Question

In the given figure, if AB = c, AC = b and AD 1 BC, then AD =

• A. \(\frac{bc}{\sqrt{b^2+c^2}}\)
• B. \(\frac{bc}{b^2+c^2}\)
• C. \(\frac{b^2c^2}{\sqrt{b^2+c^2}}\)
• D. None of these

Answer 1

The Correct Option is A

Step 1: Use the formula for the altitude of a triangle.

The area of \( \triangle ABC \) can be expressed in two ways:

• Using the base \( BC \) and height \( AD \):
• Using Heron's formula or directly from the sides \( AB = c \), \( AC = b \), and \( BC = a \):

Equating the two expressions for the area:

\[ \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot b \cdot c \cdot \sin A. \]

Simplify:

\[ BC \cdot AD = b \cdot c \cdot \sin A. \]

Step 2: Express \( \sin A \) in terms of the sides of the triangle.

From the Pythagorean theorem, the hypotenuse of \( \triangle ABC \) is:

\[ BC = \sqrt{b^2 + c^2}. \]

The sine of angle \( A \) is given by:

\[ \sin A = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{c}{\sqrt{b^2 + c^2}}. \]

Step 3: Substitute \( \sin A \) into the equation for \( AD \).

Substitute \( \sin A = \frac{c}{\sqrt{b^2 + c^2}} \) and \( BC = \sqrt{b^2 + c^2} \) into \( BC \cdot AD = b \cdot c \cdot \sin A \):

\[ \sqrt{b^2 + c^2} \cdot AD = b \cdot c \cdot \frac{c}{\sqrt{b^2 + c^2}}. \]

Simplify:

\[ AD = \frac{b \cdot c \cdot c}{b^2 + c^2}. \]

\[ AD = \frac{bc}{\sqrt{b^2 + c^2}}. \]

Final Answer: The length of \( AD \) is \( \mathbf{\frac{bc}{\sqrt{b^2 + c^2}}} \), which corresponds to option \( \mathbf{(1)} \).