What if we wish to combine two squares of 'different' sizes to make a large square whose area is the sum of the two smaller squares?

Question

Homi · Accepted Answer

Understand the Question

If we have two squares of different side lengths $a$ and $b$ , their areas are $a^2$ and $b^2$ .
A large square whose area is the sum of these two squares will have an area of $c^2 = a^2 + b^2$ , where $c$ is its side length.
This relationship matches the Pythagoras theorem ( $a^2 + b^2 = c^2$ ), meaning the side of the new square $c$ is the hypotenuse of a right-angled triangle with perpendicular sides (legs) of lengths $a$ and $b$ .

Step 1 · Relate the Areas of the Squares

Let the side lengths of the two smaller squares be $a$ and $b$ . $\text{Area of first square} = a^2$ $\text{Area of second square} = b^2$

Let the side length of the new large square be $c$ . $\text{Area of large square} = c^2$

Since the area of the large square is the sum of the two smaller squares: $c^2 = a^2 + b^2$ Diagram 1

Step 2 · Apply the Pythagoras Theorem

The relation $c^2 = a^2 + b^2$ represents the Pythagoras theorem for a right-angled triangle. Diagram 2

In a right-angled triangle with legs $a$ and $b$ and hypotenuse $c$ : $a^2 + b^2 = c^2$

Therefore, the side length $c$ of the new square is the hypotenuse of a right-angled triangle whose perpendicular legs are $a$ and $b$ .

Step 3 · Construct the New Square

To construct the large square geometrically:

Draw a line segment of length $a$ .
Draw another segment of length $b$ perpendicular ( $90^\circ$ ) to the first segment.
Connect the endpoints to form the hypotenuse of length $c$ .
Construct a square with side length $c$ .

The area of this constructed square is: $c^2 = a^2 + b^2$

Answer

The side length of the new large square is the hypotenuse ( $c$ ) of a right-angled triangle with legs equal to the side lengths ( $a$ and $b$ ) of the two smaller squares, satisfying $c^2 = a^2 + b^2$ .

Common Mistakes

Adding Side Lengths Directly: Mistakenly assuming the new side length is $a + b$ instead of $\sqrt{a^2 + b^2}$ . The areas add up ( $a^2 + b^2$ ), not the perimeter or linear lengths.
Ignoring the Right Angle: Forgetting that segments $a$ and $b$ must meet at exactly $90^\circ$ for the connecting segment to equal $\sqrt{a^2 + b^2}$ via the Pythagoras theorem.

Question 15

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