Question

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

• A. 10
• B. \(6\sqrt{2}\)
• C. \(8\sqrt{2}\)
• D. 5

Answer 1

The Correct Option is A

 

From the figure of the right-angled triangle, let the two perpendicular sides be \( a \) and \( b \), and the hypotenuse be \( 20 \) units.

Step 1: Pythagoras Theorem 

Using the Pythagorean identity, we have: \[ a^2 + b^2 = 20^2 = 400 \tag{1} \]

Step 2: Expression for AP

The length of segment \( AP \) (from vertex A perpendicular to hypotenuse) is given by: \[ AP = \frac{ab}{20} \tag{2} \]

Step 3: Maximizing ab

To maximize \( AP \), we need to maximize the product \( ab \), subject to equation (1).

Applying the Arithmetic Mean – Geometric Mean inequality (AM ≥ GM): \[ \frac{a^2 + b^2}{2} \ge ab \] Substitute from equation (1): \[ \frac{400}{2} \ge ab \Rightarrow ab \le 200 \]

Step 4: Final Calculation

The maximum possible value of \( ab \) is \( 200 \). Therefore, the maximum value of \( AP \) is: \[ AP = \frac{ab}{20} = \frac{200}{20} = 10 \]

Conclusion:

The maximum value of AP is 10 units.