sorry, the answer is as follows: 10/square root 13 and 15/square root 13

Thank you for the correction. Let's go through the process to arrive at the given solution.

### Problem Recap:
- A right-angled triangle has a hypotenuse of 5 meters.
- The other two sides are in a given ratio.
- We need to find the exact lengths of the other two sides.

### Let's Solve:

Given:
- Hypotenuse \( c = 5 \) meters.
- The sides are in the ratio 2:3.

Let the sides be \( a = 2x \) and \( b = 3x \).

Using the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
Substituting the expressions for \( a \), \( b \), and \( c \):
\[
(2x)^2 + (3x)^2 = 5^2
\]
\[
4x^2 + 9x^2 = 25
\]
\[
13x^2 = 25
\]
\[
x^2 = \frac{25}{13}
\]
\[
x = \frac{5}{\sqrt{13}}
\]

Now, substituting back to find the lengths of the sides:
\[
a = 2x = 2 \times \frac{5}{\sqrt{13}} = \frac{10}{\sqrt{13}}
\]
\[
b = 3x = 3 \times \frac{5}{\sqrt{13}} = \frac{15}{\sqrt{13}}
\]

### Final Answer:
The lengths of the other two sides are:
- \( a = \frac{10}{\sqrt{13}} \) meters
- \( b = \frac{15}{\sqrt{13}} \) meters

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How can the lengths \( \frac{10}{\sqrt{13}} \) and \( \frac{15}{\sqrt{13}} \) be rationalized?
2. What would the triangle's area be using these side lengths?
3. How do you determine the sine, cosine, and tangent of the angles in this triangle?
4. How would the side lengths change if the hypotenuse were increased to 10 meters?
5. How does this ratio-based approach compare to using trigonometric functions to solve for the sides?

**Tip:** Rationalizing a denominator involves multiplying the numerator and denominator by the square root in the denominator to eliminate the square root from the denominator.

Explore how to solve a right-angled triangle problem with a hypotenuse of 5 meters using the Pythagorean theorem. Detailed step-by-step solution provided, including calculations for side lengths in a 2:3 ratio. Suitable for Grades 9-12.

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