To express the length of the hypotenuse $h$ as a function of the perimeter $p$ in a right triangle, let's first understand the relationship between these quantities.

Given:

• The altitude (height) of the right triangle is $10 \, \text{cm}$.
• The triangle is a right triangle, so the Pythagorean theorem applies.
• Let $h$ be the length of the hypotenuse, and $p$ be the perimeter of the triangle.

Let the legs of the right triangle be $a$ and $b$, with one leg being the altitude $a = 10 \, \text{cm}$. Thus, we have: $a = 10$ $b = b \, \text{(unknown leg)}$ $h = \, \text{hypotenuse}$

Perimeter equation:

The perimeter $p$ is the sum of the three sides: $p = a + b + h$ Substitute $a = 10$: $p = 10 + b + h$

Pythagorean theorem:

From the Pythagorean theorem, we know: $a^2 + b^2 = h^2$ Substitute $a = 10$: $10^2 + b^2 = h^2$ $100 + b^2 = h^2$ Solve for $b^2$: $b^2 = h^2 - 100$ Thus, $b = \sqrt{h^2 - 100}$.

Perimeter equation in terms of $h$:

Now, substitute $b = \sqrt{h^2 - 100}$ into the perimeter equation: $p = 10 + \sqrt{h^2 - 100} + h$ Solve for $h$ in terms of $p$: $p - 10 = \sqrt{h^2 - 100} + h$ Isolate the square root: $\sqrt{h^2 - 100} = p - 10 - h$ Square both sides: $h^2 - 100 = (p - 10 - h)^2$ Now expand the right-hand side: $h^2 - 100 = (p - 10)^2 - 2h(p - 10) + h^2$ Cancel $h^2$ on both sides: $-100 = (p - 10)^2 - 2h(p - 10)$ Solve for $h$: $2h(p - 10) = (p - 10)^2 + 100$ $h = \frac{(p - 10)^2 + 100}{2(p - 10)}$

Thus, the length of the hypotenuse $h$ as a function of the perimeter $p$ is: $h(p) = \frac{(p - 10)^2 + 100}{2(p - 10)}$

Would you like any more details on how this formula was derived or how it can be used?

Here are five related questions:

1. What is the value of $h$ when $p = 30$?
2. How would this formula change if the altitude was not 10 cm?
3. Can you find the area of the triangle given the hypotenuse and perimeter?
4. What are the side lengths of the triangle when the perimeter is known?
5. How does the formula for $h(p)$ change if the triangle is isosceles?

Tip: Always check units to ensure consistent results when working with equations like these!