Area of a rhombus from its diagonals

What to assume they know

• Base times height for a parallelogram, obtained by dissection — Area of a parallelogram, by turning it into a rectangle
• Half base times height for a triangle, with any side available as the base — Why the area of a triangle is half base times height
• That the diagonals of a rhombus bisect each other at right angles
• That an isosceles triangle can be remade as a rectangle of equal area — the construction set on Part II p.164, carried by Why the area of a triangle is half base times height
• Adding and multiplying fractions, and collecting a common factor out of a sum
• Optional, for the consistency check in section 9: the Baudhāyana-Pythagoras relation from Part II printed Chapter 2

What they should be able to do

• State why a rhombus needs no separate area formula, and why it is given one anyway
• Name the two properties of a rhombus that the diagonal derivation uses, and say which is used where
• Split a rhombus along one diagonal into two isosceles triangles on a common base
• Remake each of those triangles as a rectangle of equal area, and join the two rectangles into one
• Read off the sides of the resulting rectangle in terms of the two diagonals, and derive the formula from them
• Derive the same formula the second way, by adding the areas of two triangles that share a base
• Compute a rhombus's area from its two diagonals
• Reconcile the two available formulae for one rhombus, and use the reconciliation to find its height
• State the weakest condition under which half the product of the diagonals gives a quadrilateral's area

Where it usually goes wrong

• "Half the product of the diagonals works for any quadrilateral." It needs the diagonals to cross at right angles. For a general quadrilateral the diagonals do not, and the formula is simply wrong.
• "It works for any parallelogram." A leaning parallelogram's diagonals are not perpendicular, and the formula fails. This is the error the chapter's ordering makes most likely, because the rhombus section opens by saying a rhombus is a parallelogram.
• "The two formulae are alternatives — use whichever you have numbers for." They must agree, and making them agree tells you something new. Section 9 is where that becomes a technique rather than a slogan.
• "The diagonals of a rhombus are equal." Equal diagonals make it a square. In the worked case they are 20 cm and 15 cm.
• "The diagonals of a rhombus are its sides' lengths." Both diagonals are shorter than two sides, and the longer diagonal always exceeds a side — it is at least s√2, since for vertex angle θ the diagonals are 2s·sin(θ/2) and 2s·cos(θ/2) and the larger of those is minimised at θ = 90°. But do not say both exceed a side: the shorter diagonal can be far shorter than a side in a squashed rhombus, and at θ = 20° it is only 0.35 s. In the worked case the two diagonals, 20 cm and 15 cm, do both exceed the 12.5 cm side — which is a fact about that rhombus, not a general rule.
• "The rectangle in the dissection has the diagonals as its sides." It has one whole diagonal and half the other. Getting this wrong doubles the answer, and it is the most likely slip in reproducing the derivation.
• "The dissection only works if the rhombus is drawn point-up." The chapter draws it point-up on Part II p.164 and at a lean on Part II p.165, in the same argument, precisely so this does not stick.
• "Perpendicular diagonals mean bisecting diagonals." They are independent conditions. A kite-shaped quadrilateral has one and not the other, and the formula still holds — which is the content of section 10.

