Area of a parallelogram, by turning it into a rectangle

What to assume they know

• Area of a rectangle, and area as a count of unit squares — Area as a count of unit squares
• Half base times height for a triangle, and that height means perpendicular distance — Why the area of a triangle is half base times height
• The RHS congruence test for right-angled triangles
• Properties of a parallelogram: opposite sides equal and parallel
• That co-interior angles at a transversal of two parallels total two right angles
• Division of a whole number by a one-decimal-place number

What they should be able to do

• Construct a height of a given parallelogram and say why the article is "a" rather than "the"
• Cut a parallelogram into a triangle and a trapezium along that height
• Identify the triangle that would complete the trapezium into a rectangle, by construction rather than by guessing
• Verify with RHS that the completing triangle is congruent to the removed one, naming the three facts used
• Define dissection, and state what it preserves
• Show that the rectangle's longer side equals the parallelogram's base, using the common segment argument
• Compute a parallelogram's area from a base and its matching height, including when the base drawn is not horizontal
• Recover an unknown height from an area and a side, and explain why the shorter side carries the taller height
• Explain why a family of parallelograms with one base and one height have one area and many perimeters
• Compare a rectangle and a parallelogram built on the same two sidelengths, and justify which is larger

Where it usually goes wrong

• "The height is the other side." True only for a rectangle, which is why the chapter puts the 5 cm by 4 cm comparison in the same exercise set. In items (iii) and (iv) on Part II p.163 the height and the neighbouring side are visibly different lengths.
• "A parallelogram has one height." It has two, one for each pair of sides, and Part II p.162 makes a point of it. On Part II p.163 item 2(iii) the marked height 4.8 cm belongs to the 5 cm side, and the height belonging to the other pair of sides is a different number entirely. (Item 2 has sub-parts (i) to (iv); item 3 is the separate "Find QN" problem, so do not call 2(iii) the page's third item.)
• "Same sidelengths, same area." The 5 cm by 4 cm pair settles it. This is the misconception that most reliably survives the whole chapter.
• "A leaning parallelogram has less area than an upright one." It has less than the rectangle on the same sides, and exactly the same as the rectangle on the same base and height. Those are two different comparisons and students collapse them. Draw both comparisons in the same shot, or the confusion is guaranteed.
• "Cut and slide — obviously it fits." The chapter spends most of Part II p.161 refusing that answer. If the explanation shows the slide without the congruence check, it has taught a craft skill and no mathematics.
• "Cutting along any perpendicular works." For a strongly sheared parallelogram the foot of the perpendicular from A can land outside segment DC, and then ∆AXD is not the piece the printed argument describes. The escape is the question Part II p.162 asks next: use the other pair of sides as base. Worth one line, because a student who draws a very slanted example will hit it.
• "The rectangle obviously has the same width as the parallelogram's base." It does, but by the common-segment argument, not by inspection. That is the second place rigour is being taught.
• "The taller height belongs to the longer side." The other way round. Part II p.163 item 3 gives 6 cm against 12 cm and about 9.47 cm against 7.6 cm.

