A triangle is half a parallelogram

What to assume they know

• From rectangle to parallelogram: area survives rearrangement — area of a rectangle and of a parallelogram, and the two properties of area those arguments use
• Congruent triangles, and that congruent figures have equal area
• Alternate angles, and the test for two lines being parallel
• Midpoint of a segment
• Perpendicular from a vertex to the opposite side, including the case where the foot lands outside that side

What they should be able to do

• Derive the area of a right-angled triangle as half the enclosing rectangle
• Derive the general case by dropping a perpendicular, splitting the base into two parts and boxing each part
• Identify the configuration in which that split fails, and describe how to handle it
• Derive the same formula by fitting two congruent copies of the triangle into a parallelogram, and give the angle reason the copies close up
• State the median theorem and prove it from the area formula
• Explain why two triangles of equal area need not be congruent, and produce an example
• Apply the theorem to figures where the answer is a ratio of 1 : 1 and the reason is a shared base and height
• State what the chapter reveals and withholds about cutting one of the two half-triangles into pieces that cover the other
• Explain why doubling every side of a triangle multiplies its area by four, and whether four copies of the original then fit inside

Where it usually goes wrong

• "Half base times height only works for right-angled triangles." The chapter does the right-angled case first because it is easiest, and then does the general case twice. Students who see only the first case will hunt for a right angle that is not there.
• "The height is the shortest side, or a side at all." The height belongs to a chosen base and is measured perpendicular to it. In Fig. 6.22 the chapter draws the height outside the triangle, which is the clearest possible refutation.
• "Every triangle has one height." It has three, one per choice of base, and all three give the same area. Worth showing once.
• "The two triangles a median makes must be congruent, or the areas would not match." They match because area sees only base and height. The chapter says in as many words that the two are in general not congruent, and calls the result a surprise. An explanation that presents the theorem as obvious has thrown away the lesson.
• "Equal areas means one is a rearrangement of the other, so it is basically the same shape." Same area, different shape — and it is still true that one can be cut up to cover the other, which is a much more interesting statement than "same shape".
• "Among rectangles of a fixed perimeter there is a smallest area." There is not. The area approaches zero and never gets there. The chapter asks about both extremes and only one of them exists; the honest answer is more instructive than a manufactured one.
• "Doubling the sides doubles the area." It quadruples it. Q15 is built to force the point, and then asks the sharper question of whether four copies literally fit.

Questions to check understanding

• Area of a triangle from base and height, and the reverse
• Area of a triangle inside a rectangle where the apex position is irrelevant
• Prove that a median divides a triangle into two of equal area
• Prove equal areas for two triangles sharing a base with apexes on a common parallel
• Ratio-of-areas questions inside a square or parallelogram with an interior point
• "What fraction is shaded" on a figure whose sides are divided into equal parts
• Given a scale factor on the sides, state the factor on the area, and say whether that many copies of the original tile the enlargement
• Extremal question: among rectangles of fixed perimeter, find the largest area and discuss whether a smallest exists

