Question

warm up 11/7/25

1. \\(\sqrt{25}=\\)
2. \\(5^2 = \\)
3. \\((\sqrt{25})^2 = \\)
4. what is the area of the square with sides of 8cm?
5. what is the length of the diagonal of the square with sides of 8cm?

Explanation:

Step1: Solve $\sqrt{25}$

The square root of a number \( x \) is a value \( y \) such that \( y^2 = x \). For \( \sqrt{25} \), we know that \( 5^2 = 25 \), so \( \sqrt{25} = 5 \).

Step2: Solve \( 5^2 \)

Squaring a number means multiplying it by itself. So \( 5^2 = 5\times5 = 25 \).

Step3: Solve \( (\sqrt{25})^2 \)

First, we found that \( \sqrt{25} = 5 \) from Step 1. Then we square that result: \( 5^2 = 25 \), so \( (\sqrt{25})^2 = 25 \).

Step4: Find the area of the square

The formula for the area of a square is \( A = s^2 \), where \( s \) is the length of a side. Here, \( s = 8 \, \text{cm} \), so \( A = 8^2 = 8\times8 = 64 \, \text{cm}^2 \).

Step5: Find the diagonal of the square

For a square with side length \( s \), the diagonal \( d \) can be found using the Pythagorean theorem. In a square, the diagonal forms a right triangle with two sides, so \( d^2 = s^2 + s^2 = 2s^2 \). Taking the square root of both sides, \( d = s\sqrt{2} \). Substituting \( s = 8 \, \text{cm} \), we get \( d = 8\sqrt{2} \, \text{cm} \) (or approximately \( 11.31 \, \text{cm} \)).

Answer:

1. \( \boldsymbol{5} \)
2. \( \boldsymbol{25} \)
3. \( \boldsymbol{25} \)
4. \( \boldsymbol{64 \, \text{cm}^2} \)
5. \( \boldsymbol{8\sqrt{2} \, \text{cm}} \) (or approximately \( \boldsymbol{11.31 \, \text{cm}} \))