We are given a right triangle with side lengths $a = 5 \, \text{cm}$ and $b = 12 \, \text{cm}$, along with errors in these measurements of $\Delta a = 0.1 \, \text{cm}$ and $\Delta b = 0.2 \, \text{cm}$. Our goal is to use linear approximations to estimate the maximum errors in the area and hypotenuse calculations.

Step 1: Area of the Triangle

The formula for the area $A$ of a right triangle is:

$A = \frac{1}{2} \cdot a \cdot b$

The linear approximation for the error in the area $\Delta A$ is:

$\Delta A \approx \left| \frac{\partial A}{\partial a} \right| \Delta a + \left| \frac{\partial A}{\partial b} \right| \Delta b$

First, calculate the partial derivatives: $\frac{\partial A}{\partial a} = \frac{1}{2} b, \quad \frac{\partial A}{\partial b} = \frac{1}{2} a$

Substitute the values $a = 5 \, \text{cm}$, $b = 12 \, \text{cm}$:

$\frac{\partial A}{\partial a} = \frac{1}{2} \times 12 = 6, \quad \frac{\partial A}{\partial b} = \frac{1}{2} \times 5 = 2.5$

Now, estimate the maximum error in the area:

$\Delta A \approx |6| \cdot 0.1 + |2.5| \cdot 0.2 = 0.6 + 0.5 = 1.1 \, \text{cm}^2$

So, the maximum error in the area is:

$\boxed{1.10} \, \text{cm}^2$

Step 2: Hypotenuse of the Triangle

The length of the hypotenuse $c$ is given by the Pythagorean theorem:

$c = \sqrt{a^2 + b^2}$

The linear approximation for the error in the hypotenuse $\Delta c$ is:

$\Delta c \approx \left| \frac{\partial c}{\partial a} \right| \Delta a + \left| \frac{\partial c}{\partial b} \right| \Delta b$

First, calculate the partial derivatives: $\frac{\partial c}{\partial a} = \frac{a}{\sqrt{a^2 + b^2}}, \quad \frac{\partial c}{\partial b} = \frac{b}{\sqrt{a^2 + b^2}}$

Substitute $a = 5 \, \text{cm}$, $b = 12 \, \text{cm}$, and $c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13 \, \text{cm}$:

$\frac{\partial c}{\partial a} = \frac{5}{13}, \quad \frac{\partial c}{\partial b} = \frac{12}{13}$

Now, estimate the maximum error in the hypotenuse:

$\Delta c \approx \left| \frac{5}{13} \right| \cdot 0.1 + \left| \frac{12}{13} \right| \cdot 0.2 = \frac{5}{13} \cdot 0.1 + \frac{12}{13} \cdot 0.2$

Calculating:

$\Delta c \approx \frac{5}{13} \times 0.1 + \frac{12}{13} \times 0.2 = 0.0385 + 0.1846 = 0.2231 \, \text{cm}$

So, the maximum error in the hypotenuse is:

$\boxed{0.22} \, \text{cm}$

Would you like any further details or explanations on this problem?

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Here are 5 related questions for further practice:

1. How would the error change if the side lengths were measured more accurately?
2. What would be the maximum error in the perimeter of the triangle?
3. If the side lengths were $8 \, \text{cm}$ and $15 \, \text{cm}$, what would be the errors in the area and hypotenuse?
4. How can the method of linear approximation be applied to volume calculation for a right prism?
5. How would a 5% increase in each side length affect the area of the triangle?

Tip: The linear approximation technique is useful for estimating small changes in functions when exact error propagation formulas are complex.