A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with right triangles. At what rate is the length of the person's shadow changing when the person is 16 ft from the lamppost?
In similar triangles, both the two triangles must satisfy the two properties. One is the side proportional, and the other is equal in angles. There are three criteria in similarity. They are AA similarity, SSS similarity, and SAS similarity. The below one satisfies the AA similarity.

The measurement or size of something from end to end is referred to as its length. To put it another way, it is the greater of the higher two or three dimensions of a geometric shape or object. For instance, the length and width of a rectangle define its dimensions.

According to data in the given question,

We have the given information:

The height of the person is 7 ft.

The person is walking away from the post at a rate of 5ft/sec.

The height of the lamppost is 20ft.

Let the person's distance from the bottom of the light post be x ft.

And his shadow's length is y ft.

Form the similar triangles,

[tex]\frac{x+y}{20}=\frac{y}{7}\\[/tex]

7(x+y) = 20y

7x+7y = 20y

20y-7y = 7x

13y = 7x

y = [tex]\frac{7}{13}x[/tex]

Now, we will differentiating wrt t,

[tex]\frac{dy}{dt}=\frac{7}{13}\frac{dx}{dt}................(1)[/tex]

We know that,

[tex]\frac{dx}{dt}=5\frac{ft}{sec}[/tex]

Putting the value of [tex]\frac{dx}{dt}[/tex] in equation (1),

[tex]\frac{dy}{dt}=\frac{7}{13}.5=\frac{35}{13}=2.69\frac{ft}{sec}[/tex]

Therefore, the length of the shadow is changing rate at 2.69 [tex]\frac{ft}{sec}[/tex].

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