A man of height `'h'`is walking away from a street lamp with a constant speed`'v'`.The height of the street lamp is `3h`The rate at which of the length of man's shadow is increasingwhen he is at a distance `10h`from the base of the street lamp is:

Options

A. `v//2`
B. `v//3`
C. `2v`
D. `v//6`

Explanation

To solve the problem, we will use similar triangles and related rates. Let's break it down step by step.

### Step 1: Understand the Geometry
We have a street lamp of height \( 3h \) and a man of height \( h \). The man is walking away from the lamp at a speed \( v \). Let \( x \) be the distance of the man from the base of the lamp, and \( y \) be the length of the shadow of the man.

### Step 2: Set Up the Relationship Using Similar Triangles
From the geometry of the situation, we can set up a relationship using similar triangles:
- The triangle formed by the street lamp and the tip of the shadow has a height of \( 3h \) and a base of \( x + y \).
- The triangle formed by the man and his shadow has a height of \( h \) and a base of \( y \).

Using the property of similar triangles, we have:
\[ \frac{3h}{x + y} = \frac{h}{y} \]

### Step 3: Simplify the Equation
Cross-multiplying gives:
\[ 3h \cdot y = h \cdot (x + y) \]
Dividing both sides by \( h \) (assuming \( h \neq 0 \)):
\[ 3y = x + y \]
Rearranging gives:
\[ 2y = x \quad \Rightarrow \quad y = \frac{x}{2} \]

### Step 4: Differentiate with Respect to Time
Now, we differentiate both sides with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{1}{2} \frac{dx}{dt} \]

### Step 5: Substitute Known Values
We know that the man is walking away from the lamp at a speed \( v \), so:
\[ \frac{dx}{dt} = v \]
Substituting this into the differentiated equation gives:
\[ \frac{dy}{dt} = \frac{1}{2} v \]

### Step 6: Find the Rate of Change of the Shadow Length
Thus, the rate at which the length of the man's shadow is increasing is:
\[ \frac{dy}{dt} = \frac{v}{2} \]

### Final Answer
The rate at which the length of the man's shadow is increasing when he is at a distance \( 10h \) from the base of the street lamp is:
\[ \frac{v}{2} \]

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