Isosceles right triangle and √2

Take a unit square and slice it along one diagonal. The cut produces two triangles, each with a right angle where two sides of the square meet, and two legs of equal length $1$ — an isosceles right triangle. The diagonal of the square becomes the hypotenuse of each triangle. Apply the Baudhayana–Pythagoras relation with both legs equal to $1$:

where: $c$ is the hypotenuse and the two equal legs have length $1$. Since no whole number or simple fraction squares to exactly $2$, $\sqrt2$ is the first irrational length most students meet — yet it is perfectly real and measurable, sitting right there as the diagonal of a unit square. The same fact can be seen without any algebra at all, purely through areas. Build a square on each side of the triangle: the square on either leg has area $1^2=1$, while the square on the hypotenuse turns out to have exactly twice that area — area $2$, so its side is $\sqrt2$. This doubling-of-area picture is the geometric heart of the Baudhayana–Pythagoras theorem for this special triangle, and it is why ancient geometers could discover the relation by literally counting unit squares, long before algebraic notation existed. Scaling up from the unit square, any isosceles right triangle with equal legs of length $a$ obeys the same pattern:

where: $a$ is the length of each equal leg and $c$ is the hypotenuse. Since $\sqrt2\approx1.414$, the hypotenuse is always a little less than one-and-a-half times a leg — a quick mental check for whether an answer is in the right ballpark. The angles tell the same story from a different direction. Because the two legs are equal, the two base angles opposite them are equal too. The three angles of a triangle sum to $180°$, one of them is the right angle ($90°$), so the remaining $90°$ splits evenly into $45°$ and $45°$. Every isosceles right triangle is therefore a "45-45-90 triangle," and cutting a square along its diagonal always produces exactly this shape. This is exactly the picture that opens the chapter The Baudhayana-Pythagoras Theorem. Rather than starting with an arbitrary right triangle, the textbook deliberately chooses the simplest one — equal legs — so that a student can discover the relation $a^2+b^2=c^2$ by drawing squares and counting unit cells, before the general theorem is stated for triangles with unequal legs. Seen this way, the isosceles right triangle is not just one example among many; it is the doorway through which the whole theorem is first met, and $\sqrt2$ is the very first irrational number most students encounter by name.

Problem: An isosceles right triangle has legs of length $5\text{ cm}$ each. Find the hypotenuse, correct to two decimal places, and then find the side of a square garden whose diagonal path is $10\text{ m}$ long.

Step 1. For the triangle, identify the equal legs: $a=b=5\text{ cm}$, so $a^2+b^2=c^2$ gives $25+25=c^2$, hence $c^2=50$.

Step 2. Use the leg-form shortcut instead of simplifying $\sqrt{50}$ directly: since both legs are equal, $c=a\sqrt2=5\sqrt2$. With $\sqrt2\approx1.414$, $c\approx5\times1.414=7.07\text{ cm}$.

Step 3. For the garden, the diagonal of a square is the hypotenuse of the isosceles right triangle formed by two of its sides, so $c=10\text{ m}$ and $c=a\sqrt2$ gives $a=\dfrac{10}{\sqrt2}=\dfrac{10\sqrt2}{2}=5\sqrt2\approx7.07\text{ m}$.

Step 4. Notice both answers come out to the same number, $7.07$, even though one problem started from a leg and the other from a diagonal — they describe the identical geometric picture viewed from opposite ends.

The hypotenuse of the $5\text{ cm}$ triangle is $\approx7.07\text{ cm}$, and the garden's side is $\approx7.07\text{ m}$.

Competitive stretch

Problem: The square built on the hypotenuse of an isosceles right triangle has area $32\text{ cm}^2$. Find the length of each leg.

Step 1. The area of the square on the hypotenuse is $c^2$, so $c^2=32$, giving $c=\sqrt{32}=4\sqrt2\text{ cm}$.

Step 2. Use the doubling-of-area idea in reverse: the square on the hypotenuse has twice the area of the square on one leg, so the leg-square has area $32\div2=16\text{ cm}^2$, and the leg is $a=\sqrt{16}=4\text{ cm}$.

Step 3. Cross-check with $c=a\sqrt2$: $c=4\times1.414\approx5.66\text{ cm}$, and indeed $4\sqrt2\approx5.66\text{ cm}$, matching Step 1. Each leg is $4\text{ cm}$ long.

• Multiplying and dividing by $\sqrt2$ backwards. The hypotenuse is always the longer side, so go from leg to hypotenuse by multiplying by $\sqrt2$ (a number bigger than $1$), and from hypotenuse to leg by dividing.
• Rounding $\sqrt2$ too early. Using $1.4$ or $1.5$ partway through a multi-step problem can shift the final answer; keep at least three decimal places ($1.414$) until the last step.
• Applying $c=a\sqrt2$ to a triangle that is not isosceles. The shortcut only works when both legs are equal; for a general right triangle use $a^2+b^2=c^2$ with the actual leg lengths.
• Doubting that $\sqrt2$ is a "real" length. Because it has no terminating decimal, some students treat it as not measurable — but it is exactly the diagonal of a unit square, drawable and measurable like any other length.
• Confusing the leg-square area with the hypotenuse-square area. Remember the doubling rule runs one way: the hypotenuse-square is twice a leg-square, never the other way round — so to go from hypotenuse area to leg area you halve, not double.

Ask the tutor: Want to see why $\sqrt2$ is irrational, or try more practice problems on isosceles right triangles — ask the AI tutor for a step-by-step walkthrough.