The two shorter sides (i.e. not the hypotenuse) of a right angle triangle are measured to be (5±0.1)
 cm and (12±0.2)
 cm, where ±0.1
 and ±0.2
 represent the size of the errors made in each measurement.


If we use values 5cm and 12cm as the side lengths, the area of the triangle and the length of the hypotenuse can be computed. Use a linear approximation to estimate the maximum error made in computing the area and hypotenuse length based on the errors in the side length measurements.

Error in area (enter a positive value rounded to two decimal places, don't include units)

Answer = ?


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?

---

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.

The two shorter sides (i.e. not the hypotenuse) of a right angle triangle are measured to be (5±0.1) cm and (12±0.2) cm, where ±0.1 and ±0.2 represent the size of the errors made in each measurement. If we use values 5cm and 12cm as the side lengths, the area of the triangle and the length of the hypotenuse can be computed. Use a linear approximation to estimate the maximum error made in computing the area and hypotenuse length based on the errors in the side length measurements.

This problem involves using linear approximations to estimate the maximum errors in calculating the area and hypotenuse of a right triangle with sides 5 cm and 12 cm. Learn how to apply the Pythagorean theorem and partial derivatives to solve this geometry problem. Suitable for Grades 10-12 students.

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