In triangle ABC `m_(1),m_(2),m_(3)` are the lenghts of the medians through A,B and C respectively . If `C=(pi)/(2)`, then `(m_(1)^(2)+m_(2)^(2))/(m_(3)^(2))`=

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem step by step, we will use the properties of medians in a triangle, particularly when one angle is a right angle. Given that angle C = 90 degrees, we can denote the sides opposite to vertices A, B, and C as a, b, and c respectively, where c is the hypotenuse.

### Step-by-Step Solution

1. **Identify the sides of the triangle**:
- Let \( a \) be the length of side BC,
- Let \( b \) be the length of side AC,
- Let \( c \) be the length of side AB (hypotenuse).

2. **Use the formula for the lengths of the medians**:
- The formula for the length of the median from vertex A (denoted as \( m_1 \)) is:
\[ m_1 = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} \]
- The formula for the length of the median from vertex B (denoted as \( m_2 \)) is:
\[ m_2 = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2} \]
- The formula for the length of the median from vertex C (denoted as \( m_3 \)) is:
\[ m_3 = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} \]

3. **Apply the Pythagorean theorem**:
- Since angle C is 90 degrees, we have:
\[ a^2 + b^2 = c^2 \]

4. **Calculate \( m_1^2 \) and \( m_2^2 \)**:
- For \( m_1^2 \):
\[ m_1^2 = \left(\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\right)^2 = \frac{1}{4} (2b^2 + 2c^2 - a^2) \]
- Substitute \( c^2 = a^2 + b^2 \) into \( m_1^2 \):
\[ m_1^2 = \frac{1}{4} (2b^2 + 2(a^2 + b^2) - a^2) = \frac{1}{4} (2b^2 + 2a^2 + 2b^2 - a^2) = \frac{1}{4} (3a^2 + 4b^2) \]

- For \( m_2^2 \):
\[ m_2^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}\right)^2 = \frac{1}{4} (2a^2 + 2c^2 - b^2) \]
- Substitute \( c^2 = a^2 + b^2 \) into \( m_2^2 \):
\[ m_2^2 = \frac{1}{4} (2a^2 + 2(a^2 + b^2) - b^2) = \frac{1}{4} (2a^2 + 2a^2 + 2b^2 - b^2) = \frac{1}{4} (4a^2 + b^2) \]

5. **Calculate \( m_3^2 \)**:
- For \( m_3^2 \):
\[ m_3^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}\right)^2 = \frac{1}{4} (2a^2 + 2b^2 - c^2) \]
- Substitute \( c^2 = a^2 + b^2 \):
\[ m_3^2 = \frac{1}{4} (2a^2 + 2b^2 - (a^2 + b^2)) = \frac{1}{4} (a^2 + b^2) = \frac{c^2}{4} \]

6. **Combine the results**:
- Now we need to find \( \frac{m_1^2 + m_2^2}{m_3^2} \):
\[ m_1^2 + m_2^2 = \frac{1}{4} (3a^2 + 4b^2) + \frac{1}{4} (4a^2 + b^2) = \frac{1}{4} (7a^2 + 5b^2) \]
- Therefore,
\[ \frac{m_1^2 + m_2^2}{m_3^2} = \frac{\frac{1}{4} (7a^2 + 5b^2)}{\frac{c^2}{4}} = \frac{7a^2 + 5b^2}{c^2} \]
- Since \( c^2 = a^2 + b^2 \):
\[ \frac{7a^2 + 5b^2}{a^2 + b^2} \]

7. **Final Calculation**:
- This simplifies to:
\[ \frac{7 \cdot \frac{a^2}{c^2} + 5 \cdot \frac{b^2}{c^2}}{1} = 5 \]

Thus, the final answer is:
\[ \frac{m_1^2 + m_2^2}{m_3^2} = 5 \]