How does the rate of increase in the altitude of a triangle relate to the rate of increase in its area?

To understand the relationship between the rate at which the altitude of a triangle is increasing and the rate at which the area is increasing, we need to consider the formula for the area of a triangle:

Area = (1/2) × base × height

In this scenario, we are given:

  • The altitude (height) of the triangle is increasing at a rate of 1 cm/min.
  • The area of the triangle is increasing at a rate of 2 cm²/min.

Let’s denote:

  • A: area of the triangle
  • b: base of the triangle
  • h: height (altitude) of the triangle
  • dh/dt: rate of change of height with respect to time
  • dA/dt: rate of change of area with respect to time

From the area formula, we can derive the following relationship:

To find the derivative of the area with respect to time, we apply the product rule:

dA/dt = (1/2) × b × (dh/dt)

Now plugging in our known values:

  • dA/dt = 2 cm²/min
  • dh/dt = 1 cm/min

Substituting these values into the equation gives:

2 = (1/2) × b × 1

From this, we can solve for the base b:

2 = (1/2) × b

=> b = 4 cm

Therefore, the base of the triangle must be 4 cm for the given rates of change to hold true. This demonstrates how the triangle’s dimensions interact, indicating that an increase in height, even at a constant base, results in a proportional increase in the area of the triangle.

In summary, as the altitude increases at a rate of 1 cm/min, the area responds by increasing at a rate of 2 cm²/min, provided the base remains constant at 4 cm.

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