Why the area of a triangle is half base times height

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What to assume they know

Area as a count of unit squares, and the rectangle formula — Area as a count of unit squares
That a rectangle's diagonal produces two congruent triangles, each half of it — Area as a count of unit squares
Congruence, and at least the RHS test for right-angled triangles
What a perpendicular is, and how to drop one from a point to a line
Parallel lines, and that the distance between two parallels is constant
Multiplying and dividing simple decimals and fractions

What they should be able to do

Show that a triangle on one side of a rectangle, with its apex anywhere on the opposite side, is half that rectangle
Construct the enclosing rectangle for a given triangle from a chosen base: two perpendiculars at the base's ends and a parallel through the apex
Justify that the figure so constructed is a rectangle, and that its shorter side equals the triangle's height
Compute a triangle's area from any one side and the height belonging to that side, and give the answer with a squared unit
Handle the case in which the foot of the perpendicular falls outside the chosen base, by writing the triangle as the difference of two others
Given a triangle's area and one side, recover the height belonging to that side
Use two different base-height pairs on one triangle to find an unknown altitude
Convert a given triangle into a rectangle of equal area and back again, and say which measurement each conversion preserves
Compare the areas of a triangle and a square built on the same sidelength, and justify the comparison without computing any square root

Where it usually goes wrong

"The height is the slanted side." It is the perpendicular distance from the base to the opposite vertex, and in items (ii) and (iii) of Part II p.157 those two things are visibly different. This is the single most common error on the topic.
"The height has to be inside the triangle." Part II p.154's lower figure and Part II p.158's second item both put the foot outside the base. The formula does not care; the picture does.
"½ base times height is a rule you memorise." It is a rectangle halved by its diagonal, twice over. If the explanation cannot show the two little rectangles, it has not taught the topic.
"A triangle has one base." It has three, each with its own height, and the three products all come to the same thing. Part II p.155 exists to prove it.
"Half of the rectangle means half the drawn unit squares are whole." The diagonal slices through squares. The halving comes from congruence, not from the grid being obliging.
"You need all three sides to find the area." You need one side and the height belonging to that side. Mixing a side with the wrong height is the second most common error, and Part II p.155 is where it shows up, because three numbers are given and only two of them pair.
"If the base cannot be boxed, the formula fails." Part II p.154 handles it by subtraction, and the subtraction works because the three points lie on one line in a known order.
"An equilateral triangle and a square on the same side must be close in area." The square is more than twice the triangle. The height-less-than-side argument settles it in one line.
"Remaking a shape means keeping its sides." The Śulba-Sūtras problems keep the area and change everything else. Say which quantity is being preserved each time, or the exercises look arbitrary.

Questions to check understanding

Find a triangle's area from a base and its height, with the height drawn inside and then with it drawn outside
Given the area and one side, find the height belonging to that side
Given two sides and the altitude to one of them, find the altitude to the other
Find the area of a right-angled triangle from its two legs, and say why no extra construction is needed
Decide which of two triangles drawn in congruent rectangles has the greater area, with a reason
Describe a construction that turns a given triangle into a rectangle of equal area, or of twice the area
Compare a triangle and a square built on the same sidelength, with a reason that does not compute a square root
Explain where the ½ in the triangle formula comes from — the reasoning item that separates a student who has understood the chapter from one who has memorised it

Examples worth working on the board

Items marked printed are stated or worked on the page; items marked not in the book are worked out here or an added argument on the chapter's inputs and must not be presented as something the chapter states.

