The sides of a triangle ABC are 6 , 7 , 8 and the smallest angle being C then the length of altitude from C is

Options

A. `(7)/(2)sqrt(15)`
B. `(7)/(3)sqrt(15)`
C. `(7)/(4)sqrt(15)`
D. none of these

Answer

To find the length of the altitude from vertex C in triangle ABC with sides 6, 7, and 8, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the sides and angles**:
Given the sides of triangle ABC are \( a = 6 \), \( b = 7 \), and \( c = 8 \). Since angle C is the smallest angle, we assign side \( a = 6 \) opposite to angle C.

2. **Calculate the semi-perimeter (s)**:
The semi-perimeter \( s \) of the triangle is calculated as:
\[ s = \frac{a + b + c}{2} = \frac{6 + 7 + 8}{2} = \frac{21}{2} \]

3. **Calculate the area (Δ) using Heron's formula**:
Heron's formula states:
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ \Delta = \sqrt{\frac{21}{2} \left(\frac{21}{2} - 6\right) \left(\frac{21}{2} - 7\right) \left(\frac{21}{2} - 8\right)} \]
Simplifying each term:
\[ s - a = \frac{21}{2} - 6 = \frac{21 - 12}{2} = \frac{9}{2} \]
\[ s - b = \frac{21}{2} - 7 = \frac{21 - 14}{2} = \frac{7}{2} \]
\[ s - c = \frac{21}{2} - 8 = \frac{21 - 16}{2} = \frac{5}{2} \]
Now substituting back into the area formula:
\[ \Delta = \sqrt{\frac{21}{2} \cdot \frac{9}{2} \cdot \frac{7}{2} \cdot \frac{5}{2}} = \sqrt{\frac{21 \cdot 9 \cdot 7 \cdot 5}{16}} \]

4. **Calculate the area**:
First, calculate \( 21 \cdot 9 \cdot 7 \cdot 5 \):
\[ 21 \cdot 9 = 189 \]
\[ 189 \cdot 7 = 1323 \]
\[ 1323 \cdot 5 = 6615 \]
Thus,
\[ \Delta = \sqrt{\frac{6615}{16}} = \frac{\sqrt{6615}}{4} \]

5. **Find the altitude from C (h)**:
The area can also be expressed as:
\[ \Delta = \frac{1}{2} \cdot \text{base} \cdot \text{height} \]
Here, the base is side \( a = 6 \) and height is \( h \):
\[ \Delta = \frac{1}{2} \cdot 6 \cdot h = 3h \]
Setting the two expressions for area equal:
\[ 3h = \frac{\sqrt{6615}}{4} \]
Solving for \( h \):
\[ h = \frac{\sqrt{6615}}{12} \]

6. **Simplifying \( \sqrt{6615} \)**:
Factor \( 6615 \):
\[ 6615 = 3 \cdot 5 \cdot 7 \cdot 63 = 3^2 \cdot 5 \cdot 7 \cdot 3^2 = 3^4 \cdot 5 \cdot 7 \]
Thus,
\[ \sqrt{6615} = 3^2 \cdot \sqrt{5 \cdot 7} = 9\sqrt{35} \]
Therefore,
\[ h = \frac{9\sqrt{35}}{12} = \frac{3\sqrt{35}}{4} \]

### Final Answer:
The length of the altitude from point C is:
\[ h = \frac{3\sqrt{35}}{4} \]