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Sagot :
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

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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