Solve. Leave the answers in terms of pi ?.
a)Find a polynomial for the sum of the areas of the circles shown in the figure.
b)Find the sum of the areas when r = 5 and r = 11.3.
this is what I got but I did not do it right. I needed to To write an expression for the sum, i need to put everything together by adding the three terms that i wrote below but not sure how to start.
(pi*r2)+(pi*r2)
(pi*52)+(pi*11.32)
(25pi)+(127*69pi)
(152.69pi)
Large circle: A=pi*r2, Medium circle A=pi*32, Small circle A=pi*22
Anna, I saw this post yesterday, and hoped you would realize that you can't post figures or pictures, so we can't see " the circles shown in the figure".
Do you have 2 separate circles, not overlapping ?
if so, then
for r=5 , area = π(5^2) = 25π
for r = 11.3, area = π(11.3^2) = 127.69π
so the total area = 25π + 127.69π = 152.69π
You seem to have that answer.
btw, to show exponents use the ^ key
e.g. 5^2 = 25
To find the polynomial for the sum of the areas of the circles, you'll need to add the areas of the three circles together.
a) Let's start by finding the expressions for the areas of each circle:
- Large circle: A = π*r^2 (as given)
- Medium circle: A = π*3^2 = 9π
- Small circle: A = π*2^2 = 4π
Now, to find the sum of all three circles, you add the individual areas together:
Sum = π*r^2 + 9π + 4π
This can be simplified to:
Sum = π*r^2 + 13π
b) Now, to find the sum of the areas when r = 5 and r = 11.3:
For r = 5:
Sum = π*(5^2) + 13π
= 25π + 13π
= 38π
For r = 11.3:
Sum = π*(11.3^2) + 13π
= 127.69π + 13π
= 140.69π
So, when r = 5, the sum of the areas is 38π, and when r = 11.3, the sum of the areas is 140.69π.
To find the sum of the areas of the circles in terms of pi, you need to add together the areas of the individual circles.
a) The areas of the three circles are:
Large circle: A = pi * r^2
Medium circle: A = pi * 3^2 (since the radius is 3)
Small circle: A = pi * 2^2 (since the radius is 2)
To find the polynomial for the sum of the areas, you simply add these three expressions together:
Sum of areas = pi * r^2 + pi * 3^2 + pi * 2^2
b) To find the sum of the areas when r = 5 and r = 11.3, simply substitute these values into the expression for the sum of areas:
When r = 5:
Sum of areas = pi * 5^2 + pi * 3^2 + pi * 2^2
Sum of areas = 25pi + 9pi + 4pi
Sum of areas = 38pi
When r = 11.3:
Sum of areas = pi * 11.3^2 + pi * 3^2 + pi * 2^2
Sum of areas = 127.69pi + 9pi + 4pi
Sum of areas = 140.69pi
Therefore, the sum of area when r = 5 is 38pi and when r = 11.3 is 140.69pi.