Question
Question: How can I physically generate a square with two \(\text{4x4}\) squares, three \(\text{3x3}\) squares...
How can I physically generate a square with two 4x4 squares, three 3x3 squares, four 2x2 squares and 4 1x1 squares?
Solution
In this question we will first write the total area we have by using the area of the square formula which is a=s2 where a is the area of the square and s is the length of the side of the square. We will then find the total area which we have from all the given squares and then try to fit a square which can have all or some squares in it.
Complete step by step solution:
We have the squares given as two 4x4 squares, three 3x3 squares, four 2x2 squares and four 1x1 squares?
Now on using the formula of the area of a square and then multiplying it with the number of squares present, we will get the total area.
Therefore, the area is:
⇒2×(4×4)=32 units
⇒3×(3×3)=27 units
⇒4×(2×2)=16 units
⇒4×(1×1)=4 units
Now the total area is:
⇒32+27+16+4=79 units.
Now we know that a square with the area of 79 units cannot be made using the provided set of squares.
The greatest possible number which has an integer square root which is lesser than 79 is a square with 64 square units. Now a square of 64 units has a length of 8 units.
Now even though it is a perfect square, there exists no possible combination of squares such that they create another greater square.
The only square that can be created from the given sets of squares is a square of 49 units which has a side of 7 units. The square can be made up as:
Where the yellow square represents the 4x4 square, blue squares represent 3x3 squares, green squares represent 2x2 squares and red squares represent 1x1 squares.
Note: It is to be remembered that we have not used some of the squares provided to us because upon inserting them, we would have not got a perfect square. It is to be noted that the area of the square is 49 units therefore the area unused is 79−49=30 units.