How To Really Solve This Fun Geometry Puzzle?

I recently came across this fun geometry puzzle from geometry puzzle specialist, Catriona Agg. Since the puzzle managed to engage me, I thought I’d write about it.

This geometry puzzle is essentially made of three parts. For the first part, you have two identical squares stacked on top of each other. Then, two differently-sized squares join the first part from the left to form the second part.

Finally, a straight line connects the right-hand intersection of the squares on the left with the right-hand intersection of the squares on the right.

The length of the line that connects the two intersections is 4 units. Given this information, your challenge is to compute the area of the (pink) shaded region, that is, the combined area of the differently-sized squares on the left.

Do you think you can figure this out?

Hint:

This puzzle appears challenging on the surface. But don’t let it fool you. Once you see the trick to it, it turns out to be a really simple puzzle.

Spoiler Alert:

Beyond this section, I will be explicitly discussing solutions to this puzzle. So, if you wish to give this puzzle a try on your own, I suggest that you pause reading this essay at this point.

Once you are done with your attempt, you may return to this essay and continue reading to compare your approach with mine.

This essay is supported by Generatebg

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Setting Up the Problem

In order to solve this puzzle, all we need to do is compute the area of the two differently-sized squares on the left. Since we do not know the dimensions of these squares, let us assign a couple of variables to unknowns that will help us out.

Let the side length of the lower square on the right be ‘x’ units, and the height of the triangle be ‘y’ units. This might appear insufficient now, but these two variables are sufficient for us to solve this problem.

Now that we have a couple of variables to work with, let us see how we may proceed.

Applying the Pythagorean Theorem to the Fun Geometry Puzzle

Since the lower square has a side length of ‘x’ units, the triangle on the right has a base that is ‘x’ units long as well. In fact, all the sides of the upper square are also ‘x’ units long. Furthermore, since we are dealing with squares here, the vertical side of the triangle is perpendicular to its base.

We are dealing with a right-angled triangle here. Consequently, we may conveniently apply the Pythagorean theorem. If we do so, we arrive at the following equation:

x² + y² = 4²

x² + y² = 16 →(1)

For convenience, let us register this as equation 1. As you can see, this equation relates our two unknowns with a number. We are making progress. Let us see what more we can do with our variables.

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Solving the Fun Geometry Puzzle by taking Advantage of Squares

The key to solving this Puzzle is realising that we are dealing with squares here. If you were perceptive enough, you would have noticed it already. The side length of the big square on the bottom left is (x + y) units. Similarly, the side length of the smaller square above it is (x — y) units.

To solve this geometry puzzle, we need to compute the sum of the areas of these two squares. Given the fact that we know the relations for their side lengths, why don’ we try and compute the sum of their areas?

Let us register this as equation 2. We now know that the shaded area amounts to 2(x² + y²). But from equation 1, we know that x² + y² = 16. When we plug this result into equation 2, we get the following result:

There we go. The area of the shaded region is 32 units².

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Further reading that might interest you: How To Really Solve This Tricky Algebra Problem? and How To Really Solve 1ˣ = -1?

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