A man of 2 metres height walks at a uniform speed of 6 km/hr away from a lamp post of 6 metres high. Find the rate at which the length of his shadow increases.

Answer

To solve the problem, we need to find the rate at which the length of the man's shadow increases as he walks away from the lamp post. We will use related rates and similar triangles to derive the solution step by step.

### Step 1: Understand the Geometry of the Problem
We have a lamp post of height 6 meters and a man of height 2 meters walking away from it. Let:
- \( Y \) be the distance from the lamp post to the man.
- \( X \) be the length of the man's shadow.

### Step 2: Set Up the Relationship Using Similar Triangles
From the geometry, we can set up a relationship using similar triangles:
1. The triangle formed by the lamp post, the ground, and the tip of the shadow.
2. The triangle formed by the man, the ground, and the tip of his shadow.

From the similar triangles, we have:
\[ \frac{6}{X + Y} = \frac{2}{X} \]

### Step 3: Cross-Multiply to Find a Relationship
Cross-multiplying gives us:
\[ 6X = 2(X + Y) \]
Expanding this, we have:
\[ 6X = 2X + 2Y \]
Rearranging gives:
\[ 4X = 2Y \quad \Rightarrow \quad Y = 2X \]

### Step 4: Differentiate with Respect to Time
Now, we differentiate both sides of the equation \( Y = 2X \) with respect to time \( t \):
\[ \frac{dY}{dt} = 2 \frac{dX}{dt} \]

### Step 5: Substitute Known Values
We know the speed of the man walking away from the lamp post, which is given as:
\[ \frac{dY}{dt} = 6 \text{ km/hr} \]
Substituting this into the differentiated equation gives:
\[ 6 = 2 \frac{dX}{dt} \]

### Step 6: Solve for \(\frac{dX}{dt}\)
Now, we can solve for \(\frac{dX}{dt}\):
\[ \frac{dX}{dt} = \frac{6}{2} = 3 \text{ km/hr} \]

### Final Answer
Thus, the rate at which the length of the man's shadow increases is:
\[ \frac{dX}{dt} = 3 \text{ km/hr} \]