14.1 Rectangles and Squares

Let us begin with a puzzle. In how many ways can a square be cut into 4 pieces that all have equal area?

The first answers that come to mind are the obvious ones, two straight cuts crossing at the centre, or the two diagonals. Yet there are actually infinitely many solutions. Begin with any equal division, then reshape each part: shrink it slightly along one edge while lengthening it by the same amount along another edge. As long as the loss and the gain balance out, every part keeps its original area.

Math Talk Invent your own surprising ways to split a square into 4 equal-area pieces. Can you do it using only curved cuts? Can you arrange it so that no two of the four pieces have the same shape?

Counting unit squares

Think of mosaic tiling, where a floor is filled evenly with coloured chips. Suppose two rectangular patches are to be filled, one measuring 9\text{ cm} \times 5\text{ cm}, the other 11\text{ cm} \times 4\text{ cm}. Which patch swallows more chips?

To decide, count how many non-overlapping unit squares (each 1\text{ cm} \times 1\text{ cm}) pack into each rectangle.

  • The 9 \times 5 rectangle holds 9 \times 5 = 45 unit squares.
  • The 11 \times 4 rectangle holds 11 \times 4 = 44 unit squares.

So the 9\text{ cm} \times 5\text{ cm} patch covers a touch more space and takes a few more chips.

The Key Idea The area of a region is the number of unit squares (whole or fractional) required to tile it. For a rectangle that count is simply the product of the two sides:

\textbf{Area of a rectangle} = \text{length} \times \text{width}.

Areas carry squared units, for example 45\text{ sq.\,cm} or 45\text{ cm}^2.

A diagonal of a rectangle cuts it into two congruent triangles, so each triangle covers exactly half the unit squares. In the 9 \times 5 rectangle, each triangle has area \tfrac12 \times 9 \times 5 = 22.5\text{ cm}^2. Remember this picture, it is the germ of the entire triangle formula.

Why perimeter cannot measure area

Why bother counting unit squares? Couldn’t we just read off the perimeter, the length of the boundary, and use it to compare areas?

We cannot. Perimeter and area are separate ideas, and neither dictates the other. Two regions can share the same perimeter yet differ in area, and a region with a bigger perimeter can even enclose a smaller area.

Math Talk Hunt for two rectangles where one has the bigger perimeter but the smaller area. (A long skinny rectangle against a chunkier one is a good place to look.) For instance, a 1 \times 14 rectangle has perimeter 30 and area 14, while a 4 \times 6 rectangle has perimeter 20 but area 24, bigger perimeter, smaller area. Can you build a non-rectangular pair that shows the same twist?

Worked Example

Compare a 2 \times 12 rectangle with a 7 \times 7 square.

  • Rectangle: perimeter = 2(2+12) = 28, area = 2 \times 12 = 24.
  • Square: perimeter = 4 \times 7 = 28, area = 7 \times 7 = 49.

Here both shapes share the same perimeter (28), yet the square wraps up far more area (49 vs 24). Equal boundary, unequal interior, proof that perimeter says nothing certain about area.

Figure it Out: Rectangles and Paths

Practice

  1. Recover the missing side lengths.
    1. A large rectangle is divided by one horizontal cut and one vertical cut into four small rectangles. Their areas are 24\text{ in}^2 (top-left), 30\text{ in}^2 (top-right), 40\text{ in}^2 (bottom-left) and 50\text{ in}^2 (bottom-right). The top row is 6 in tall and the left column is 4 in wide. Find the unknown lengths.
    2. A rectangle of area 60\text{ m}^2 is split into two pieces of areas 36\text{ m}^2 and 24\text{ m}^2, with one shared side of 6 m. Find the unknown lengths.
  2. A walking strip (shaded) runs around a rectangular lawn EFGH, set inside a larger rectangle ABCD. (i) Which measurements do you need to find the strip’s area? (ii) If only the strip’s width is given, is that enough? (iii) Does the strip’s area change if the outer rectangle is shifted while the inner lawn stays put inside it?
  3. A field 16\text{ m} \times 11\text{ m} has a cross-path cutting across it both ways. What extra measurement is needed for the path’s area? Write a formula.
  4. Find the area of a coiled hose of constant width by uncoiling it into one long straight tube of the same area.
  5. If the side of a square is doubled, by what factor do the areas of three regions 1, 2 and 3 marked inside it grow? Give reasons.
  6. Cut a square into 4 parts with two perpendicular cuts, then reassemble the pieces into a larger square with a hole in the middle.
  1. (i) Apply area = length \times width to each small piece. The top-left piece is 24\text{ in}^2 with height 6 in, so its width = 24 \div 6 = \mathbf{4\text{ in}}. The top-right piece is 30\text{ in}^2 at the same height 6 in, so its width = 30 \div 6 = \mathbf{5\text{ in}}. The bottom-left piece is 40\text{ in}^2 with width 4 in, so its height = 40 \div 4 = \mathbf{10\text{ in}}. Check the bottom-right: width 5 in \times height 10 in = 50\text{ in}^2 ✓. The rule throughout is: a missing side = area \div known side. (ii) The 36\text{ m}^2 piece with one side 6 m has the other side 36 \div 6 = 6\text{ m}; the 24\text{ m}^2 piece sharing the same 6 m side has length 24 \div 6 = 4\text{ m}. Again: missing side = area \div known side.
  2. (i) You need the outer rectangle’s length and width and the inner lawn’s length and width. Then strip area = area of ABCD - area of EFGH. For example, outer 22 \times 18 = 396\text{ m}^2, inner 16 \times 12 = 192\text{ m}^2, so strip = 396 - 192 = 204\text{ m}^2. (ii) The width w alone is enough only if the inner lawn’s dimensions are also known: split the strip into four rectangles, or use (outer) - (inner) where outer length = inner length + 2w and outer width = inner width + 2w. (iii) No, the strip area is always (outer area) - (inner area), and neither changes when the outer rectangle merely slides around the fixed inner lawn.
  3. You need the width of the cross-path (call it w). A cross-path that runs the full way in both directions has area = 16w + 11w - w^2 (the two strips, minus the overlap square counted once). With w = 1\text{ m}: 16 + 11 - 1 = 26\text{ m}^2.
  4. Picture the hose unwound into a straight line. The straightened length is the sum of all the straight runs; multiply that total length by the constant width to get the area. (For the figure shown, adding the run lengths gives the length of the equivalent straight tube; area = that length \times width.)
  5. Doubling the side multiplies every area by 2 \times 2 = \mathbf{4}. Each region swells to four times its old area, because both length and width double and area depends on their product.
  6. Slice the square along the two perpendicular lines into 4 pieces, then slide them outward and turn them so their straight edges trace a bigger square outline with a square gap at the centre. Total area is preserved, so the larger square minus the hole equals the original square.