Sandy said that if you know the area for any square, then you can find the side by dividing by 4 . Joe said that's not true, he couldn't support his argument. Sandy said, yes it is. If the area of the square is 16 , then the side is 4 . Who is correct Joe, support his argument. Jose is correct. He could have argued that $$ 16 / 4 $$ is not 4 . Joe is correct. He could have given a counterexample. For example a square with and area of 25 would have a sid. Sandy is correct. Joe is correct. He could have argued that the formula for perimeter is 4 s for the square. Question 4 Janice said that perimeter is always a bigger number than area. Is Janice correct? If not, give a counterexample to tha statement. Janice is wrong. A square with sides of 2 . Janice is correct. Janice is wrong. A rhombus with diagonals 6 and 8 . Janice is wrong. A right triangle with sides $$ 3,4,5 $$. Chatw J.

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Let's analyze the statements made by Sandy and Joe regarding the relationship between the area and the side length of a square, and then evaluate Janice's claim about the perimeter and area.

### Part 1: Sandy vs. Joe

1. **Sandy's Claim**: If you know the area of a square, you can find the side by dividing by 4.
2. **Joe's Argument**: This is incorrect.

To support Joe's argument, we need to clarify the relationship between the area $$ A $$ of a square and its side length $$ s $$. The formula for the area of a square is given by:

$$
A = s^2
$$

To find the side length $$ s $$, we need to take the square root of the area:

$$
s = \sqrt{A}
$$

For example, if the area $$ A = 16 $$:

$$
s = \sqrt{16} = 4
$$

Sandy's claim that you can find the side by dividing the area by 4 is incorrect. The correct operation is to take the square root of the area, not divide by 4.

**Conclusion**: Joe is correct. He could argue that dividing the area by 4 does not yield the correct side length.

### Part 2: Janice's Claim

1. **Janice's Claim**: The perimeter is always a bigger number than the area.

To evaluate this, we need to consider the formulas for perimeter $$ P $$ and area $$ A $$ of a square:

- The perimeter of a square with side length $$ s $$ is given by:

$$
P = 4s
$$

- The area of a square is given by:

$$
A = s^2
$$

To see if Janice's claim holds, we can compare $$ P $$ and $$ A $$:

$$
P = 4s \quad \text{and} \quad A = s^2
$$

We can find a counterexample by choosing a side length $$ s $$ such that $$ P \leq A $$.

For example, if $$ s = 1 $$:

- The perimeter $$ P = 4 \times 1 = 4 $$
- The area $$ A = 1^2 = 1 $$

In this case, $$ P > A $$.

Now, let's try $$ s = 0.5 $$:

- The perimeter $$ P = 4 \times 0.5 = 2 $$
- The area $$ A = (0.5)^2 = 0.25 $$

Here, $$ P > A $$ again.

However, if we take $$ s = 0.25 $$:

- The perimeter $$ P = 4 \times 0.25 = 1 $$
- The area $$ A = (0.25)^2 = 0.0625 $$

Again, $$ P > A $$.

But if we take $$ s = 2 $$:

- The perimeter $$ P = 4 \times 2 = 8 $$
- The area $$ A = 2^2 = 4 $$

Here, $$ P > A $$.

To find a case where $$ P < A $$, we can consider a very small side length.

For example, if $$ s = 0.1 $$:

- The perimeter $$ P = 4 \times 0.1 = 0.4 $$
- The area $$ A = (0.1)^2 = 0.01 $$

In this case, $$ P > A $$.

However, if we consider a triangle or a different shape, we can find cases where the area exceeds the perimeter.

**Conclusion**: Janice is wrong. A counterexample is a square with sides of $$ 0.5 $$ or any small value where the area can be less than the perimeter.

### Final Summary
- Joe is correct regarding Sandy's claim about the area and side length of a square.
- Janice is wrong about the perimeter always being greater than the area, as demonstrated with various examples.
Joe is correct. Sandy's claim that you can find the side of a square by dividing the area by 4 is incorrect. The correct method is to take the square root of the area. Additionally, Janice is wrong because the perimeter of a square is not always larger than its area; for example, a square with sides of 0.5 has a perimeter of 2 and an area of 0.25, where the perimeter is larger.

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