Two triangles, two identical rectangles (Part II p.153, first pair of figures). Rectangle ABCD with A top left, B top right, C bottom right, D bottom left. In the left copy X sits on AB nearer A; in the right copy Y sits on AB near the middle. The question asks which of ∆XDC and ∆YDC has the greater area. Printed: dropping the altitudes from X and from Y makes it clear that each triangle is exactly half of rectangle ABCD — so neither is greater.
The same question with the apex on another side (Part II p.153, second pair of figures). In the left copy X is on AB near B; in the right copy Y is on AD, on the left-hand side. The comparison asked for is ∆XDC against ∆YBC. Printed: the same altitude argument settles it; both are half the rectangle. Note: the two triangles do not even share a base, and they are still equal. That is worth pausing on, because it is the seed of Triangles with the same base and height have the same area.
Fig. 7.1 (Part II p.153, the chapter's only numbered figure). Rectangle ABCD, X on AB, Y the foot of the perpendicular from X down to DC. The marked lengths are 4 along the left side from D up to A, and 5 along DC. The printed instruction is to find the area of ∆XDC. Not in the book: ½ × 5 × 4 = 10 sq. units. Printed: the chapter then asks what measurements a triangle's area needs, and answers that the sidelengths of the outer rectangle are what is wanted.
Building the outer rectangle (Part II p.153, bottom strip of three small figures). Given ∆ABC with BC at the bottom: first drop perpendiculars to BC at B and at C; then draw the line l through A parallel to BC, meeting those two perpendiculars at E and D. Printed as three stages with the parallel marked l ∥ BC.
The derivation (Part II p.154). Printed: BCDE is a rectangle, with the question "how?" left to the student; its sides are taken as the base and the height; BXAE is also a rectangle, so the rectangle's height equals the triangle's height; and the boxed result is that a triangle's area is half its base times its height. Not in the book, and the argument the explanation exists to make: the altitude AX splits ∆ABC into ∆ABX and ∆AXC; ∆ABX is half of rectangle BXAE and ∆AXC is half of rectangle XCDA, because each is cut off by that rectangle's own diagonal; add the two halves and you have half of BCDE. The ½ in the formula is the ½ from Part II p.150, used twice.
The awkward triangle (Part II p.154, lower figure). ∆ABC drawn so that the foot D of the perpendicular from A falls outside the segment BC, to the left of B; the height is marked h. Printed: ∆ABC is what is left when ∆ADB is taken away from ∆ADC, and either of those two may be boxed as in Fig. 7.1; the printed three-line algebra is ½h·DC − ½h·DB, then ½h(DC − DB), then ½h·BC, with the conclusion that the formula holds for every kind of triangle. Note what carries the argument: DC − DB = BC only because D, B and C lie on one line in that order.
Two base-height pairs on one triangle (Part II p.155, under the subheading "Some Applications of the Area Formula"). ∆ABC with X the foot of the altitude from A to BC and Y the foot of the altitude from B to AC. The marked lengths are 3 for AX, 5 for BC and 4 for AC, and the instruction is to find BY. Printed: the area is ½ × AX × BC = 15/2 sq. units; the same area written with the other pair is ½ × BY × AC = ½ × 4 × BY = 2·BY; setting the two equal gives 2·BY = 15/2 and BY = 15/4 = 3.75 units. This is the strongest single piece of evidence in the chapter that a triangle has one area and three ways of computing it.
Three triangles to measure (Part II p.157, Figure it Out 1). (i) ∆ABC with E the foot of the altitude from A; the altitude AE is 3 cm and the side BC is 4 cm. (ii) ∆DEF with N the foot of the perpendicular from D onto EF; that perpendicular DN is 3.2 cm and the side EF is 5 cm — so this one asks for a slanted side to be used as the base. (iii) ∆NAT with the right angle at A, the leg NA 4 cm and the leg AT 3 cm. Not in the book: 6 cm², 8 cm² and 6 cm². Items (i) and (iii) look nothing alike and come out the same, which is the point of setting them together.
An altitude to find, with the foot outside (Part II p.158, Figure it Out 2). ∆ABC with X the foot of the perpendicular from A to line BC, lying to the left of B; AX is 4 units, BC is 6 units, AC is 8 units, and Y is the foot of the perpendicular from B to AC. Not in the book: the area is ½ × 6 × 4 = 12 sq. units, so ½ × 8 × BY = 12 and BY = 3 units. This item is the obtuse case and the two-pairs trick in one figure.
The Śulba-Sūtras framing (Part II p.158). Printed: the chapter explains that these ancient Indian geometric texts deal with altar construction, that an altar had to have both a prescribed shape and a prescribed area, and that this is what generates problems asking for one shape to be remade as another of the same area; it adds that Euclid's Elements poses and solves problems of the same kind. Two such items follow: item 4, remake a rectangle as a triangle of equal area, and item 5, remake a triangle as a rectangle of equal area. Not in the book for item 5: keep the base and halve the height, or halve the base and keep the height — and the cut that does it physically is to slice along the mid-height line and swing the two upper corner pieces outward.
More of the same, in the parallelogram exercise set (Part II p.163 item 5; Part II p.164 items 6, 7 and 8). Printed: a rectangle of twice a given triangle's area, with the follow-up question of how many different methods the student can find; a rectangle whose area matches a given triangle's; the note that an isosceles triangle can be remade as a rectangle more simply still, with ∆ABC drawn on a base BC and D the foot of the altitude from A, and the printed hint that ∆ADB and ∆ADC can each be made into half of a rectangle; and the reverse problem, a rectangle remade as an isosceles triangle.
Triangle against square on one sidelength (Part II p.164, Figure it Out 9). Printed: the item asks whether an equilateral triangle beats a square built on the same side, then whether two of those triangles together beat it, and it asks for reasons both times. Not in the book, and no square roots required: the equilateral triangle's height is a leg of a right triangle whose hypotenuse is the side, so the height is strictly less than the side. Therefore the triangle's area, half the side times that height, is less than half the square's area — so the square wins, and even two of the triangles together come to less than the square. This is the argument to show; computing √3/4 buys nothing at this class and hides the reason.

