14.4 Special Shapes by Dissection

For a parallelogram, a rhombus and a trapezium we can write short, easy-to-remember formulas. Each one springs from the same powerful idea: cut the shape into pieces and rearrange them into a rectangle of equal area. This cut-and-shuffle technique is called dissection, and many such constructions appear in the ancient Indian Śulba-Sūtras.

The parallelogram

Take a parallelogram ABCD with AB \parallel DC. Drop a perpendicular AX from A down to the base DC, this is a height of the parallelogram. Slice off the right triangle \triangle AXD and slide it over to the opposite end; it fits perfectly (by the RHS congruence of \triangle AXD with the triangle that completes the figure) and reshapes the parallelogram into a rectangle of equal area. That rectangle has width equal to the base DC and height equal to AX. Hence:

\boxed{\;\textbf{Area of a parallelogram} = \text{base} \times \text{height}.\;}

base height

Note that the height is the perpendicular gap between the two parallel sides, not the slanting side. Any side can act as the base, as long as you pair it with its own perpendicular height; the area always comes out the same.

Math Talk Picture sliding the top edge of a parallelogram sideways, further and further, while holding its base and height fixed. The area never budges (\text{base} \times \text{height} stays put), but the slanting sides stretch longer and longer, so the perimeter grows without bound! Same area, ever-larger perimeter.

Worked Example

In parallelogram PQRS the base SR = 15\text{ cm}, the height to that base is 8\text{ cm}, and the side PS = 10\text{ cm}. The height drawn onto side PS is QN. Find QN.

Area, using base SR and its height 8:

\text{Area} = 15 \times 8 = 120 \text{ cm}^2.

The same area, using base PS = 10 and height QN:

120 = 10 \times QN \quad\Rightarrow\quad QN = \frac{120}{10} = 12 \text{ cm}.

Figure it Out: Parallelograms

Practice

  1. Several parallelograms are drawn on the same base with the same height. (i) What can you say about their areas? (ii) What about their perimeters, which is largest, which smallest?
  2. Find the areas of these parallelograms: (i) base 8\text{ cm}, height 5\text{ cm}; (ii) base 6\text{ cm}, height 4\text{ cm}; (iii) base 5.5\text{ cm}, height 6\text{ cm}.
  3. In parallelogram PQRS, base SR = 15\text{ cm}, height 8\text{ cm}, side PS = 10\text{ cm}. Find the height QN to side PS.
  4. A rectangle and a parallelogram both have side lengths 6\text{ cm} and 4\text{ cm}. Which has the greater area?
  5. Describe a method to build a rectangle whose area is twice that of a given triangle.
  6. [Śulba-Sūtra] Describe a method to build a rectangle of the same area as a given triangle.
  7. [Śulba-Sūtra] Show how an isosceles triangle can be dissected into a rectangle.
  8. [Śulba-Sūtra] Turn a rectangle into an isosceles triangle by dissection.
  9. Which has the greater area, an equilateral triangle or a square of the same side length? And, two such triangles together, or that one square?
  1. (i) All have equal area, since area = base \times height and both are shared. (ii) Their perimeters differ: the most slanted (most stretched) parallelogram has the largest perimeter, while the upright rectangle on that base has the smallest.
    1. 8 \times 5 = \mathbf{40\text{ cm}^2}. (ii) 6 \times 4 = \mathbf{24\text{ cm}^2}. (iii) 5.5 \times 6 = \mathbf{33\text{ cm}^2}.
  3. Area = 15 \times 8 = 120\text{ cm}^2; then QN = 120 \div 10 = \mathbf{12\text{ cm}}.
  4. The rectangle has the greater area. On the shared base 6\text{ cm}, the rectangle’s height is the full 4\text{ cm}, giving 24\text{ cm}^2. The parallelogram’s perpendicular height is less than its slanting side 4\text{ cm}, so its area is below 24\text{ cm}^2.
  5. Build a rectangle on the same base as the triangle with the same height as the triangle: its area = b \times h, which is twice the triangle’s \tfrac12 bh.
  6. Reverse the direction: on the triangle’s base b, build a rectangle of height \tfrac{h}{2} (half the triangle’s height); area = b \times \tfrac{h}{2} = \tfrac12 bh, equal to the triangle. Use the midline dissection to physically rearrange the pieces.
  7. Drop the altitude AD from the apex of the isosceles triangle to the base, splitting it into two congruent right triangles \triangle ADB and \triangle ADC. Each is half of a rectangle; reflect one across AD and set the two right triangles together to form a rectangle of width \tfrac12 \times \text{base} and height AD.
  8. Reverse the previous dissection: cut the rectangle along a line through suitable midpoints and reassemble the two right triangles base-to-base into an isosceles triangle of the same area.
  9. A square of side s has area s^2. An equilateral triangle of side s has area \tfrac{\sqrt3}{4}s^2 \approx 0.433\,s^2, which is less than the square. Two such triangles give \tfrac{\sqrt3}{2}s^2 \approx 0.866\,s^2, still less than the square. So the square wins both contests.

