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The altitude (i.e., height) of a triangle is increasing at a rate of 1.5 cm/minute while the area of the triangle is increasing at a rate of 1.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 82 square centimeters?

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Answer: The base of the triangle is decreasing at the rate of 2.16cm per minute

Explanation:

The area of a triangle is = 1/2 base × height.

We would make use of the product rule to find the rate of change of the base of this triangle.

Let us use A to represent the area, h to represent height, and b to represent base.

dA/dt = [(1/2) × (dh/dt) × b] + [(1/2) × (db/dt) × h]

Where:-

dh/dt = 1.5, dA/dt = 1.5, h = 10, db/dt = the unknown.

Before proceeding further, we must find the base of the triangle. Since we already know that the height = 10cm and the area = 82 square centimeters. We can find the base.

Formula for area of a triangle is 1/2 × b × h = A.

We substitute accordingly: 1/2 × b × 10 = 82.

10b/2 = 82

10b = 164

b = 16.4cm

Since the base is 16.4cm, we can calculate the rate of change of the base of the triangle (db/dt) using the product rule formula stated earlier on.

1.5 = [(1/2) × (1.5) × (16.4)] + [(1/2) × (db/dt) × (10)]

1.5 = (1.5 × 8.2) + (db/dt × 5)

1.5 = (12.3) + (db/dt × 5)

db/dt × 5 = 1.5 - 12.3

db/dt × 5 = -10.8

db/dt = -10.8/5

db/dt = -2.16cm/min

Therefore, the base of the triangle is decreasing at the rate of 2.16cm per minute

User John Pavley
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