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. 6.20A (p. 132), three panels sharing one caption. Left: rectangle ABCD with A, D on top and B, C below, and the right-angled triangle ABC shaded inside it, base b along BC and height h up AB. Right: rectangle HFGI with the acute triangle EFG shaded inside it, apex E on the top edge, and the base split at J into b₁ = FJ and b₂ = JG, with the chapter noting b = b₁ + b₂ and the height h marked from E down to the base.
• Fig. 6.20B (p. 132). A triangle EFG with E up and to the left of F, so that the angle at F opens past a right angle and the perpendicular from E falls outside FG. No enclosing rectangle is drawn. The chapter asks the reader to work out what to do here.
• What to do, which the chapter leaves open. Extend GF beyond F to the foot of the perpendicular. The big triangle on the extended base minus the extra triangle outside FG leaves triangle EFG, and the two halves-of-rectangles subtract instead of adding. Verified algebraically: ½(b + e)h − ½*e**h* = ½bh, where e is the overhang. This argument is added here.
• Fig. 6.21A and Fig. 6.21B (p. 132). Two congruent triangles drawn apart with a visible gap between them, labelled ABC and A′B′C′; then the same pair fitted together as parallelogram ABCD, the shared edge shown as a dashed diagonal.
• The angle reason, as the chapter frames it (p. 133). It directs the reader to compare a pair of angles across the join and to keep in mind the criterion for two lines being parallel. The clean version when explaining it: the copy is the original turned through half a turn about the midpoint of one side; a half-turn sends every line to a parallel line, so the images of the other two sides come out parallel to those sides, and the four sides pair off into two parallel pairs.
• Fig. 6.22 (p. 133). Triangle with apex A at the top, B and C at the ends of the base, D the midpoint of BC, the median AD dashed, tick marks showing BD = DC, and the base labelled BC = 2a units. The height h is drawn to the right of the triangle, as a vertical segment from the horizontal line through A down to C, with a right-angle mark at the top — so the height is measured outside the figure, which is worth pointing out because it is exactly the move section 4 needed.
• The median theorem as printed (p. 133). Both halves have base a and the same height h, so both have area ah/2, and the chapter states as a Theorem that a median divides a triangle into two triangles of equal area. It then says outright that the two are in general not congruent, and that this ought to be a surprise.
• A concrete instance. Base BC = 8 cm with D at 4 cm, apex A placed 6 cm above the line but only 1 cm horizontally from B. Verified: both halves have area 12 cm²; AB = √37 ≈ 6.08 cm while AC = √85 ≈ 9.22 cm, so the two triangles are visibly different and identical in area. Use numbers this lopsided — a nearly isosceles picture hides the whole point.
• Think and Reflect (p. 133). Can one of the two halves be cut by straight cuts into pieces that exactly cover the other? The chapter then says it is possible and deliberately refuses to say how few pieces are needed, inviting students to find out.
• Think and Reflect (p. 134), first half. The same question for any two polygons of equal area, with three cases to try: a square against a non-square rectangle of equal area; two differently shaped triangles of equal area; a triangle against a square of equal area. Students are asked to form their own conjecture.
• The answer to that conjecture, which the chapter does not give. Any two polygons of equal area can be cut into finitely many pieces that reassemble into one another. This is a real theorem and its Class 9 form is: it is always possible. The explanation may say so, must flag it as an addition, and should mention that the corresponding statement for solids is false — which is why the chapter is right to leave it as a conjecture rather than a fact to be assumed.
• Think and Reflect (p. 134), second half. Rectangles of perimeter 40 units, sides not required to be whole numbers. How many are there; is there one of largest area, with its dimensions; is there one of smallest area, with its dimensions, and is either answer a surprise? Verified: infinitely many; the largest is the 10 × 10 square with area 100; and there is no smallest — the area can be made as close to zero as you like without ever reaching it, so the second surprise is that the question has no answer. Say that plainly. An explanation that invents a smallest rectangle here teaches a falsehood.
• Exercise Set 6.2 Q1 (p. 142), Fig. 6.31. A rectangle with A top-left, B top-right, C bottom-right, D bottom-left, measuring 10 cm across the bottom and 8 cm up the right side, with a point E on side BC and triangle ADE shaded. Find the area of triangle ADE. Verified: 40 cm², base AD = 8 cm and height 10 cm — and the answer does not depend on where E sits on BC, which is the reason the question is first in the set.
• Exercise Set 6.2 Q9 (p. 143), Fig. 6.32. D is the midpoint of BC in triangle ABC, the median AD is drawn, and P is any point on AD; show the areas of triangles ABP and ACP agree. Verified: apply the median theorem twice, to ABC and to PBC, and subtract. The printed figure colours the two triangles green and red.
• Exercise Set 6.2 Q10 (p. 143), Fig. 6.33. In square ABCD with A and D on top and B, C below, a point P is taken inside and joined to all four corners; the two triangles on one pair of opposite sides are coloured red and the two on the other pair green. Find the ratio of red area to green area. Verified: 1 : 1 — each pair sums to half the square, because the two heights in each pair add to the side.
• Exercise Set 6.2 Q11 (p. 143), Fig. 6.34. Take AB halved at D in a triangle ABC; put P somewhere along BC; then place Q on AB so that CQ comes out parallel to PD. Prove that triangle BPQ has half the area of ABC. Verified: triangles QCD and QCP share the base QC and have their apexes on a line parallel to it, so they are equal in area; substituting one for the other turns triangle BPQ into triangle BDC, which the median theorem makes half of ABC. In the printed figure Q lies between A and D, and the bookkeeping depends on that.
• End-of-chapter Q15 (pp. 150–151), three parts. A rectangle of sides a, b against one of sides 2a, 2b; a triangle of sides a, b, c against one of sides 2a, 2b, 2c; and the same triangle against one of sides 3a, 3b, 3c. Each part asks for the area multiple and then whether that many copies of the small figure actually fit inside the large one. Verified: multiples 4, 4 and 9; and yes in all three cases — the triangle splits into four by joining the midpoints of its sides, and into nine by trisecting them.
• End-of-chapter Q20 (p. 152), Fig. 6.48. Lines drawn from one vertex to the two points trisecting the opposite side, cutting the triangle into three; the leftmost is shaded blue and the rightmost red. Show the two shaded areas are equal, then cut one up to cover the other. Verified: equal thirds — the three sub-triangles have equal bases and the same height, the median theorem's argument with three parts instead of two.
• End-of-chapter Q16 (p. 151), Figs. 6.43 and 6.44. Two "what fraction is shaded?" figures with tick marks indicating equal divisions of the sides. In Fig. 6.43 the shaded piece inside the triangle is four-sided, not a triangle: one corner of the outer triangle together with three marked points on its sides. In Fig. 6.44 the shaded piece is a tilted quadrilateral that is itself a square. Both are median-theorem reasoning applied several times; the tick marks are the only data, and they are inside the artwork.

