Answer is: option4

\( h < 1.5 \)

Solution:

Let:

• \( b \) be the base of the triangle (cm)
• \( h \) be the height of the triangle (cm)
• \( A \) be the area of the triangle (\(\text{cm}^2\))

Given rates:

• \(\frac{db}{dt} = -0.2\) cm/sec (since the base is decreasing)
• \(\frac{dh}{dt} = 0.1\) cm/sec (since the height is increasing)

The formula for the area of a triangle is:

\[ A = \frac{1}{2} bh \]

Differentiate both sides with respect to time \( t \):

\[ \frac{dA}{dt} = \frac{1}{2} \left( b \frac{dh}{dt} + h \frac{db}{dt} \right) \]

Substituting the given values

Given \( \frac{db}{dt} = -0.2 \), \( \frac{dh}{dt} = 0.1 \), and \( b = 3 \):

\[ \frac{dA}{dt} = \frac{1}{2} \left( (3)(0.1) + h(-0.2) \right) \]

\[ \frac{dA}{dt} = \frac{1}{2} (0.3 - 0.2h) \]

For the area to be increasing, we need \( \frac{dA}{dt} > 0 \):

\[ \frac{1}{2} (0.3 - 0.2h) > 0 \]

\[ 0.3 - 0.2h > 0 \]

\[ 0.3 > 0.2h \]

\[ \frac{0.3}{0.2} > h \]

\[ 1.5 > h \]

Final Answer - \( h < 1.5 \)