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 10 cm and the area is 120 cm^2?

The base is decreasing at 2 cm/min.

Step-by-step explanation:

The area (A) of a triangle is given by:

  (1)

Where:

b: is the base

h: is the altitude = 10 cm

If we take the derivative of equation (1) as a function of time we have:

We can find the base by solving equation (1) for b:

Now, having that dh/dt = 1 cm/min, dA/dt = 2 cm²/min we can find db/dt:

Therefore, the base is decreasing at 2 cm/min.

I hope it helps you!  

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

step-by-step explanation:

1