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In a triangle, the base is increasing at a constant rate of 2.8 cm per second, while the height is decreasing at a constant rate of 0.7 cm per second. At the instant when the base of the triangle is 6 cm and the height of the triangle is 10 cm, what is the rate of change of the area of the triangle?

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abidemiokin

Answer:

Step-by-step explanation:

Area of a triangle = 1/2bh

b is the base

h is the height

dA/dt = dA/db * db/dt + dA/dh * dh/dt

h = 10cm

b = 6cm

dA/db = h/2 = 10/2

dA/db = 5cm

dA/dh = b/2

dA/dh = 6/2 = 3cm

db/dt = 2.8cm/s

dh/dt = 0.7cm/s

Substitutte into the formula;

dA/dt = 5(2.8) + 3(0.7)

dA/dt = 14 + 2.1

dA/dt = 16.1cm²/s

Hence the rate of change of the area of the triangle is 16.1cm²/s

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