Constructing Squares and Exploring Dot Grids

A square on squared paper does not have to follow the printed horizontal and vertical lines. Its vertices can lie on grid points while its sides slope between them. Such a square may cover 2 or 5 square units even though neither side is a whole number of grid spacings. This is another place where thinking about square areas is more useful than trying to read a side length with a ruler.

We will first use the theorem to construct new squares whose areas are three times, five times, or a specified amount less than a given area. We will then return to the dot grid and connect each tilted side to a small right triangle.

For every construction, distinguish a drawing that looks plausible from an explanation that proves it works. The explanation should establish both the square shape and the required area.

Constructing a Square with Three Times a Given Area

Let the original square have side s and area A = s². You already know how to construct a square with area 2A: build it on the original diagonal. Its side has square 2s².

Now form a right triangle whose perpendicular sides are the original side s and the original diagonal. By the theorem, the square on the new hypotenuse has area s² + 2s² = 3s². That is three times the original area.

This construction uses known lengths without requiring a decimal approximation to the diagonal. Transfer the original diagonal length with a compass or a paper strip. Draw it perpendicular to a copy of the original side, join their free endpoints, and construct a square on the joining segment.

Constructing a Square with Five Times a Given Area

There are different ways to obtain five copies of the same area. One convenient choice is A + 4A. A square of side 2s has area 4s², so a right triangle with perpendicular sides s and 2s connects areas A and 4A.

The square on that triangle’s hypotenuse has area s² + (2s)² = s² + 4s² = 5s². Thus its side is s√5. Notice that doubling a side is now useful because we intentionally want four times the area for one part of the sum.

Another route is to first construct squares with areas 2A and 3A, then combine those using their side lengths as perpendicular sides of a right triangle. Their areas add to 5A. Different constructions can reach the same required square area.

Constructing a Square from the Difference of Two Areas

Area addition uses the hypotenuse as the new side. For an area difference, reverse the relationship. Suppose the larger square has side L and the smaller has side s, where L > s. We want a side x satisfying x² = L² − s².

Rearrange this as s² + x² = L². It describes a right triangle with hypotenuse L and one perpendicular side s. Its remaining perpendicular side x is the required square’s side.

A ruler-and-compass construction can produce that length directly. Draw a segment AC of length s. At A draw a perpendicular ray. With centre C and radius L, draw an arc meeting the ray at B. Then BC = L and AC = s, with a right angle at A. Build the required square on AB.

The radius must exceed AC so that a positive height is possible. If the two starting areas were equal, the difference would be zero, giving no square of positive side length.

Drawing Tilted Squares on a Dot Grid

Assume neighbouring grid points are 1 unit apart horizontally and vertically. A side from one grid point to another may involve moving p units horizontally and q units vertically. Those two movements form perpendicular sides of a right triangle.

The straight joining segment is that triangle’s hypotenuse. Its squared length is p² + q². A square built on the segment therefore has area p² + q² square units, even when the segment length itself is not an integer.

To make the next side of the square, turn the step pattern through a right angle. For example, “2 right and 1 up” becomes “1 left and 2 up”. Repeat the turns to complete four equal, perpendicular sides. This creates a square, whereas choosing four arbitrary grid points might create only a general quadrilateral.

Investigating Areas of 2, 3, 4, and 5 Square Units

For area 2, choose horizontal and vertical displacements 1 and 1. The side has squared length 1² + 1² = 2. Four such diagonal grid segments form a tilted square.

For area 4, use a side that moves 2 units horizontally and 0 vertically. The square is aligned with the grid, and its area is 2² + 0² = 4. Allowing a zero displacement is necessary to include these ordinary grid squares.

For area 5, the 2-and-1 step pattern works. Area 3 is different. If either displacement were at least 2, its square alone would already be at least 4. So both displacements would have to be either 0 or 1. Their squared sum can then be only 0, 1, or 2, never 3.

Therefore a square of area 3 cannot have all its vertices at integer grid points on this unit grid. This does not mean that a square of area 3 cannot exist. Such a square has side √3; it simply cannot satisfy this particular grid-vertex condition.

Which Integer Areas Are Possible?

An integer area is possible on this grid exactly when it can be written as p² + q² for non-negative integers p and q, not both zero. The necessity comes from the horizontal and vertical displacements of a side. The sufficiency comes from rotating that step pattern to build the other three sides.

Some possibilities are 1 = 1² + 0², 2 = 1² + 1², 4 = 2² + 0², 5 = 2² + 1², 8 = 2² + 2², 9 = 3² + 0², 10 = 3² + 1², and 13 = 3² + 2². Not every positive integer appears.

For any small target area N, you can investigate all whole-number displacements whose squares do not exceed N. Check their square sums systematically. You do not need a more advanced classification rule to solve the small grid problems here.