Figures to have open

• Fig. 6.20A (p. 132) in both its panels, redrawn and able to be shown moving, with the base split visible.
• Fig. 6.20B (p. 132), redrawn, together with the extension of the base and the subtraction — the second half is added here and the printed figure stops short of it.
• Fig. 6.21A → Fig. 6.21B (p. 132) as one movement: two triangles rotating into a parallelogram. Essential.
• Fig. 6.22 (p. 133), redrawn with a deliberately lopsided apex so the two halves are obviously different shapes. The printed figure is already lopsided; keep it that way.
• Fig. 6.31 (p. 142), with E draggable along BC and the area readout staying at 40 cm². This is an added extension of a printed figure and it is the cheapest way to make the thesis land.
• Fig. 6.33 (p. 143), the square with the interior point and the four triangles in two colours. Note that all the shading information is in colour, not in labels.
• Figs. 6.43 and 6.44 (p. 151) if section 11 is extended; their tick marks must be redrawn, as they carry the only data.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics (NCF-SE 2023), Chapter 6, §6.8 "Area of a Triangle" (pp. 132–134), including the printed Theorem and the subheading on the median (p. 133), and the two Think and Reflect boxes on pp. 133 and 134.
• Figures 6.20A, 6.20B, 6.21A, 6.21B (all p. 132) and 6.22 (p. 133).
• Exercise Set 6.2 Q1 with Fig. 6.31 (p. 142); Q9 with Fig. 6.32, Q10 with Fig. 6.33 and Q11 with Fig. 6.34 (p. 143).
• End-of-chapter exercises Q15 (pp. 150–151), Q16 with Figs. 6.43 and 6.44 (p. 151) and Q20 with Fig. 6.48 (p. 152).
• Chapter Summary (p. 154) carries the half-base-times-height bullet but not the median theorem.
• The parallelogram formula this topic depends on is §6.7 (pp. 130–132), handled in From rectangle to parallelogram: area survives rearrangement; the alternative side-based formula is §6.8.1 (pp. 134–137), handled in Heron's formula: area from the three sides alone.