The rhombus

A rhombus is a parallelogram, so base \times height still applies. But a rhombus carries a bonus property, its diagonals are perpendicular bisectors of each other, which gives a tidy diagonal formula.

Cut the rhombus ABCD along its diagonals into four right triangles and rearrange them (a Śulba-Sūtra dissection) into a rectangle WXYZ whose width is one full diagonal AC and whose height is half the other diagonal BD. So:

\text{Area of rhombus} = AC \times \tfrac{BD}{2} = \tfrac12 \times AC \times BD.

\boxed{\;\textbf{Area of a rhombus} = \tfrac12 \times (\text{product of the diagonals}).\;}

You can also see this by slicing the rhombus along one diagonal into two triangles and summing:

\text{Area} = \tfrac12 \times AO \times BD + \tfrac12 \times CO \times BD = \tfrac12 \times BD \times (AO + CO) = \tfrac12 \times BD \times AC,

since AO + CO = AC. Same formula.

Worked Example

Find the area of a rhombus whose diagonals measure 24\text{ cm} and 18\text{ cm}.

\text{Area} = \tfrac12 \times 24 \times 18 = \tfrac12 \times 432 = 216 \text{ cm}^2.

The trapezium

A trapezium has exactly one pair of parallel sides. Take trapezium WXYZ with WX \parallel ZY, the parallel sides of lengths a and b, and the perpendicular gap between them h (the height). Drop perpendiculars WM and XN onto ZY. The middle piece WXNM is a rectangle, flanked by a triangle on each side:

\text{Area} = \tfrac12 \times h x + a h + \tfrac12 \times h y = h\!\left(\tfrac{x}{2} + a + \tfrac{y}{2}\right),

where x and y are the two triangle bases. Since the long parallel side is b = x + a + y, we have x + y = b - a, and so

\text{Area} = \tfrac12 h (b - a + 2a) = \tfrac12 h (a + b).

\boxed{\;\textbf{Area of a trapezium} = \tfrac12 \times \text{height} \times (\text{sum of parallel sides}).\;}

a b h

Here is a charming second proof: take two copies of the trapezium, spin one through 180°, and butt them together along the slanting side. Because the co-interior angles add up to 180°, the two copies lock into a single parallelogram of base (a+b) and height h. The trapezium is half of it:

\text{Area of trapezium} = \tfrac12 \times \text{Area of parallelogram} = \tfrac12 h (a + b).

The Śulba-Sūtras and the Craft of Dissection The Śulba-Sūtras (“rules of the measuring cord”) are ancient Indian texts of geometry, composed to guide the exact building of fire altars. Because an altar had to follow a fixed shape and a fixed area, the builders constantly faced tasks like “reshape this rectangle into a triangle of the same area” or “reshape this rhombus into a rectangle.” Their solution was dissection, slice a figure into parts and put them back together as a different figure of equal area. The same family of problems surfaces, many centuries afterward, in Euclid’s Elements. The parallelogram, rhombus and trapezium formulas of this chapter are all descendants of that ancient cut-and-rebuild craft.

Worked Example

A trapezium has parallel sides 28\text{ m} and 40\text{ m}, with the perpendicular distance between them 16\text{ m}. Find its area.

\text{Area} = \tfrac12 \times 16 \times (28 + 40) = \tfrac12 \times 16 \times 68 = 8 \times 68 = 544 \text{ m}^2.

Worked Example

A window pane is shaped like a trapezium with parallel edges 18\text{ cm} and 30\text{ cm} and a height of 10\text{ cm}. A second pane is a rhombus whose diagonals are 16\text{ cm} and 9\text{ cm}. Which pane has the larger area, and by how much?

