In a triangle ABC , right angle at B, AC=5 cm, median `AL=(3sqrt5)/(2)`. Find the length of median CM.

Answer

To solve the problem step by step, we will use the properties of triangles and the median.

### Step 1: Understand the Triangle and Given Information
We have a right triangle ABC with a right angle at B. The length of side AC is given as 5 cm, and the length of median AL is given as \( \frac{3\sqrt{5}}{2} \).

### Step 2: Set Up the Right Triangle
Since triangle ABC is a right triangle at B, we can denote:
- \( AB = c \)
- \( BC = a \)
- \( AC = 5 \) cm (hypotenuse)

According to the Pythagorean theorem:
\[ AB^2 + BC^2 = AC^2 \]
This gives us:
\[ c^2 + a^2 = 5^2 \]
\[ c^2 + a^2 = 25 \quad \text{(Equation 1)} \]

### Step 3: Use the Median AL
Since L is the midpoint of side BC, we can express the length of median AL using the formula for the median in a triangle:
\[ AL = \frac{1}{2} \sqrt{2AB^2 + 2AC^2 - BC^2} \]
Substituting the known values:
\[ \frac{3\sqrt{5}}{2} = \frac{1}{2} \sqrt{2c^2 + 2(5^2) - a^2} \]
Multiplying both sides by 2:
\[ 3\sqrt{5} = \sqrt{2c^2 + 50 - a^2} \]
Squaring both sides:
\[ (3\sqrt{5})^2 = 2c^2 + 50 - a^2 \]
\[ 45 = 2c^2 + 50 - a^2 \]
Rearranging gives us:
\[ 2c^2 - a^2 = 45 - 50 \]
\[ 2c^2 - a^2 = -5 \quad \text{(Equation 2)} \]

### Step 4: Solve the System of Equations
Now we have two equations:
1. \( c^2 + a^2 = 25 \) (Equation 1)
2. \( 2c^2 - a^2 = -5 \) (Equation 2)

From Equation 1, we can express \( a^2 \) in terms of \( c^2 \):
\[ a^2 = 25 - c^2 \]
Substituting this into Equation 2:
\[ 2c^2 - (25 - c^2) = -5 \]
This simplifies to:
\[ 2c^2 - 25 + c^2 = -5 \]
\[ 3c^2 - 25 = -5 \]
\[ 3c^2 = 20 \]
\[ c^2 = \frac{20}{3} \]

Now substituting back to find \( a^2 \):
\[ a^2 = 25 - \frac{20}{3} = \frac{75}{3} - \frac{20}{3} = \frac{55}{3} \]

### Step 5: Find the Length of Median CM
Now, we can find the length of median CM using the median formula:
\[ CM = \frac{1}{2} \sqrt{2BC^2 + 2AC^2 - AB^2} \]
Substituting in our values:
\[ CM = \frac{1}{2} \sqrt{2\left(\frac{55}{3}\right) + 2(25) - \left(\frac{20}{3}\right)} \]
Calculating:
\[ CM = \frac{1}{2} \sqrt{\frac{110}{3} + 50 - \frac{20}{3}} = \frac{1}{2} \sqrt{\frac{110 + 150 - 20}{3}} = \frac{1}{2} \sqrt{\frac{240}{3}} = \frac{1}{2} \sqrt{80} = \frac{1}{2} \cdot 4\sqrt{5} = 2\sqrt{5} \text{ cm} \]

### Final Answer
The length of median CM is \( 2\sqrt{5} \) cm.