Questions to check understanding

• Find a rhombus's area from its two diagonals
• Find one diagonal from the area and the other diagonal
• Find a rhombus's side from its two diagonals, and then its height
• Given a rhombus's side and height, find the product of its diagonals
• Decide whether half the product of the diagonals may be used for a given quadrilateral, and say why
• Describe a dissection that turns a rhombus into a rectangle of equal area, naming the two properties it uses
• Describe a dissection that turns a rectangle into an equal-area rhombus
• Derive the diagonal formula by adding the areas of two triangles, and say which property of the rhombus each step needed — 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 opening concession (Part II p.164, under the subheading "Rhombus"). Printed: because a rhombus is a parallelogram, the parallelogram formula already covers it; but its extra properties allow a different dissection into a rectangle holding the same amount, and the chapter attributes that second method to one of the Śulba-Sūtras. It then sets the whole dissection as work to be done by the student, with the printed instruction to try it.
• The rhombus as drawn (Part II p.164, first figure). Rhombus ABCD with A at the top, B at the left, D at the right and C at the bottom, and O at the centre where the diagonals cross. So AC is the upright diagonal and BD the crosswise one. The right angle at O is marked; single tick marks on the four sides AB, AD, BC and CD mark those equal, and double tick marks on BO and OD mark the two halves of diagonal BD equal. The two mark types carry the two different facts the derivation needs, and "half-sides" names nothing in the figure. The upper triangle ABD is tinted one colour and the lower triangle CBD another.
• The four pieces (Part II p.164, second and third figures, continuing onto Part II p.165). Printed: the two tinted triangles are each split along the upright diagonal into two right triangles, and the four right triangles are then drawn separated, each with the right angle at O marked and the equal half-diagonals tick marked.
• The reassembly (Part II p.165, top strip). Printed: each of the two isosceles triangles becomes an equal-area rectangle, and those two rectangles are then joined into one, drawn with the upper half in one tint and the lower half in the other. Printed reasoning: ABCD being a rhombus, its four sides match and each diagonal squarely halves the other, which makes ∆ABD and ∆CBD isosceles; each of those two becomes an equal-area rectangle, and the pair join into one rectangle, labelled WXYZ, holding as much as the rhombus does. In the printed figure X and Y are the upper corners and W and Z the lower ones.
• The sides of that rectangle (Part II p.165). Printed: XW equals the length of diagonal AC, and WZ is one half of the remaining diagonal BD. The printed chain then runs: the rhombus's area equals the rectangle's, which is XW × WZ, which is AC × BD/2, which is half of AC × BD. Printed boxed result: a rhombus's area is half the product of its diagonals. Not in the book: the half in that formula is not a mysterious extra factor — it is sitting in the width of the rectangle. Each of the two half-rectangles is one half-diagonal tall and half the other diagonal wide, so stacking them adds the heights and leaves the width alone. Say that out loud, because the algebra hides it.
• The second route (Part II pp.165–166). ∆ADB and ∆CDB, drawn on a rhombus ABCD set at a lean with A upper left, B upper right, D lower left and C lower right, and the right angle at O marked. Printed: the area of ∆ADB is half of AO times BD and the area of ∆CDB is half of CO times BD, and the rhombus is their sum; the chapter then instructs the student to simplify and confirm that the same formula comes out. Not in the book simplification: take out the common half and the common BD to get half of BD times (AO + CO), and AO + CO is the whole diagonal AC, so the result is half of AC times BD as before.
• Why the second route is stronger (not in the book; the chapter does not remark on it). Look at what the second derivation actually used. It took BD as the base of both triangles and AO and CO as their heights — and that step is legitimate for the single reason that AC meets BD at a right angle. It never used the equal sides, and it never used the fact that O is the midpoint of BD. So the same argument gives half the product of the diagonals for any quadrilateral whose diagonals cross squarely, bisecting or not. This is the chapter's own working, one observation further on, and it is the best answer to a student who asks what the formula is really about.
• Diagonals given (Part II p.169, Figure it Out 1). Printed: the diagonals of a rhombus are given as 20 cm and 15 cm, and its area is asked for. Not in the book: half of 20 × 15 is 150 cm².
• The consistency check (not in the book; the chapter sets nothing of the kind). The same rhombus must also satisfy base times height. Its half-diagonals are 10 cm and 7.5 cm, and they meet at a right angle, so each side is 12.5 cm — a scaled 3, 4, 5 triangle, which is why the numbers come out clean. Then 12.5 × height = 150, so the height is 12 cm. Two formulae, one figure, and the agreement pins down a length that neither formula was asked for. In general, for a rhombus of side s the height must be the product of the diagonals divided by twice s.
• Rectangle into rhombus (Part II p.169, Figure it Out 2). Printed: the item asks for a dissection turning a rectangle into an equal-area rhombus. Not in the book: the whole construction of Part II pp.164–165 run in reverse — halve the rectangle across, and remake each half-rectangle as an isosceles triangle on the long side, which is the problem set on Part II p.164 as item 8.
• A third way to see the formula (not in the book; not in the chapter). Draw the rectangle whose sides are parallel to the two diagonals and just contain the rhombus: its sides are the two diagonals, so its area is their product. Each of the rhombus's four right triangles has a congruent partner in the corner of that box outside the rhombus. So the rhombus is exactly half of the box, and the formula reads directly off the picture with no algebra at all. This is the version to leave a student with.

Figures to have open

• The rhombus with its diagonals, both properties marked (Part II p.164). Needs the right angle at the centre and the four tick marks; the argument uses both.
• The four-piece explosion and the reassembly into WXYZ (Part II pp.164–165). This is the printed figure most worth following closely, including the two tints, because the colours are what let a student see which half of WXYZ came from which triangle. Redraw as a movement rather than as the book's static strip.
• The leaning rhombus for the second route (Part II p.165, lower right), with BD marked as the shared base and AO and CO marked as the two heights.
• The bounding box whose sides are the diagonals. Not in the chapter; needed for the third derivation in section 10. Draw the four outside corner triangles tinted to match their partners inside.
• A quadrilateral with perpendicular but non-bisecting diagonals, beside a leaning parallelogram whose diagonals are not perpendicular. Not in the chapter; needed for section 10's contrast.
• 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 "Rhombus", Part II pp.164–166. The subheading begins in the lower half of Part II p.164, the dissection figures run across Part II pp.164–165, the boxed formula is on Part II p.165, and the second derivation ends with the simplification instruction at the head of Part II p.166.
• Part II p.164, Figure it Out items 7 and 8, for the isosceles-triangle dissection that section 4 depends on and for its reverse. Those items are printed in the parallelogram exercise set and are carried by Why the area of a triangle is half base times height.
• Part II p.169, Figure it Out items 1 and 2. The remaining items of that block belong to Area of a trapezium, derived two different ways.
• Part II p.158 for the Śulba-Sūtras framing, owned by Why the area of a triangle is half base times height.
• Part II p.160, closing line, where the chapter announces the three special formulae.
• Part II p.171, the chapter SUMMARY, fourth bullet, for the rhombus formula.
• Backward pointers: Area of a parallelogram, by turning it into a rectangle for base times height and for the meaning of dissection; Why the area of a triangle is half base times height for half base times height and for the isosceles construction.
• Part II printed Chapter 2, "The Baudhayana-Pythagoras Theorem", supplies the relation used only in section 9's consistency check, to get a side of 12.5 cm from half-diagonals of 10 cm and 7.5 cm.