Figures to have open

Fig. 7.1 (Part II p.153) with the numbers 4 and 5 and the right angle at Y. This is the chapter's anchor figure and must be redrawn accurately.
The three-stage construction of the outer rectangle (Part II p.153, bottom). Redraw as three panels; the third must show the parallel line marked and the two right angles at the base.
The exploded box (Part II p.154 plus an addition made here). The rectangle BCDE with the altitude AX drawn, the two sub-rectangles tinted differently, and the two triangle halves shown lifting out. The chapter does not draw this; it is the figure the whole derivation needs.
The obtuse triangle with the external foot (Part II p.154, lower figure), with ∆ADC and ∆ADB outlined separately so the subtraction is visible.
The two-altitude triangle (Part II p.155), with 3, 5 and 4 in place and both altitudes drawn. Must show clearly which number pairs with which.
The three exercise triangles (Part II p.157, item 1). Redraw all three with their labels intact; in (i) and (ii) the marked side is the whole base, which the printed label placement does not make obvious.
The isosceles-triangle dissection (Part II p.164, item 7) as a schematic with the two half-pieces shown swinging into place.
No photograph is needed anywhere in this topic.

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 "Triangles", Part II pp.153–154, for the two comparison questions, Fig. 7.1, the construction of the outer rectangle, the boxed formula and the obtuse case.
Part II p.155, under the smaller-tier subheading "Some Applications of the Area Formula", for the two-base-height-pairs worked example. The rest of that page — the four triangles cut by a rectangle's diagonals, and the printed general statement about a vertex joined to a midpoint — belongs to Triangles with the same base and height have the same area.
Part II pp.157–158, Figure it Out items 1 to 5, and the printed paragraph on Part II p.158 about the Śulba-Sūtras and Euclid's Elements. Items 6, 7 and 8 of that block belong to Triangles with the same base and height have the same area and to Any polygon is a pile of triangles.
Part II p.163 item 5 and Part II p.164 items 6 to 9, which sit inside the parallelogram exercise set but are triangle-to-rectangle problems and are carried here. Area of a parallelogram, by turning it into a rectangle notes the handover.
Part II p.171, the chapter SUMMARY, first bullet, for the triangle formula, and the boxed statement on Part II p.154 where it first appears. The SUMMARY box shares Part II p.171 with the last of the unit-conversion questions.
Backward pointer: Area as a count of unit squares for the count-of-unit-squares definition and for the half-a-rectangle fact on Part II p.150 that this whole derivation is built on.