere on l. Which one has the smallest perimeter? Since BC is common to every one of these triangles, minimising the perimeter really just means minimising the sum of the other two sides, AB + AC, over every choice of point A on l. Treat line l as a mirror, and reflect point C across it to get a point C'. Because reflection preserves distance, AC = AC' for any point A on l, so minimising AB + AC is exactly the same problem as minimising AB + AC'. But AB + AC' is smallest exactly when A lies on the straight line segment from B to C' (a straight line is always the shortest path between two points), so the minimum-perimeter triangle is the one where A is the specific point where segment BC' crosses line l. This single argument, using nothing but a mirror-image reflection, pins down the answer completely, with no measuring or guessing required, and it turns out this special point A always lies exactly on the perpendicular bisector of BC.

A quadrilateral's area can be found by drawing just one diagonal, splitting it into two triangles whose areas add up to the whole. A pentagon needs two such cuts, and in general, any polygon at all, however many sides it has, can always be divided into triangles. Since a triangle's area is already fully solved, 1/2 x base x height, this single fact means the area of literally any polygon can be found: split it into triangles any way that works, compute each triangle's area separately, and add. A particularly elegant special case of this fact answers its own natural follow-up question: is there a way to build a NEW quadrilateral with exactly half the area of a given one? Join the midpoints of all four sides of any quadrilateral, in order, and the resulting inner quadrilateral always has exactly half the area of the original one, for any quadrilateral at all, convex or not -- a genuinely surprising, clean result built from nothing more than the midpoint and median facts already established.

Take parallelogram ABCD and construct AX perpendicular to DC, calling AX a height of the parallelogram. Cutting along AX splits the parallelogram into triangle AXD and trapezium ABCX. Extend XC to the right and drop a perpendicular from B to meet it at Y: triangle BYC is formed, completing ABCX into rectangle ABYX. Is triangle AXD congruent to this new triangle BYC? Yes, by the RHS congruence rule: BY = AX (opposite sides of rectangle ABYX), both have a right angle, and BC = AD (opposite sides of the original parallelogram). Since they're congruent, triangle AXD fits exactly into the space of triangle BYC, so the parallelogram and the rectangle ABYX cover exactly the same area. This process, cutting a figure into pieces and rearranging those same pieces into a different figure of equal area, is called dissection, and it is the single technique this entire chapter reuses again and again. Since Area(ABYX) = AX x XY, and since DX = CY (both are the 'overhang' cut off from each end), adding the shared middle part XC to both shows DC = XY exactly, so Area of parallelogram ABCD = base (DC) x height (AX) = base x height, using DC as the base. Any other side of the parallelogram, together with its own corresponding perpendicular height, works exactly as well, since the same cut-and-fit argument applies regardless of which side is chosen as the base.

A rhombus is a parallelogram, so base x height already applies to it, but its extra property, all four sides equal with diagonals that are perpendicular bisectors of each other, unlocks a second, often more convenient formula. Since the diagonals AC and BD cross at right angles and bisect each other, triangles ABD and CBD are both isosceles, and each can be dissected into a rectangle exactly as before. Joining those two resulting rectangles produces one larger rectangle WXYZ with the very same total area as the rhombus, and its side lengths turn out to be exactly XW = AC (the full length of one diagonal) and WZ = half of BD (half the other diagonal). So Area of rhombus ABCD = Area of rectangle WXYZ = AC x (BD/2) = 1/2 x AC x BD, i.e., half the product of the diagonals. The same result also drops out from simply adding the two triangles directly: Area(ADB) = 1/2 x AO x BD and Area(CDB) = 1/2 x CO x BD (using the diagonals' perpendicularity), and since AO + CO = AC, adding these two areas together gives exactly 1/2 x AC x BD once again, confirming the formula two genuinely different ways.

A trapezium WXYZ with WX parallel to ZY can be split by dropping two perpendiculars, WM and XN, down to ZY, creating rectangle WXNM in the middle with a triangle on each side. Labelling MZ = x, WM = XN = h, WX = a, NY = y, the total area is 1/2 x h x MZ + (WX x WM) + 1/2 x h x NY = 1/2 h(x+a+y) + 1/2 h(a) ... which simplifies to 1/2 x h x (x+y+2a). Writing b for the OTHER parallel side ZY, note b = x+y+a, so x+y = b-a, and substituting gives Area = 1/2 h(b-a+2a) = 1/2 h(a+b): Area of a trapezium = 1/2 x height x sum of the two parallel sides. This same formula also holds for a trapezium shaped so that one perpendicular's foot falls outside the trapezium entirely, verified the same way as the general triangle case earlier, by working with a difference of areas instead of a sum. A second, quite different way to reach the exact same formula uses rotation rather than cutting: take two identical copies of the trapezium and rotate the second one 180 degrees, then join the two along one non-parallel side. The angles along the join add to exactly 180 degrees (since WX is parallel to ZY in both copies), which forces the resulting 6 corners to collapse into just 4, meaning the joined shape is a plain parallelogram, with base (a+b) and the same height h as either trapezium. Since the parallelogram is built from exactly two copies of the original trapezium, Area of the trapezium = half of Area of the parallelogram = 1/2 x h x (a+b), the same formula, reached without cutting the shape into pieces at all.

The area formulas built in this chapter apply to whatever unit of length is used, but real-world areas are compared across very different units: a sheet of A4 paper (21 cm x 29.7 cm), a classroom floor (usually metres), a plot of farmland (sometimes acres), or a city (square kilometres). Converting between units of LENGTH is direct (1 inch = 2.54 cm, 1 foot = 12 inches), but converting AREA units needs the conversion factor applied TWICE, once for each dimension: since 1 in = 2.54 cm, 1 square inch = 2.54 x 2.54 = 6.4516 square cm exactly, not simply 2.54. The same doubling applies going the other way, from square feet up to the large-scale unit of an acre (1 acre = 43,560 sq ft), and to the square kilometre used for whole cities or districts. India additionally uses many local, region-specific units for land area, such as bigha, gaj, katha, dhur, cent, and ankanam, each with its own conversion depending on the region -- a reminder that the mathematics of area is universal, but the units people actually use to talk about it are not.