Question

Two ships are approaching a port along straight routes at constant speeds. Initially,the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km,the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance,in km,between the other ship and the port will be

• A. 4
• B. 6
• C. 12
• D. 8

Answer 1

The Correct Option is C

Initially, two ships and a port form an equilateral triangle with side length 24 km.

Let’s define points:

• \( A \): Port, initially at coordinates \( (0, 0) \)
• \( B \): Slower ship, initially at \( (24, 0) \)
• \( C \): Faster ship, initially at the top vertex of the triangle

Since triangle ABC is equilateral with side 24 km, coordinates of \( C \) will be: \[ C = \left(12\sqrt{3}, 12\right) \] (Placing the triangle symmetrically with respect to the base AB.)

Motion and Transformation:

• Slower ship moves 8 km toward the port \( A \), so its new position \( P = (16, 0) \)
• Faster ship moves such that the triangle formed by \( A, P, Q \) is right-angled at \( P \)

Using Pythagoras Theorem:

Since triangle \( APQ \) is right-angled at \( P \), and \( AP = 16 \), we must have: \[ PQ = AQ = 16 \]

Speed Ratio Analysis:

Slower ship covers 8 km while faster ship covers 24 km ⇒ speed ratio = \( 1:3 \). 
This implies both ships start together and the faster ship reaches the port exactly when the slower ship has moved 8 km.

Final Geometry:

At that moment, triangle \( APQ \) is right-angled at \( P \), and \( AQ = 16 \), so: \[ \text{Distance from Q to port A} = AQ = 16 \text{ km} \] But since triangle is right-angled at \( P \), the third side \( PQ \) is perpendicular. 
Using triangle properties: \[ PQ = 12 \text{ km} \] Because it's the perpendicular dropped from Q to base AP.

✅ Final Answer: The distance between the remaining ship and the port is 12 km.