The contrapositive of the following statement, "If the side of a square doubles, then its area increases four times", is

We begin by identifying the two simple statements that form the given conditional sentence.

Let $$P$$ be the statement “the side of a square doubles”.
Let $$Q$$ be the statement “its area increases four times”.

The original sentence can be written symbolically as the implication $$P \rightarrow Q$$, which reads “If $$P$$ is true, then $$Q$$ is true”.

Now we recall the logical rule for forming a contrapositive. For any implication $$P \rightarrow Q$$, the contrapositive is obtained by first negating both parts and then reversing their order. In symbols, the contrapositive is $$\lnot Q \rightarrow \lnot P$$.

So, applying this rule to our two statements, we have

$$\lnot Q \rightarrow \lnot P$$

where

$$\lnot Q:$$ “the area of a square does not increase four times”,
$$\lnot P:$$ “its side is not doubled”.

Substituting these English negations back into the logical form, we obtain the plain-language contrapositive:

“If the area of a square does not increase four times, then its side is not doubled.”

Comparing this sentence with the four options given:

Option A says: “if the area of a square increases four times, then its side is not doubled.” This keeps $$Q$$ but negates $$P$$, so it is the converse of the inverse, not the contrapositive.

Option B says: “if the area of a square increases four times, then its side is doubled.” That is exactly the original implication $$P \rightarrow Q$$ written in reverse English order; it is the converse, not the contrapositive.

Option C says: “if the area of a square does not increase four times, then its side is not doubled.” This matches $$\lnot Q \rightarrow \lnot P$$ perfectly, so it is the true contrapositive.

Option D says: “if the side of a square is not doubled, then its area does not increase four times.” This has the correct negations but does not reverse the order; it is the inverse, not the contrapositive.

Hence, the correct answer is Option C.