Can you see why the method works in the case where the two squares are the same size? Does it agree with the method we used earlier to combine two same sized squares into a bigger square?

Question

Homi · Accepted Answer

Understand the Question

When two squares are of equal size with side length $s$ , each has an area of $s^2$ .
Arranging them along the legs of a right-angled triangle forms an isosceles right-angled triangle with legs of length $s$ .
By the Pythagoras theorem, the square on the hypotenuse has an area equal to the sum of the two squares: $s^2 + s^2 = 2s^2$ .
This gives a side length of $s\sqrt{2}$ , which completely agrees with the earlier method of combining two squares by summing their areas.

Step 1 · Analyze the Case of Equal Squares Using Pythagoras Theorem

Let the two initial squares have side length $s$ .

$\text{Area A} = s \times s = s^2$ $\text{Area B} = s \times s = s^2$

Since the squares are built on the legs of a right-angled triangle, both legs are equal to $s$ , forming an isosceles right-angled triangle. Diagram 1

Let $c$ be the hypotenuse (the side length of the square on the hypotenuse).

By Pythagoras theorem

\begin{aligned} c^2 &= s^2 + s^2 \\ c^2 &= 2s^2 \\ c &= \sqrt{2s^2} \\ &= \sqrt{2} \times \sqrt{s^2} \\ c &= s\sqrt{2} \end{aligned}

Thus, the method works because the area of the square on the hypotenuse is the sum of the areas of the two smaller squares.

Step 2 · Compare with the Earlier Area Addition Method

In the earlier method, combining two identical squares of area $s^2$ gives a total combined area of: $\text{Total Area} = s^2 + s^2 = 2s^2$

If $X$ is the side length of the combined square

\begin{aligned} X^2 &= 2s^2 \\ X &= \sqrt{2s^2} \\ X &= s\sqrt{2} \end{aligned}

Both methods produce the exact same side length $s\sqrt{2}$ .

Answer

Yes, the method works for equal-sized squares because it represents the Pythagoras theorem for an isosceles right-angled triangle, giving a side length of $s\sqrt{2}$ . This completely agrees with the earlier method of combining two squares of area $s^2$ to get a total area of $2s^2$ with side length $\sqrt{2s^2} = s\sqrt{2}$ .

Common Mistakes

Adding Side Lengths Directly: Incorrectly assuming the side of the combined square is $s + s = 2s$ instead of adding areas ( $s^2 + s^2 = 2s^2 \implies \text{side} = s\sqrt{2}$ ).
Forgetting to Take the Square Root: Identifying the new area as $2s^2$ but forgetting that the side length requires taking the square root: $\sqrt{2s^2} = s\sqrt{2}$ .

Question 17

More questions in IT