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Your iris controls the amount of light that enters your eye by changing the size of your pupil. An eye is shown resembling a large circle. The pupil of the eye is marked resembling a smaller circle inside the large circle. The iris of the eye surrounding the pupil is also marked. The radius of the large circle is labeled 6 millimeters. The distance between the large circle and the small circle is labeled x. a. Write a polynomial in standard form that represents the area of your pupil. Write your answer in terms of $\pi$ . Area = mm $^2$ b. The width $x$ of your iris decreases from 4 millimeters to 2 millimeters when you enter a dark room. How many times greater is the area of your pupil after entering the room than before entering the room?

Sagot :

MrRoyal

Answer:

[tex](a)\ Area = \pi (6-x)^2[/tex]

(b) 4 times

Step-by-step explanation:

The question is illustrated with the attached figure

Solving (a): The area of the pupil

First, we need to calculate the radius (r) of the pupil.

Since the smaller circle is the pupil, then

[tex]r = 6 - x[/tex]

Area is then calculated as:

[tex]Area = \pi r^2[/tex]

[tex]Area = \pi (6-x)^2[/tex]

Solving (a): The area of the pupil

First, we need to calculate the radius (r) of the pupil.

Since the smaller circle is the pupil, then

[tex]r = 6 - x[/tex]

Area is then calculated as:

[tex]Area = \pi r^2[/tex]

[tex]Area = \pi (6-x)^2[/tex]

Solving (b): How many times greater is the area of the pupil, if x decreases from 4 to 2

Initially,

[tex]x = 4[/tex]

So, the area of the pupil is:

[tex]Area = \pi (6-x)^2[/tex]

[tex]Area = \pi(6 - 4)^2[/tex]

[tex]Area = \pi(2)^2[/tex]

[tex]Area = 4\pi[/tex]

When x reduces to 2, the area of the pupil is:

[tex]Area = \pi (6-x)^2[/tex]

[tex]Area = \pi (6-2)^2[/tex]

[tex]Area = \pi (4)^2[/tex]

[tex]Area = 16\pi[/tex]

Calculate the ratio (r) of both areas:

[tex]r = \frac{16\pi}{4\pi}[/tex]

[tex]r = 4[/tex]

Hence, it is 4 times greater

View image MrRoyal
facundo3141592

We want to answer different things regarding circles, we will see that the area of the pupil is:

A = pi*(6mm -x)^2

And that the area of the pupil is 4 times larger after entering the dark room.

Working with circles:

We know that the radius of the pupil is R = 6mm - x

The area of the pupil is given by:

A = pi*(6mm - x)^2

This is the polynomial we wanted.

Now we know that the width of the iris decreases from 4mm to 2mm when you enter a dark place, so x goes from 4mm to 2mm, then the two areas are:

  • A = pi*(6mm - 4mm)^2 = 3.14*(2mm)^2 = 12.56 mm^2
  • A' = pi*(6mm - 2mm)^2  =3.14*(4mm)^2 =  50.24 mm^2

The quotient between the areas is:

A'/A = (50.24 mm^2)/(12.56 mm^2) = 4

So the area of the pupil is 4 times larger after entering the room than before entering the room.

If you want to learn more about circles, you can read:

https://brainly.com/question/25306774

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