The base of the triangle is decreasing at a rate of 3.60 centimeters/minute.
According to the statement
we have to find that the rate at which the base of triangle changing.
So, For this purpose,
Area of triangle: A = (1/2)(b * h)
Related Rates - Derivative with respect to time:
(d/dt)[A = (1/2)(b * h)]
Then
dA/dt = (1/2)[b * dh/dt + h * db/dt]
Given: h = 11 centimeters, A = 91 square centimeters, b = ??
A = (1/2)(b * h)
Then it becomes
b = 2A/h
substitute the value then
b = 2(91)/11
b = 16.54 centimeters
Then
h = 11 centimeters, b = 16.54 centimeters, dA/dt = 5 square centimeters/minute, dh/dt = 3 centimeters/minute, db/dt = ?? centimeters/minute
So,
(d/dt)[A = (1/2)(b * h)]
Then it become
dA/dt = (1/2)[b * dh/dt + h * db/dt]
5= (1/2)[16.54 * 3 + 11 * db/dt]
5= 49.62/2 + 11/2 * db/dt
11/2 * db/dt = 5- 49.62/2
5.5*db/dt = 5 - 24.81
db/dt = -19.81/5.5
db/dt = -3.60 centimeters/minute
So, The base of the triangle is decreasing at a rate of 3.60 centimeters/minute.
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