Questions to check understanding

• Find a parallelogram's area from a base and its matching height, with the base drawn horizontal and then drawn slanted
• Given an area and one side, find the height belonging to that side
• Given one base-height pair and the other side, find the second height
• Decide which of a rectangle and a parallelogram with the same two sidelengths has the greater area, with a reason
• State what all the parallelograms in a given family have in common and what differs
• Describe the dissection that converts a parallelogram into a rectangle, and name the congruence test that justifies it
• Explain why either pair of sides may be used as the base
• Explain why the base-times-height rule for a parallelogram is not a separate fact from the rectangle rule — the reasoning item this subsection 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 plan, stated up front (Part II p.161, under the subheading "Parallelogram"). Printed: the chapter announces that it will get a formula by remaking a parallelogram as a rectangle that holds the same amount, and asks the student for a method, suggesting a cut-out be used.
• The cut (Part II p.161, first pair of figures). Parallelogram ABCD with A top left, B top right, C bottom right, D bottom left, leaning right so that A sits to the right of D. Printed: construct AX at right angles to CD, written in the chapter's short notation; that segment is called a height of the parallelogram, with the word set in bold; the parallelogram is then cut into ∆AXD and the trapezium ABCX. The second figure shows the two pieces pulled apart with a curved arrow indicating that the triangle is to be moved to the right-hand end.
• The completing triangle (Part II p.161, second pair of figures). Printed: the chapter's own method for checking the fit is to identify the triangle that would complete ABCX into a rectangle and then test whether it is congruent to ∆AXD. It observes that the angle at X is a right angle and so the angle at A is too, because AB is parallel to XC, and asks in passing what the remaining angle must be. A third right angle is then produced by extending XC rightward and dropping a perpendicular through B, meeting the extension at Y; ∆BYC is the triangle that completes ABCX. In the printed figure the removed triangle is tinted one colour and ∆BYC another, with matching tick marks on the two hypotenuses.
• The congruence (Part II p.161). Printed: the three facts are BY = AX, because ABYX is a rectangle; the angle at Y equals the angle at X and both are right angles; and BC = AD, because ABCD is a parallelogram. By the RHS congruency criterion the two triangles are congruent, so the removed piece drops exactly onto the gap. Not in the book: the third fact is the one doing the real work, and it is the only one that uses the parallelogram property — a student who skips it has proved nothing, because two right triangles with one matching leg are not otherwise congruent.
• Dissection, defined (Part II p.161, last line, term set in bold). Printed: the chapter's name for cutting a shape up and reassembling the pieces into a different shape of the same area. Printed elsewhere: the tradition it belongs to is introduced on Part II p.158, where the chapter explains that the Śulba-Sūtras deal with altar construction and therefore with remaking one shape as another of the same area, and notes that Euclid's Elements poses problems of the same kind.
• Reading the sides back (Part II p.162, top). Printed: the parallelogram's area equals rectangle ABYX's area, which is AX × XY; AX is the height. The chapter then asks how XY relates to DC and answers it: DX = CY, so adding the common segment XC to each gives DC = XY. Since DC is the base, the boxed result is that a parallelogram's area is its base times its height. Not in the book: the common segment step is the second place the argument could quietly fail, and it is where a student learns that "the rectangle looks the same width" is not a reason.
• The other base (Part II p.162, middle figure). Parallelogram ABCD drawn leaning, with the left side AD marked as the base and a perpendicular from Z on AD across to C marked as the height. Printed: the chapter asks whether the area can be found this way too, asks whether the parallelogram can be cut along CZ and rearranged into a rectangle, and answers that it can — so any side together with its matching height will do.
• Seven parallelograms (Part II p.162, Figure it Out 1). Seven parallelograms labelled (a) to (g) drawn on a square grid, four in an upper row — (a) to (d) — and three in a lower row. The lean does not increase across the figure, and a redraw built on that will be wrong. Measured on the printed page, as horizontal offset of the top side relative to the bottom, in grid squares: (a) +0.33, (b) +1.17, (c) +2.48, then (d) −0.58, (e) −1.01, (f) −2.01, (g) −2.81. So the shear grows through (a) to (c), then reverses direction and grows again through (d) to (g) — (c) is more than four times as sheared as (d), which follows it. Printed questions: (i) what can be said about their areas, and (ii) what can be said about their perimeters, with which looks greatest and which least. Not in the book, and now measured rather than left open: all seven are drawn on a base of 5 grid squares with a height of 3 grid squares, so every one has an area of 15 square grid units. Only the shear offsets fall on half-squares, which is why they do not land on lattice intersections. The perimeters differ, growing with the magnitude of the shear regardless of its direction, so (a) is least (offset 0.33) and (g) greatest (2.81) — the answers stand, but for that reason and not because the lean rises steadily left to right. Mark the base and height identically on all seven; they are now numbered.
• Four parallelograms to measure (Part II p.163, Figure it Out 2). (i) base 7 cm along the bottom, height 4 cm drawn as a dashed perpendicular inside from the top-left corner. (ii) base 5 cm along the bottom, height 3 cm dashed inside, and a diagonal drawn across the figure. (iii) drawn leaning left with the right-hand side marked 5 cm and a dashed perpendicular of 4.8 cm running to it from the opposite side. (iv) the same arrangement with the right-hand side marked 2 cm and the dashed perpendicular 4.4 cm. Not in the book: 28 cm², 15 cm², 24 cm² and 8.8 cm². Items (iii) and (iv) exist to make the student use a slanted side as the base, and an explanation that draws all four with a horizontal base has destroyed the item. Not in the book for the diagonal in (ii): it cuts the parallelogram into two triangles of 7.5 cm² each.
• An unknown height (Part II p.163, Figure it Out 3). Parallelogram PQRS with P top left, Q top right, S bottom left and R bottom right; S, M and R lie on the bottom line with SR marked 12 cm; QM is the perpendicular from Q down to SR, marked 6 cm; PS is marked 7.6 cm; N is the foot of the perpendicular QN from Q onto PS, and QN is what is asked for. Not in the book: the area is 12 × 6 = 72 cm², so 7.6 × QN = 72 and QN = 72 ÷ 7.6 ≈ 9.47 cm. Two things to say about that number: it is deliberately not tidy, and it is larger than 6 — the shorter side always carries the taller height, because the product is fixed.
• Same sides, different area (Part II p.163, Figure it Out 4). Printed: a rectangle and a parallelogram both have sidelengths 5 cm and 4 cm, and the question is which has the greater area; the printed hint is to imagine building them on the same base, and the figure draws them overlapping on a shared 5 cm base with the 4 cm side marked on each. Not in the book: the rectangle wins. On the shared base the rectangle's height is its 4 cm side, while the parallelogram's height is a leg of a right triangle whose hypotenuse is its own 4 cm side, so it is strictly shorter. The rectangle is 20 cm²; the parallelogram is 5 cm times something less than 4 cm. Both have perimeter 18 cm, which makes this the sharpest instance in the chapter of the point Why perimeter cannot stand in for area argues for.
• Handed to other topics. Part II p.163 item 5 and Part II p.164 items 6 to 9 are triangle-and-rectangle transformation problems and are carried by Why the area of a triangle is half base times height. They are printed inside this exercise set, so anyone working from the printed page will meet them here.

