If the ratio of sides of A `Delta ` are 4 : 5 : 6 inradius = 3 cm . Find the largest altitude

Answer

To find the largest altitude of a triangle with sides in the ratio 4:5:6 and an inradius of 3 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Define the sides of the triangle**:
Let the sides of the triangle be represented as:
- \( a = 4x \)
- \( b = 5x \)
- \( c = 6x \)

2. **Calculate the semi-perimeter (s)**:
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} = \frac{4x + 5x + 6x}{2} = \frac{15x}{2} \]

3. **Use the formula for inradius (r)**:
The formula for the inradius \( r \) is:
\[ r = \frac{Area}{s} \]
Given that \( r = 3 \) cm, we can express the area \( A \) in terms of \( x \):
\[ A = r \cdot s = 3 \cdot \frac{15x}{2} = \frac{45x}{2} \]

4. **Calculate the area using Heron's formula**:
Heron's formula states:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
We need to calculate \( s - a \), \( s - b \), and \( s - c \):
- \( s - a = \frac{15x}{2} - 4x = \frac{15x - 8x}{2} = \frac{7x}{2} \)
- \( s - b = \frac{15x}{2} - 5x = \frac{15x - 10x}{2} = \frac{5x}{2} \)
- \( s - c = \frac{15x}{2} - 6x = \frac{15x - 12x}{2} = \frac{3x}{2} \)

Now substituting these values into Heron's formula:
\[ A = \sqrt{\frac{15x}{2} \cdot \frac{7x}{2} \cdot \frac{5x}{2} \cdot \frac{3x}{2}} \]
Simplifying this:
\[ A = \sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3 \cdot x^4}{16}} = \frac{\sqrt{1575} \cdot x^2}{4} \]

5. **Equate the two expressions for area**:
We have two expressions for the area:
\[ \frac{45x}{2} = \frac{\sqrt{1575} \cdot x^2}{4} \]
Cross-multiplying gives:
\[ 180x = \sqrt{1575} \cdot x^2 \]
Dividing both sides by \( x \) (assuming \( x \neq 0 \)):
\[ 180 = \sqrt{1575} \cdot x \]
Solving for \( x \):
\[ x = \frac{180}{\sqrt{1575}} = \frac{180 \cdot \sqrt{1575}}{1575} = \frac{180 \cdot \sqrt{63}}{315} = \frac{12 \cdot \sqrt{63}}{21} = \frac{4 \cdot \sqrt{63}}{7} \]

6. **Calculate the sides**:
Now substituting \( x \) back to find the sides:
- \( a = 4x = \frac{16 \cdot \sqrt{63}}{7} \)
- \( b = 5x = \frac{20 \cdot \sqrt{63}}{7} \)
- \( c = 6x = \frac{24 \cdot \sqrt{63}}{7} \)

7. **Find the largest altitude**:
The altitude corresponding to the smallest base will be the largest. The smallest side is \( a = \frac{16 \cdot \sqrt{63}}{7} \). Using the area formula:
\[ A = \frac{1}{2} \cdot base \cdot height \]
We can find the height \( h_a \) corresponding to side \( a \):
\[ h_a = \frac{2A}{a} = \frac{2 \cdot \frac{45x}{2}}{a} \]
Substitute \( A \) and \( a \):
\[ h_a = \frac{45x}{\frac{16 \cdot \sqrt{63}}{7}} = \frac{315x}{16 \cdot \sqrt{63}} \]

Finally, substituting \( x \):
\[ h_a = \frac{315 \cdot \frac{4 \cdot \sqrt{63}}{7}}{16 \cdot \sqrt{63}} = \frac{315 \cdot 4}{7 \cdot 16} = \frac{1260}{112} = 11.25 \text{ cm} \]

### Final Answer:
The largest altitude of the triangle is **11.25 cm**.