`Delta ABC` is an isosceles triangle and `bar(AB) = bar(AC) = 2a `unit, `bar(BC)` = a unit. Draw `AD_|_BC`, and find the length of `bar(AD)` .

Options

A. `sqrt(15)` a unit
B. `(sqrt15)/2` a unit
C. `sqrt(17)` a unit
D. `(sqrt17)/(2)`a unit

Answer

To solve the problem, we need to find the length of \( AD \) in the isosceles triangle \( ABC \) where \( AB = AC = 2a \) units and \( BC = a \) unit. We will draw \( AD \) perpendicular to \( BC \).

### Step-by-Step Solution:

1. **Identify the Triangle and Points**:
- We have triangle \( ABC \) with \( AB = AC = 2a \) and \( BC = a \).
- Let \( D \) be the foot of the perpendicular from \( A \) to \( BC \).

2. **Set Up the Coordinate System**:
- Place point \( B \) at \( (0, 0) \) and point \( C \) at \( (a, 0) \).
- Since \( AB = AC \) and both are equal to \( 2a \), point \( A \) will be directly above the midpoint of \( BC \).

3. **Find the Coordinates of Point A**:
- The midpoint \( M \) of \( BC \) is at \( \left( \frac{a}{2}, 0 \right) \).
- The height \( AD \) will be vertical from \( A \) to \( D \).
- Let the coordinates of point \( A \) be \( \left( \frac{a}{2}, h \right) \).

4. **Use the Distance Formula**:
- We know \( AB = 2a \):
\[ AB = \sqrt{\left( \frac{a}{2} - 0 \right)^2 + (h - 0)^2} = 2a \]
- Squaring both sides:
\[ \left( \frac{a}{2} \right)^2 + h^2 = (2a)^2 \]
\[ \frac{a^2}{4} + h^2 = 4a^2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ h^2 = 4a^2 - \frac{a^2}{4} \]
\[ h^2 = \frac{16a^2}{4} - \frac{a^2}{4} \]
\[ h^2 = \frac{15a^2}{4} \]

6. **Finding the Length of \( AD \)**:
- Taking the square root:
\[ h = AD = \sqrt{\frac{15a^2}{4}} = \frac{\sqrt{15}}{2} a \]

### Final Answer:
The length of \( AD \) is \( \frac{\sqrt{15}}{2} a \).