Chapter 11.3 — Exercise 11.2 — Parallelograms

Seeing "Same Base, Same Parallels" Directly

Both parallelograms sketched above share the base AB and both fit entirely between the same two dashed parallel lines, even though one leans noticeably more than the other — and their areas are still exactly equal. This is precisely what the first theorem guarantees: base × height determines a parallelogram's area completely, and both the base and the perpendicular distance between the parallels are shared by any two figures drawn this way, regardless of how far either one leans sideways.

Sliding the top edge of a parallelogram sideways while keeping it running along the same upper parallel line is really the physical picture worth holding onto here: the shape stretches and leans differently at every position, but the base never moves and the perpendicular gap to the top line never changes either, so the area — base times that fixed gap — stays locked no matter how far the slide goes.

Recovering a Height From a Known Area

Parallelogram ABCD has area 36 cm² and AB = 4.2 cm; find the height of parallelogram ABEF, which shares base AB with ABCD and sits between the same parallels. Since area = base × height, 4.2 × height = 36, giving height ≈ 8.57 cm for ABCD. Since ABCD and ABEF share the same base and the same parallels, the first theorem guarantees their areas match — and since they also share the same base length, their heights must match too, so ABEF's height is also 8.57 cm, recovered without ever needing ABEF's own area stated separately at all. This is a genuinely different kind of "equal height" argument than simply measuring both figures directly — it leans entirely on the theorem's guarantee that equal base and equal area together force equal height, rather than on any direct geometric measurement of ABEF ever being taken.

The Same Parallelogram, Two Different Bases

Parallelogram ABCD has AB = 10 cm with AE ⊥ DC (AE = 8 cm), and CF ⊥ AD (CF = 12 cm); find AD. Using AB as the base: area = AB × AE = 10 × 8 = 80 cm². Using AD as the base instead: area = AD × CF, so 80 = AD × 12, giving AD ≈ 6.67 cm. A single parallelogram has exactly one area, but it has as many base-height pairs as it has sides — this problem exploits that directly by computing the same fixed area twice over, once from each available base-height pair, purely to recover a length that was never actually given directly in the original statement at all.

Splitting a Parallelogram's Interior Point in Two Ways

P and Q lie on sides DC and AD of parallelogram ABCD; prove area of △APB = area of △BQC. Taking AB as base: △APB and parallelogram ABCD share base AB and sit between the same parallels AB and DC, so area of △APB = ½ × area of ABCD. Taking BC as base instead: △BQC and parallelogram ABCD share base BC and sit between the same parallels AD and BC, so area of △BQC = ½ × area of ABCD too. Since both triangles equal the identical half of the same parallelogram's area, area of △APB = area of △BQC.

Neither triangle needs comparing directly to the other at any point in this proof — both are separately measured against the same fixed reference, the parallelogram's own area, which makes them equal to each other purely by both being equal to that same third quantity in turn — a genuinely useful proof pattern worth naming explicitly: whenever two things each need comparing against a common reference rather than against each other directly, proving both separately equal to that shared reference is often the cleanest available route to the final result.

Deriving the Trapezium Formula From Scratch

Trapezium ABCD has parallel sides AB = a and DC = b, with perpendicular distance h between them; prove its area is ½(a + b)h. Diagonal BD cleanly splits ABCD into △ABD and △BCD, both sharing the exact identical height h (the fixed perpendicular distance between the two parallel sides). Area of ABCD = area of △ABD + area of △BCD = ½ × a × h + ½ × b × h = ½h(a + b).

This derivation is worth comparing against the general quadrilateral formula from the introduction, ½ × diagonal × (h₁ + h₂): here the two triangles' heights are equal (both simply h, the distance between the parallels) rather than two separate perpendiculars onto the diagonal, which is exactly what collapses the more general formula down into this much simpler, trapezium-specific version instead.

A Farmer's Field, Split Into Equal Thirds

Parallelogram-shaped field PQRS has A as the midpoint of RS, joined to P and Q; the field splits into △PSA, △PAQ, △QAR. Since △PAQ and PQRS share base PQ between the same parallels, area of △PAQ = ½ × area of PQRS — meaning △PSA and △QAR together make up the other half, so area of △PSA + area of △QAR = area of △PAQ exactly. If the farmer wants to sow groundnuts equal to the combined pulses-and-paddy area, groundnuts belong in △PAQ alone, with pulses and paddy carefully splitting the two remaining triangles evenly between themselves instead.

Four Right Triangles Inside a Rhombus

Rhombus ABCD has diagonals AC = d₁ and BD = d₂ crossing at O; prove its area is ½d₁d₂. Since a rhombus's own diagonals always bisect each other at O, OA = OC = ½d₁ and OB = OD = ½d₂. The whole rhombus splits neatly into four triangles — AOB, BOC, COD, DOA — each with area ½ × ½d₁ × ½d₂ = ⅛d₁d₂.

This is the promised proof behind the rhombus formula stated without derivation back in the introduction — a rhombus's diagonals cross at exactly 90°, so each of the four small triangles is right-angled with its two legs being the two half-diagonals, which is precisely what makes ½ × leg × leg the correct area formula for each individual piece, before all four pieces get carefully summed back together into the final result.

From Parallelograms to Triangles Directly

Every theorem here compared a parallelogram to another parallelogram, or a parallelogram to a triangle sharing its base. Exercise 11.3 closes the chapter by comparing two triangles directly against each other, on the same base and between the same parallels, without any parallelogram at all acting as the reference point in between the two.