From rectangle to parallelogram: area survives rearrangement

What to assume they know

• Area of a rectangle as ab, from Class 8, at the level of "I can use it"
• Congruence of triangles, and that congruent figures have equal area
• Perpendicular distance from a point to a line, and the foot of a perpendicular
• Opposite sides of a parallelogram being equal and parallel
• Perimeter as a walk around the border, and why perimeter-to-side ratios are fixed — the idea of a special case, met with the square and rectangle perimeters

What they should be able to do

• State what a unit of area is and why area is always reported relative to one
• Explain why a rectangle of sides a and b has area ab, in terms of counting unit squares
• State the two properties of area the chapter's arguments rely on, and identify where each is used
• Carry out the cut-and-slide transformation from parallelogram to rectangle and state the resulting formula
• Identify the case in which the printed construction fails, and describe the repair the chapter offers
• Explain why the repair terminates rather than going on forever
• Distinguish the height of a parallelogram from its sides, and identify the height for a stated choice of base
• Explain why two side lengths do not determine a parallelogram's area, using a hinged model or the angle argument the chapter hints at
• Deduce that parallelograms and triangles on the same base with apex on a fixed parallel line have equal area, and use it on the chapter's exercises
• Derive the trapezium formula by splitting into a parallelogram and a triangle

Where it usually goes wrong

• "Area is length times width, for everything." It is for a rectangle. For a parallelogram it is base times height, and the height is not a side of the figure. This is the single most common error on this material.
• "The slanted side is the height." The height is measured perpendicular to the base. Fig. 6.17 marks it with a dashed perpendicular for exactly this reason.
• "Cutting and moving a piece might change the area." It cannot, and saying so out loud is not pedantry — it is the entire justification for the formula. Make the property explicit before using it, and the parallelogram, the triangle and the circle all become one idea.
• "The construction in Fig. 6.17 always works." It does not; the chapter itself raises the thin case on the same page and supplies a repair. A student who has met the gap will trust the formula more, not less.
• "Two shapes with the same four side lengths have the same area." A hinged parallelogram is a counterexample you can build from four strips of card, and the chapter builds it with a rhombus in Fig. 6.27 (p. 137). This is also the hinge on which Brahmagupta's formula, and Heron's as the case where a side vanishes turns.
• "Same area implies same perimeter, or the other way round." Neither. But for rectangles, having both the same does force congruence — end-of-chapter Q10. Get the order of quantifiers right and this stops being confusing.
• "The trapezium formula is a new thing to memorise." It is a parallelogram plus a triangle, and the chapter gives three separate ways of seeing it. Derive it; do not present it.

Questions to check understanding

• Area of a parallelogram from base and height, and the reverse: height from area and base
• Given a parallelogram with two named bases, state the corresponding heights
• Area of a trapezium from its parallel sides and height, with the derivation asked for
• Prove that two triangles or two parallelograms on the same base and between the same parallels have equal area
• Ratio-of-areas questions in which the answer is 1 : 1 and the reason is a shared height
• Explain why the side lengths of a parallelogram do not determine its area
• Decide whether two rectangles with equal area and equal perimeter must be congruent, with justification
• Draw an area model for a given algebraic identity

Examples worth working on the board

Inputs, not answers. Values marked Verified are worked out here; the chapter prints no answers and this volume has no appended answer key.

