## Challenging 9 squares puzzle: Relate the areas of the other 8 squares to the small black square

9 squares puzzle: Find area of rectangle with 9 gradually smaller squares inside it and the side length of smallest square 3 units. Time 30 minutes.

### The puzzle

In figure below the nine squares together form the outer rectangle. The side length of the smallest black shaded square is 3 (in length unit).

Find the area of the rectangle (shaded blue covering all the nine squares).

**Recommended time:** 30 minutes.

### Solution to the 9 Squares puzzle: Find area of rectangle

At first, finding the area of the outer rectangle seems to be an involved and complicated task. You are right. It is complicated, especially

- if you don't have earlier experience of solving this type of puzzle and more importantly,
- if you THINK it is complicated.

Realizing that **quick and elegant solution** of this apparently difficult puzzle will depend on a *fully systematic step by step reason based approach*,

- we start asking relevant questions one after the other,
- analyze the puzzle figure to find the answer and then
- make conclusions based on the answer.

We won't deviate at all from approach of step by step reasoning using

question, analysis and answer methodfinally homing in to the solution.In the process of raising important questions and resulting analysis we'll discover key patterns that would be of sure help for solution.

#### Solution Stage 1: 9 Squares puzzle: Find area of rectangle - What do we know and want to know for solving the puzzle?

Answering this itself is the first step (for solving any problem for that matter). As answer, the first conclusion is made,

Conclusion 1:As it must be possible to get the area of the whole rectangle from the area of the smallest square embedded deep into the rectangle, this smallest square must be related in a simple way gradually from the squares adjacent to it to all the other 8 squares.

This is an obvious conclusion and nothing great about it.

But this realization helps to raise the next question, relevant and more focused,

Question 2:How are the areas of the adjacent squares related to the area of the smallest square?

We realize that answer to this question in all its details should give us the first breakthrough. Notice that as promised to keep things simple,

Instead of trying to get all the details of the relations between areas of the squares we would go forward in small easy steps—the immediately visible steps.

This approach won't waste time and *load our mind with the demands of a complex and heavy task.*

*Two simple observations* would make problem solving easier,

- As area of a square is square of its side-length that are same for all sides of the square, instead of the inter-relations between areas we would think in terms of the inter-relations between side-lengths of the squares.
- We would assume the
**side-length of the smallest square as 1 instead of 3**, because ultimately side-lengths of all the other squares must be a multiple of the side-length of the smallest square.

After we evaluate the height and breadth of the rectangle, we would simply multiply both by 3 and take their product to get the answer (in fact the side-length of the smallest square can be any real number, it won't disturb the nature of inter-relations between the side-lengths of the 9 squares at all).

#### Solution Stage 2: 9 Squares puzzle: Find area of rectangle -Finding relations between side-lengths of the adjacent squares by walking along the sides

This is the time to label the other squares. The following graphic shows the side-length relations between a few squares adjacent to the smallest square,

True to our promise of small easy steps we have labeled all eight other squares but **drawn conclusions on side-length relations between only four**.

Keeping the first data small would help understanding the way the side-lengths are related with each other easier (there is much to say for small incremental steps, but think it over yourself).

**How did we get the relations between side 'a' with sides 'b', 'c' or 'd'?**

It is simple. If you walk upwards from the junction point of upper horizontal side of square 'a' and right vertical side of square 'b' **up to the top right corner** of square 'b', a length of just the side-length of the smallest square, you get the side-length of square 'b'.

$\Rightarrow$ 'b' would be just 'a' with 1 added to it.

This is walking or traversing along a side in VERTICAL direction.

To get the side-length of 'c' again traverse a length of 1 horizontally right and you would get the side-length of 'c' by adding 2 to 'a'.

So these are the

primary and essentially similar actionsby which you should get finally side-lengths of all the squares in terms of 'a'.

True, it seems even then we would have the job of dealing with the single unknown value of side-length 'a', but the job of solving the apparently complicated puzzle is simplified greatly by this discovery of the **first key pattern,**

All side-lengths are inter-related by traversing vertically or horizontally along the common side of two squares by incrementally and possibly known fixed lengths.

**Note:** Any Geometry puzzle on squares you can solve by using this mechanism.

The actions needed clearer and inter-relations between the few squares clearly defined, **next question** that comes up automatically is,

Question 2:How is the side-length of 'e' related to those of 'b' and 'c'?

Yes, while looking at the squares 'a', 'b', 'c', 'd' and 'e', you have kept your eyes open and discovered **the second key pattern,**

What length is $(c-d)$? It seems to be nearly equal to 'e'.

Not exactly. To this difference just add 1, the portion of 'c' down below beyond the intersection of lower horizontal side of 'c' and right vertical side of 'd', and you will get 'e'.

Aha, we have got the exact length of square 'e'. It is just,

$e=c-(d-1)=(c-d)+1=(a+2)-(a-1)+1=4$.

This is the **first major breakthrough.** And we have achieved it without any sweat.

The following figure now shows the rest of the side-lengths in terms of 'a' using the value of $e=4$.

Side-lengths of 'f' and 'g' are easy to get—just by adding 4 to 'c' first and then adding another 4 to 'f'. But you met your first difficulty in getting 'h' in the same way to express it in terms of 'a'.

For side-length of 'h' you had to add 'a' with 'd', the direction and way of getting the side-length different from any other previous squares.

Naturally you wonder,

**Question 3:** Why getting side-length of 'h' is different from any other squares?

Trying to answer the question, you find the answer simply as,

Because the small segment $(g-h)$ is of unknown length. In all cases earlier, you have added or subtracted known lengths to the previous side.

You sense, **in this unknown length of the small section the key to the solution is hidden.**

So you try to answer the question,

**Question 4**: What exactly is the length of $(g-h)$?

Oh well, it is fully unknown, $(g-h)=11-a$.

No this is not the way. The exact length of this section is not so easy to get.

**Then, what is the way forward?**

When you look at the figure again, thinking of getting the height and breadth that are needed for calculating the area of the rectangle, you notice the lower horizontal side to be $4a$ and the right vertical side to be $3a+9$.

*This is the point when you look at the top horizontal side of the rectangle and immediately get your solution.*

As the *two horizontal sides of the rectangle must be equal,*

$\text{Breadth}=f+g=2a+16=a+b+h=4a$,

Or, $2a=16$,

Or, $a=8$.

$\text{Breadth}=f+g=2a+16=32$ and,

$\text{Height}=g+h=3a+9=33$.

$\text{Area of rectangle}=32\times{33}\times{9}$ square units.

Not so difficult after all, you will surely agree.

### End note

There is a satisfaction in solving a not so simple problem just by taking one simple step after another, asking an important question, analyzing the problem for answer and making conclusions.

You may yourself summarize the steps into a method, if you wish.

This is the method of *systematic problem solving that helps to solve potentially any problem.*

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