1. ## concentric circles with radii, outside and inside areas

i need help with this.

2. Originally Posted by aaaaaa1234
i need help with this

Sounds like the area between every 2nd consecutive circle is shaded.

Therefore my suggestion is

$Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)$

$Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]$

Should be fairly easy from there!

3. Originally Posted by pickslides
Sounds like the area between every 2nd consecutive circle is shaded.

Therefore my suggestion is

$Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)$

$Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]$

Should be fairly easy from there!

im not sure i totally get it, so is the answer 1156? if not, then how do u do it? thanks. if you could solve that step by step, i'd really appreciate it. thanks you

4. Hello, aaaaaa1234!

48 concentric circles are drawn having radii 1, 2, 3, ... 48.
The area outside the circles of odd radii but inside the circles of even radii are shaded.
What is the area of the shaded region?
A circle has area: . $\pi r^2$

The shaded area is: . $A \:=\:(\pi\!\cdot\!2^2 - \pi\!\cdot\!1^2) + (\pi\!\cdot\!4^2 - \pi\!\cdot\!3^2) + (\pi\!\cdot\!6^2-\pi\!\cdot\!5^2) + \hdots + (\pi\!\cdot\!48^2 - \pi\!\cdot\!47^2)$

. . $= \;3\pi + 7\pi + 11\pi + \hdots + 95\pi \;=\; \underbrace{(3 + 7 + 11 + \hdots + 95)}_{\text{arithmetic series}}\pi$

The series has: first term $a = 3$, common difference $d = 4$, and $n = 24$ terms.

. . Its sum is: . $\tfrac{24}{2}[2(3) + 23(4)] \:=\:1176$

Therefore: . $A \;=\;1176\pi$

5. Originally Posted by pickslides
Sounds like the area between every 2nd consecutive circle is shaded.

Therefore my suggestion is

$Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)$

$Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]$

Should be fairly easy from there!

When visualising the problem you should see similar to an achery target.

Every second circle (or hoop section) coloured in. My equation calculates the area of each hoop (as the difference of the area of the outer part and inner part of each coloured circle) and summing them together to find total area. The area of the first hoop (the biggest hoop) is simply

$Area = (\pi.48^2-\pi.47^2)$

then the next biggest

$Area = (\pi.46^2-\pi.45^2)$

then the next biggest

$Area = (\pi.44^2-\pi.43^2)$

and so on until you get to the end

$Area = (\pi.2^2-\pi.1^2)$

Add all these guys together and you get

$Total Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)$

In the next line of my calculation I simply removed $\pi$ as a common factor.

$Total Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]$

I arrived at the solution $Total Area = \pi.1176 \approx 3692.64 units^2$

6. Originally Posted by pickslides
When visualising the problem you should see similar to an achery target.

Every second circle (or hoop section) coloured in. My equation calculates the area of each hoop (as the difference of the area of the outer part and inner part of each coloured circle) and summing them together to find total area. The area of the first hoop (the biggest hoop) is simply

$Area = (\pi.48^2-\pi.47^2)$

then the next biggest

$Area = (\pi.46^2-\pi.45^2)$

then the next biggest

$Area = (\pi.44^2-\pi.43^2)$

and so on until you get to the end

$Area = (\pi.2^2-\pi.1^2)$

Add all these guys together and you get

$Total Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)$

In the next line of my calculation I simply removed $\pi$ as a common factor.

$Total Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]$

I arrived at the solution $Total Area = \pi.1176 \approx 3692.64 units^2$
i think i get it, so for like pi[(48^2-47^2) you would do 48-47 and you end up with 1^2?

7. Originally Posted by aaaaaa1234
i think i get it, so for like pi[(48^2-47^2) you would do 48-47 and you end up with 1^2?
No that arithmetic isn't very robust.

$48^2-47^2 \neq (48-47)^2$

The arithmetic you should use is.

$48^2 = 2304$

$47^2 = 2209$

So

$48^2-47^2 = 2304-2209 = 95$

You would have to repeat that calculation for every hoop before summing together.

I would let ms excel do the work for me

8. would the answer be 1176pi or 1176pi unit squares?

9. 1176pi units squared is the answer