• Fig. 6.16 (p. 130). Four small figures in a row, sharing one caption. A unit square labelled 1 unit on each side with area 1 sq. unit; a 1-by-b strip with area b; a square of side a with area a²; a rectangle a by b with area ab. The sequence is the point: the strip is the bridge from one square to many.
• Fig. 6.17 (p. 131). On the left, parallelogram ABCD with A and D on top and B, C below. On the right, a copy labelled A′B′C′D′ with a green triangle EA′B′ standing to the left of it, giving rectangle EB′C′F; base b and height h are labelled inside the rectangle, and h is also marked on the left edge. The printed claim is that the rectangle and the parallelogram have the same base, the same height and therefore the same area, so the parallelogram's area is bh.
• The two properties, stated. (i) Moving a piece rigidly does not change its area. (ii) The area of a figure cut into non-overlapping pieces is the sum of the pieces' areas. Every area argument in this chapter — Fig. 6.17, Fig. 6.19, Fig. 6.20A, Fig. 6.21, Fig. 6.37 and the trapezium questions — is one of these two, or both.
• Fig. 6.18 (p. 131), inside the first Think and Reflect box. A very slanted parallelogram ABCD, A and D high on the right, B and C low on the left. The box asks what happens when the perpendicular from C to AD lands outside the segment AD, so that the cut-and-slide has nothing to slide into.
• Fig. 6.19 (p. 131). The repair. Mark two new points at equal offsets from the ends of the top side — one just inside it near D, the other beyond A on the same line produced. The figure they make with B and C is again a parallelogram, and the small triangle lost at one end is congruent to the small triangle gained at the other, so the area has not moved. The chapter says to repeat as often as needed.
• Why the repair terminates, which the chapter does not say. Each step shifts the top side along its own line by a chosen amount while keeping base and height fixed. Choose that amount to be the whole overhang and one step suffices; choose it smaller and finitely many steps still bring the foot of the perpendicular onto the base, because the overhang is a fixed finite length. This is an added argument.
• Second Think and Reflect (pp. 131–132). Can a parallelogram's area be found from its side lengths alone, as a rectangle's can? The printed hint asks what happens to the area when the angle between adjacent sides is changed with the lengths held fixed.
• The collapsing parallelogram, worked. Sides 6 cm and 4 cm. Verified: with the angle between them at 90° the area is 24 cm²; at 60° it is 24 sin 60° ≈ 20.8 cm²; at 30° it is 12 cm²; at 5° about 2.1 cm²; at 0° it is 0. The sides never change. Note: Class 9 need not use the sine — a hinged strip model or the height shrinking visibly is enough, and the numbers above are as a check, not for the student yet.
• Exercise Set 6.2 Q6 (p. 142). In a parallelogram ABCD, put two arbitrary points P and Q somewhere along the side AB, and compare the areas of triangles PCD and QCD as a ratio. Verified: 1 : 1, because both have base CD and apex on the parallel line AB, so they share a height.
• Exercise Set 6.2 Q7 (p. 142). In a parallelogram PQRS, pick any point O somewhere along the diagonal PR, and show that triangles PSO and PQO come out equal in area. Verified: the cleanest route is that S and Q lie the same perpendicular distance from line PR, so the two triangles share base PO and have equal heights.
• Fig. 6.42 and end-of-chapter Q11 (p. 150). A trapezium with parallel sides a on top and b below, height h, drawn already divided by a segment from the upper-right vertex down to the base into a parallelogram on the left and a triangle on the right. The question asks for the trapezium formula from the parallelogram formula plus this figure. Verified: parallelogram ah, triangle ½(b − a)h, total ½(a + b)h.
• End-of-chapter Q12 and Q13 (p. 150). Two more routes to the same formula: cut the trapezium into two triangles, and put two copies of it together to make a parallelogram. Verified: the two triangles have areas ½ah and ½bh on the two parallel sides with the same height; the doubled figure is a parallelogram of base a + b and height h, so the trapezium is half of (a + b)h. Three independent derivations of one formula is unusually generous.
• End-of-chapter Q10 (p. 150), starred. Given a pair of rectangles that agree both in area and in perimeter, must they be congruent? Verified: yes. Equal perimeter fixes a + b, equal area fixes ab, and a pair of numbers with a known sum and a known product is determined. This is a genuinely good closing beat, because the answer for rectangles is yes while for parallelograms it is emphatically no — which is section 8 restated.
• End-of-chapter Q1 (p. 149), Fig. 6.41. A square divided into four cells labelled a², ab, ab, b², with a and b marked along the top and down the side, presented as the area picture of (a + b)² = a² + 2ab + b². Students are then asked for two more such pictures: one for the difference of two squares factorised as (a + b)(a − b), and one for the square of a three-term sum. Worth including because it is additivity of area used to prove algebra.

Figures to have open

• Fig. 6.16 (p. 130), redrawn as four panels with the unit square emphasised.
• Fig. 6.17 (p. 131), redrawn and able to be shown moving: the cut must actually move, or the argument is only asserted. The chapter's own figure.
• Fig. 6.18 and Fig. 6.19 (p. 131), redrawn together. These two are the reason to make the explanation — most treatments of the parallelogram formula never mention the gap, and this chapter both raises it and repairs it on one page.
• A physical hinged parallelogram — four strips pinned at the corners — collapsing, with the four side lengths displayed unchanged. Not in the book; the chapter only hints at this in a Think and Reflect.
• Fig. 6.42 (p. 150), redrawn with the parallelogram and triangle in two colours.
• Fig. 6.41 (p. 149) redrawn, plus the two companion pictures the question asks for — the difference of two squares, and the square of a three-term sum — sketched as the extension.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics (NCF-SE 2023), Chapter 6, §6.6 "Area of a Rectangle" (p. 130) and §6.7 "Area of a Parallelogram" (pp. 130–132), the latter including both Think and Reflect boxes, the second of which runs over onto p. 132 with its hint.
• Figures 6.16 (p. 130), 6.17, 6.18 and 6.19 (all p. 131).
• Exercise Set 6.2 Q6 and Q7 (p. 142).
• End-of-chapter exercises Q1 with Fig. 6.41 (p. 149); Q10, Q11 with Fig. 6.42, Q12 and Q13 (p. 150).
• Forward pointer inside the chapter: the impossibility of getting a 4-gon's area from its sides is settled with Fig. 6.27 on p. 137, handled in Brahmagupta's formula, and Heron's as the case where a side vanishes; the triangle is §6.8 (pp. 132–133), handled in A triangle is half a parallelogram.
• The Chapter Summary (p. 154) does not list the parallelogram formula.