Constructing squares of given area

Take a square dot grid — dots spaced $1$ unit apart in rows and columns, like graph paper. Pick any dot as a starting point and draw a segment to another dot that is $a$ units away horizontally and $b$ units away vertically. That segment is the hypotenuse of a right triangle with legs $a$ and $b$, so by the Baudhayana–Pythagoras theorem its length is

where: $a$ and $b$ are the whole-number horizontal and vertical grid steps, and $\ell$ is the length of the tilted (slanted) segment. Now repeat that same step vector three more times, turning $90°$ at each corner. Four equal, perpendicular sides close up into a tilted square — tilted because its sides run diagonally across the grid rather than along the rows and columns. Its side length is $\ell=\sqrt{a^2+b^2}$, and because area does not care whether a square is "upright" or tilted, its area is exactly

where: the right-hand side is a whole number as long as $a$ and $b$ are whole numbers, even though the side length $\sqrt{a^2+b^2}$ itself is usually irrational. The figure below shows the tilted square built with $a=1,\ b=2$, enclosed inside an upright square so you can see how the area is checked.

Reading the picture: enclose the tilted square in the smallest upright square that contains it, then subtract the four leftover corner right triangles. That "enclose and subtract" idea is exactly the derivation below.

Problem: Construct a tilted square of area $5$ square units on a dot grid, and verify its area using the enclosing-square method.

Step 1. We need whole numbers with $a^2+b^2=5$. Trying small values: $1^2+2^2=1+4=5$. ✓ So take $a=1,\ b=2$.

Step 2. Starting at dot $(0,0)$, move $1$ unit right and $2$ units up to reach $(1,2)$ — that is the first side of the tilted square.

Step 3. Rotate the step vector $(1,2)$ by $90°$ to get $(-2,1)$; from $(1,2)$ this reaches $(-1,3)$. Rotate again to get $(-1,-2)$, reaching $(-2,1)$. Rotate once more to get $(2,-1)$, which returns you to $(0,0)$. The four corners $(0,0),(1,2),(-1,3),(-2,1)$ form the tilted square.

Step 4. Verify by enclosing: the enclosing upright square has side $a+b=1+2=3$, so its area is $3^2=9$. Each corner right triangle has legs $1$ and $2$, area $\tfrac12\times1\times2=1$; four of them total $4$.

Step 5. Subtract: $9-4=5$ square units, matching $a^2+b^2=1^2+2^2=5$ exactly. The constructed tilted square has side length $\sqrt5$ units and area exactly $5$ square units.

Competitive stretch

For area $13$: try $a=2,\ b=3$, since $2^2+3^2=4+9=13$. The enclosing square has side $2+3=5$ (area $25$), and four corner triangles of legs $2,3$ total $4\times\tfrac12\times2\times3=12$. Then $25-12=13$. ✓ So a tilted square built from step $(2,3)$, side length $\sqrt{13}$, has area exactly $13$ square units — the same enclose-and-subtract check works for any target area that can be written as a sum of two whole-number squares.

• Trying to measure the slanted side with a ruler. The whole point of the construction is that you never measure $\sqrt5$ or $\sqrt{13}$ directly — you place whole-number grid steps and let the area come out exactly right on its own.
• Rotating the step vector incorrectly. Each side must be the previous side turned by exactly $90°$; a wrong turn breaks the shape, and the four sides stop being equal and perpendicular.
• Mixing up area and side length. The area is the whole number $a^2+b^2$; the side length is the (usually irrational) square root $\sqrt{a^2+b^2}$ — do not report one when asked for the other.
• Forgetting the enclosing square's side is $a+b$, not $a$ or $b$ alone. The check only works if the upright square drawn around the tilted square has side exactly $a+b$, so that all four corner triangles have legs $a$ and $b$.

Ask the tutor: Want to try constructing a tilted square of a different area, say $17$ or $20$ square units — ask the AI tutor to help you find the right $a$ and $b$ and sketch the steps.