Figures to have open

• The four-panel dissection (Part II p.161). All four printed states are needed: the parallelogram with AX, the two pieces separated, the trapezium with the gap, and ∆BYC in place. Redraw with the two triangles tinted as the book tints them, because the tick marks and the colours are what make the congruence readable.
• The two-heights figure (Part II p.162, middle). Redraw with both bases and both heights on one parallelogram, which the printed pages show separately.
• The seven grid parallelograms (Part II p.162, item 1). Must be redrawn on a visible grid, and the base and height must be marked identically on all seven, since that is the answer to part (i): base 5 squares, height 3 squares, area 15 square grid units, on every one of the seven. Reproduce each figure's own shear offset from the measured list in Worked examples — they alternate direction between the rows.
• The four measurable parallelograms (Part II p.163, item 2). All four, in their printed orientations. Items (iii) and (iv) must keep their lean; rotating them to sit on a horizontal base removes the whole point.
• The PQRS figure with two heights (Part II p.163, item 3), with 12 cm, 6 cm and 7.6 cm marked and QN drawn but unlabelled.
• The overlapping rectangle and parallelogram (Part II p.163, item 4) on a shared base, with both 4 cm sides marked and both heights drawn — the second height is not in the printed figure and is what settles the item.
• 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 "Parallelogram", Part II pp.161–162, for the dissection, the RHS argument, the definition of dissection, the common-segment step, the boxed formula and the second choice of base.
• Part II pp.162–163, Figure it Out items 1 to 4. Item 1 is on Part II p.162; items 2 to 4 are on Part II p.163.
• Part II p.163 item 5 and Part II p.164 items 6 to 9 are printed inside the same exercise set but are triangle-and-rectangle transformation problems, and are carried by Why the area of a triangle is half base times height.
• Part II p.158 for the Śulba-Sūtras framing that section 7 needs, and for the mention of Euclid's Elements. That paragraph is owned by Why the area of a triangle is half base times height.
• Part II p.160, closing line, where the chapter announces that special formulae follow for a parallelogram, a rhombus and a trapezium.
• Part II p.171, the chapter SUMMARY, third bullet, for the parallelogram formula.
• Backward pointers: Area as a count of unit squares for the rectangle formula this argument reduces to, and Why the area of a triangle is half base times height for the perpendicular meaning of height.
• Forward pointer: Area of a rhombus from its diagonals opens by noting that a rhombus is a parallelogram, so everything proved here already applies to it.