Trapezium pane:

\text{Area} = \tfrac12 \times 10 \times (18 + 30) = \tfrac12 \times 10 \times 48 = 240 \text{ cm}^2.

Rhombus pane:

\text{Area} = \tfrac12 \times 16 \times 9 = 72 \text{ cm}^2.

The trapezium pane is larger, by 240 - 72 = 168\text{ cm}^2.

Figure it Out: Rhombus & Trapezium

Practice

  1. Find the area of a rhombus whose diagonals are 24\text{ cm} and 18\text{ cm}.
  2. Describe a method to convert a rectangle into a rhombus of equal area by dissection.
  3. Find the areas of: (i) a trapezium with parallel sides 11\text{ ft} and 17\text{ ft} and height 8\text{ ft}; (ii) a trapezium with parallel sides 28\text{ m} and 40\text{ m} and height 16\text{ m}; (iii) a rhombus with diagonals 16\text{ in} and 7\text{ in}; (iv) a trapezium with parallel sides 14\text{ ft} and 22\text{ ft} and height 9\text{ ft}.
  4. [Śulba-Sūtra] Convert an isosceles trapezium into a rectangle by dissection.
  5. Given a way to convert trapezium ABCD into a rectangle EFGH of equal area, explain how to locate the rectangle’s vertices.
  6. Construct a trapezium of area 96\text{ cm}^2 using the trapezium \leftrightarrow rectangle idea.
  7. A regular hexagon is divided into a trapezium, an equilateral triangle and a rhombus. Find the ratio of their areas.
  8. ZYXW is a trapezium with ZY \parallel WX, and A is the midpoint of XY. Show that the area of trapezium ZYXW equals the area of \triangle ZWB (for B as in the figure).
  1. \tfrac12 \times 24 \times 18 = \mathbf{216\text{ cm}^2}.
  2. Cut the rectangle along a diagonal into two right triangles, then slide and flip one so the two equal slant edges meet, the four corners regroup into a rhombus whose diagonals are the rectangle’s length and width; area is preserved. (Equivalently, a rhombus of diagonals d_1, d_2 has the same area \tfrac12 d_1 d_2 as a rectangle of sides d_1 and \tfrac{d_2}{2}.)
    1. Trapezium: \tfrac12 \times 8 \times (11 + 17) = \tfrac12 \times 8 \times 28 = \mathbf{112\text{ ft}^2}. (ii) \tfrac12 \times 16 \times (28 + 40) = \mathbf{544\text{ m}^2}. (iii) Rhombus: \tfrac12 \times 16 \times 7 = \mathbf{56\text{ in}^2}. (iv) \tfrac12 \times 9 \times (14 + 22) = \tfrac12 \times 9 \times 36 = \mathbf{162\text{ ft}^2}.
  4. Drop perpendiculars from the two top vertices of the isosceles trapezium to the longer parallel side, slicing off two congruent right triangles at the ends. Swing each end-triangle up and inward; they plug the gaps above, leaving a rectangle of width \tfrac{a+b}{2} and height h, area \tfrac12 h(a+b), the same as the trapezium.
  5. Mark the midpoints of the two slant sides and draw the midline of the trapezium (parallel to the parallel sides). Fold the two outer triangles inward about this midline: \triangle AHI \cong \triangle DGI and \triangle BEJ \cong \triangle CFJ, so they exactly fill in to form rectangle EFGH of equal area. The rectangle’s height is half the trapezium’s height (its width is the midline length).
  6. Pick any height h and parallel sides a, b with \tfrac12 h(a+b) = 96. For example h = 8, then a + b = 24 (say a = 10, b = 14): area = \tfrac12 \times 8 \times 24 = \mathbf{96\text{ cm}^2}. Construct it, then dissect into the equal-area rectangle as a check.
  7. In a regular hexagon split this way, the trapezium : equilateral triangle : rhombus areas come out in the ratio 3 : 1 : 2. (The hexagon is six unit equilateral triangles; the trapezium holds 3 of them, the rhombus 2, and the equilateral triangle 1.)
  8. Since A is the midpoint of XY, drawing ZA and extending lets us swing \triangle ZAY onto \triangle BAX (they are congruent, sharing the midpoint A and equal vertical angles). This converts the trapezium into the triangle \triangle ZWB without changing area, so Area(ZYXW) = Area(\triangle ZWB).