Question

Fill in the blanks!

Traingle: \sqrt(62) in. , 6 in. , a

a^(2) + b^(2) = c^(2)
a^(2) + _^(2) = (\sqrt(_))^(2)
a^(2) + _ = _
a^(2) = _
\sqrt(a^(2)) = \sqrt(26)
a = \sqrt(26) in.

Answer 1

To fill in the blanks for the given triangle, we will assume you're solving for the missing side \( a \) using the Pythagorean theorem. The triangle has one side of \( \sqrt{62} \) inches, another side of \( 6 \) inches, and we need to find the missing side \( a \).

We can identify \( \sqrt{62} \) as the hypotenuse and \( 6 \) as one of the legs of the triangle.

Using the Pythagorean theorem:

\[
a^2 + b^2 = c^2
\]

where:
- \( a \) is the length of one leg (unknown),
- \( b = 6 \) inches is the length of the other leg,
- \( c = \sqrt{62} \) inches is the hypotenuse.

Now we can fill in the equations step by step using the values provided:

1. **Starting with**:
\[
a^2 + b^2 = c^2
\]
Filling in the known values:
\[
a^2 + 6^2 = (\sqrt{62})^2
\]

2. **Now substitute the squares:**
\[
a^2 + 36 = 62
\]

3. **Next, solve for \( a^2 \)**:
\[
a^2 = 62 - 36
\]
\[
a^2 = 26
\]

4. **Finally, take the square root:**
\[
\sqrt{a^2} = \sqrt{26}
\]

Now, we can fill in the respective blanks:

1. a^(2) + 6^(2) = (\sqrt{62})^(2)
2. a^(2) + 6^(2) = (\sqrt{62})^(2)
3. a^(2) + 36 = 62
4. a^(2) = 26
5. \sqrt{a^2} = \sqrt{26}
6. a = \sqrt{26} in.

So the blanks can be filled as follows:

1. 6
2. 6
3. 36
4. 26
5. 26

Thus, the complete set of filled equations is:

- a^(2) + 6^(2) = (\sqrt{62})^(2)
- a^(2) + 6^(2) = 62
- a^(2) + 36 = 62
- a^(2) = 26
- \sqrt(a^(2)) = \sqrt{26}
- a = \sqrt{26} in.