We are given a right-angled triangle $MNL$ with sides $MN = x$, $ML = x+3$, and $NL = 7$. We need to find the perimeter of the triangle.

1. Problem Description

We are given a right-angled triangle $MNL$ with sides $MN = x$, $ML = x+3$, and $NL = 7$. We need to find the perimeter of the triangle.

2. Solution Steps

Since $MNL$ is a right-angled triangle, we can use the Pythagorean theorem:

$a^2 + b^2 = c^2$

where $a$ and $b$ are the lengths of the two shorter sides, and $c$ is the length of the hypotenuse. In our case, $a = x$, $b = x+3$, and $c = 7$.

Plugging these values into the Pythagorean theorem, we get:

$x^2 + (x+3)^2 = 7^2$

Expanding the equation:

$x^2 + (x^2 + 6x + 9) = 49$

$2x^2 + 6x + 9 = 49$

$2x^2 + 6x - 40 = 0$

Dividing the equation by 2:

$x^2 + 3x - 20 = 0$

Now we solve this quadratic equation for $x$. We can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a=1$, $b=3$, and $c=-20$. Plugging these values into the formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-20)}}{2(1)}$

$x = \frac{-3 \pm \sqrt{9 + 80}}{2}$

$x = \frac{-3 \pm \sqrt{89}}{2}$

Since the length of a side cannot be negative, we take the positive root:

$x = \frac{-3 + \sqrt{89}}{2} \approx \frac{-3 + 9.43}{2} \approx \frac{6.43}{2} \approx 3.215$

So, $x \approx 3.215$.

Then, $x+3 \approx 3.215 + 3 = 6.215$.

The sides of the triangle are approximately $3.215$, $6.215$, and $7$.

The perimeter is the sum of the sides:

$P = x + (x+3) + 7$

$P \approx 3.215 + 6.215 + 7$

$P \approx 16.43$

However, the problem doesn't give any units. Therefore, we write:

$P = x + (x+3) + 7 = 2x + 10$

$x = \frac{-3+\sqrt{89}}{2}$

$P = 2(\frac{-3+\sqrt{89}}{2}) + 10$

$P = -3+\sqrt{89} + 10 = 7 + \sqrt{89}$

3. Final Answer

The perimeter of the triangle is $7 + \sqrt{89}$.