To determine the values of [tex]\(x\)[/tex] such that they lie between two consecutive integers, we first start with the given right triangle.
### Step 1: Understanding the Triangle
In a right triangle:
- One perpendicular side is [tex]\(1 \)[/tex] meter (denote as [tex]\(a\)[/tex]).
- The other perpendicular side is [tex]\(x \)[/tex] meter (denote as [tex]\(x\)[/tex]).
- The hypotenuse (third side) is [tex]\(c\)[/tex] meter (denote as [tex]\(c\)[/tex]).
### Step 2: Apply the Pythagorean Theorem
For a right triangle, the Pythagorean Theorem states that:
[tex]\[ a^2 + x^2 = c^2 \][/tex]
### Step 3: Insert Given Values into the Theorem
Given [tex]\( a = 1 \)[/tex] meter, the equation becomes:
[tex]\[ 1^2 + x^2 = c^2 \][/tex]
[tex]\[ 1 + x^2 = c^2 \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Isolate [tex]\( x^2 \)[/tex]:
[tex]\[ x^2 = c^2 - 1 \][/tex]
Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \sqrt{c^2 - 1} \][/tex]
### Step 5: Determine Integer Bounds
We need to find the two consecutive integers that [tex]\( x \)[/tex] lies between. This means we need to evaluate values of [tex]\( x \)[/tex] by using different [tex]\( c \)[/tex] values.
Let's calculate it step-by-step.
- If [tex]\( c = 2 \)[/tex]:
[tex]\[ x = \sqrt{2^2 - 1} = \sqrt{4 - 1} = \sqrt{3} \approx 1.732 \][/tex]
The integers between which [tex]\(1.732\)[/tex] lies are [tex]\(1\)[/tex] and [tex]\(2\)[/tex].
- If [tex]\( c = 3 \)[/tex] (for further illustration):
[tex]\[ x = \sqrt{3^2 - 1} = \sqrt{9 - 1} = \sqrt{8} \approx 2.828 \][/tex]
The integers between which [tex]\(2.828\)[/tex] lies are [tex]\(2\)[/tex] and [tex]\(3\)[/tex].
### Conclusion:
The specific consecutive integers between which [tex]\( x \)[/tex] lies depend on different values of [tex]\( c \)[/tex]. If [tex]\( c \)[/tex] is such that it maintains the hypotenuse and side-lengths being integers relevant to √ of expressions:
- Most commonly used side-lengths would be relative to simpler calculations ensuring clarity.
However, based on our first and simpler assumption given the hypotenuse (most likely simplified scenarios like [tex]\(c=2 \)[/tex] thus):
[tex]\[ 1.732 \][/tex] remains common situational. Hence,
The two consecutive integers between which [tex]\( x \)[/tex] exists are 1 and 2.
