Question

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?

Answer 1

Answer:

$(a)\ Area = \pi (6-x)^2$

(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

$r = 6 - x$

Area is then calculated as:

$Area = \pi r^2$

$Area = \pi (6-x)^2$

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

$r = 6 - x$

Area is then calculated as:

$Area = \pi r^2$

$Area = \pi (6-x)^2$

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

Initially,

$x = 4$

So, the area of the pupil is:

$Area = \pi (6-x)^2$

$Area = \pi(6 - 4)^2$

$Area = \pi(2)^2$

$Area = 4\pi$

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

$Area = \pi (6-x)^2$

$Area = \pi (6-2)^2$

$Area = \pi (4)^2$

$Area = 16\pi$

Calculate the ratio (r) of both areas:

$r = \frac{16\pi}{4\pi}$

$r = 4$

Hence, it is 4 times greater