Given the problem:

• The perimeter of a right triangle is $60 \, \text{cm}$.
• The hypotenuse is $25 \, \text{cm}$.

We need to find the area of the triangle.

Let's denote the sides of the right triangle as $a$, $b$ (the legs), and $c$ (the hypotenuse). From the problem, we have:

1. $c = 25 \, \text{cm}$
2. The perimeter $a + b + c = 60 \, \text{cm}$

Substitute $c$ into the perimeter equation: $a + b + 25 = 60$ $a + b = 35 \quad \text{(Equation 1)}$

Since it's a right triangle, we also have: $a^2 + b^2 = c^2$ $a^2 + b^2 = 25^2$ $a^2 + b^2 = 625 \quad \text{(Equation 2)}$

To solve for $a$ and $b$, we can express $b$ in terms of $a$ using Equation 1: $b = 35 - a$

Substitute into Equation 2: $a^2 + (35 - a)^2 = 625$

Expanding this: $a^2 + (35^2 - 2 \times 35 \times a + a^2) = 625$ $a^2 + 1225 - 70a + a^2 = 625$ $2a^2 - 70a + 1225 = 625$

Combine like terms: $2a^2 - 70a + 600 = 0$

Divide by 2 to simplify: $a^2 - 35a + 300 = 0$

This is a quadratic equation. We can solve for $a$ using the quadratic formula: $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 1$, $b = -35$, and $c = 300$: $a = \frac{35 \pm \sqrt{(-35)^2 - 4 \times 1 \times 300}}{2 \times 1}$ $a = \frac{35 \pm \sqrt{1225 - 1200}}{2}$ $a = \frac{35 \pm \sqrt{25}}{2}$ $a = \frac{35 \pm 5}{2}$

So, $a = \frac{40}{2} = 20 \quad \text{or} \quad a = \frac{30}{2} = 15$

If $a = 20$: $b = 35 - 20 = 15$

If $a = 15$: $b = 35 - 15 = 20$

Thus, the legs are $a = 15$ and $b = 20$ (or vice versa).

Now, let's find the area of the triangle: $\text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 15 \times 20 = 150 \, \text{cm}^2$

So, the area of the triangle is 150 cm².

Would you like further details or have any questions?

Here are some related questions:

1. What are the other possible dimensions of a right triangle given a specific hypotenuse?
2. How can we verify the Pythagorean theorem using the dimensions found?
3. What happens to the area if we change the hypotenuse while keeping the perimeter constant?
4. How does the triangle's shape change if the perimeter is increased while keeping the hypotenuse fixed?
5. Can we derive the formula for the area of any right triangle using just its perimeter and hypotenuse?

Tip: Remember, for right triangles, the Pythagorean theorem $a^2 + b^2 = c^2$ is always a quick way to relate the sides if you know two of the three.