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The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm^2 /min. At what rate is the base of the triangle changing when the altitude is 20 cm and the area is 190 cm^2?

User Fiacobelli
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1 Answer

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First, let's recall a fundamental formula about triangles. The area of a triangle, A, is equal to one-half the base, B, times the altitude, H, of the triangle. Mathematically, this can be written as:

A = 0.5 * B * H


Now, we are given that at a particular moment, the area of the triangle is 190 cm^2 and the altitude of the triangle is 20 cm.

We can substitute these given values into our formula for the area of a triangle to solve for the base:

190 = 0.5 * B * 20
B = (2 * 190) / 20 = 19 cm

Now, we are asked for the rate of change of the base of the triangle, given that the rate of area increase is 2 cm^2/min and the rate of altitude increase is 1 cm/min. We will denote the rate of area change as dA/dt, the rate of base change as dB/dt, and the rate of altitude change as dH/dt.

Using the formula for the area again and differentiating with respect to time, we get:

dA/dt = 0.5 * B * dH/dt + H * dB/dt

We can now substitute the given rates and solve for dB/dt:

2 = 0.5 * 19 * 1 + 20 * dB/dt
dB/dt = (2 - 0.5 * 19) / 20 = -0.375 cm/min

So, the base of the triangle is decreasing at a rate of 0.375 cm/min when the altitude is 20 cm and the area is 190 cm^2.

User Astraujums
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