Understanding the Problem: Right Triangle Sides in Ratio

The problem asks us to find the perimeter of a right-angled triangle whose side lengths are in a specific ratio: \(x : (x - 1) : (x - 18)\). In a right-angled triangle, the square of the longest side (hypotenuse) is equal to the sum of the squares of the other two sides (legs). This is known as the Pythagorean theorem.

Identifying the Sides and Applying the Pythagorean Theorem

The sides of the triangle are proportional to \(x\), \(x-1\), and \(x-18\). For these to be positive side lengths of a triangle, \(x\) must be greater than 18 (so that \(x-18 > 0\) and \(x-1 > 0\)).

Among \(x\), \(x-1\), and \(x-18\), the largest value will correspond to the hypotenuse. Since \(x > x-1\) and \(x > x-18\) (assuming \(x > 1\)), \(x\) is the largest value. Therefore, the hypotenuse is proportional to \(x\), and the legs are proportional to \(x-1\) and \(x-18\).

Let the side lengths be \(k(x-18)\), \(k(x-1)\), and \(kx\) for some positive constant \(k\). Since the options for the perimeter are specific numerical values, it is most likely that the side lengths themselves are \(x-18\), \(x-1\), and \(x\), meaning \(k=1\). We will proceed with this assumption and verify if the results match the options.

Using the Pythagorean theorem with sides \(x-18\), \(x-1\), and \(x\):

\[ (x - 18)^2 + (x - 1)^2 = x^2 \]

Solving the Equation for the Value of x

Now we need to solve the equation for \(x\).

Expand the squared terms:

• \((x - 18)^2 = x^2 - 2(x)(18) + 18^2 = x^2 - 36x + 324\)
• \((x - 1)^2 = x^2 - 2(x)(1) + 1^2 = x^2 - 2x + 1\)

Substitute these back into the Pythagorean equation:

\[ (x^2 - 36x + 324) + (x^2 - 2x + 1) = x^2 \]

Combine like terms on the left side:

\[ 2x^2 - 38x + 325 = x^2 \]

Subtract \(x^2\) from both sides to get a quadratic equation:

\[ x^2 - 38x + 325 = 0 \]

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 325 and add up to -38.

The factors of 325 are \(1, 5, 13, 25, 65, 325\).

Let's consider pairs of factors. We need a negative sum, so both factors must be negative. We find that \(-13 \times -25 = 325\) and \(-13 + (-25) = -38\).

So, we can factor the quadratic equation as:

\[ (x - 13)(x - 25) = 0 \]

This gives us two possible values for \(x\):

• \(x - 13 = 0 \implies x = 13\)
• \(x - 25 = 0 \implies x = 25\)

Determining Valid Side Lengths

We assumed earlier that \(x\) must be greater than 18 for all side lengths (\(x\), \(x-1\), \(x-18\)) to be positive.

• If \(x = 13\), the sides would be \(13\), \(13-1=12\), and \(13-18=-5\). A triangle cannot have a negative side length, so \(x=13\) is not a valid solution.
• If \(x = 25\), the sides would be \(25\), \(25-1=24\), and \(25-18=7\). The side lengths are 7, 24, and 25. These are all positive, so \(x=25\) is the valid solution.

Calculating the Perimeter

The side lengths of the right-angled triangle are 7 units, 24 units, and 25 units. Let's quickly verify if these form a right triangle:

\[ 7^2 + 24^2 = 49 + 576 = 625 \]

\[ 25^2 = 625 \]

Since \(7^2 + 24^2 = 25^2\), the sides 7, 24, and 25 indeed form a right-angled triangle (it's a well-known Pythagorean triple).

The perimeter of the triangle is the sum of its side lengths:

\[ \text{Perimeter} = \text{Side } 1 + \text{Side } 2 + \text{Side } 3 \]

\[ \text{Perimeter} = 7 + 24 + 25 \]

\[ \text{Perimeter} = 56 \text{ units} \]

Conclusion

By using the Pythagorean theorem and solving for \(x\), we found the side lengths of the right-angled triangle to be 7, 24, and 25 units. The perimeter of the triangle is the sum of these side lengths, which is 56 units.

Right Triangle Sides and Perimeter Revision

Additional Information on Pythagorean Triples

The set of integers (7, 24, 25) is an example of a Pythagorean triple. A Pythagorean triple is a set of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). These triples represent the side lengths of a right-angled triangle.

Some common Pythagorean triples include:

• (3, 4, 5)
• (5, 12, 13)
• (8, 15, 17)
• (7, 24, 25)

Recognizing common Pythagorean triples can sometimes help quickly identify side lengths in problems involving right triangles.