1 x 48
2 x 24
3 x 16
4 x 12
6 x 8
9
Same thing applies to this answer as to the last answer I posted
ANSWER
![3\pi sq.\: in.](https://tex.z-dn.net/?f=3%5Cpi%20sq.%5C%3A%20in.)
EXPLANATION
Let R be the radius of the bigger circle and r, be the radius of the smaller circle.
Then their ratio is given as,
![R:r=3:1](https://tex.z-dn.net/?f=R%3Ar%3D3%3A1)
We can rewrite it as fractions to get,
![\frac{R}{r} = \frac{3}{1}](https://tex.z-dn.net/?f=%20%5Cfrac%7BR%7D%7Br%7D%20%3D%20%5Cfrac%7B3%7D%7B1%7D%20)
We make R the subject to get,
![R = 3r](https://tex.z-dn.net/?f=R%20%3D%203r)
The area of the bigger circle can be found using the formula,
![Area=\pi {r}^{2}](https://tex.z-dn.net/?f=Area%3D%5Cpi%20%7Br%7D%5E%7B2%7D%20)
This implies that,
![Area=\pi ({3r})^{2}](https://tex.z-dn.net/?f=Area%3D%5Cpi%20%28%7B3r%7D%29%5E%7B2%7D%20)
![Area=9\pi {r}^{2}](https://tex.z-dn.net/?f=Area%3D9%5Cpi%20%7Br%7D%5E%7B2%7D%20)
But it was given in the question that, the area of the bigger circle is 27π.
![27\pi=9\pi {r}^{2}](https://tex.z-dn.net/?f=27%5Cpi%3D9%5Cpi%20%7Br%7D%5E%7B2%7D%20)
We divide through by 9π to get,
![3 = {r}^{2}](https://tex.z-dn.net/?f=3%20%3D%20%7Br%7D%5E%7B2%7D%20)
This means that,
![r = \sqrt{3}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B3%7D%20)
The area of the smaller circle is therefore
![= \pi {( \sqrt{3}) }^{2}](https://tex.z-dn.net/?f=%20%3D%20%5Cpi%20%7B%28%20%5Csqrt%7B3%7D%29%20%7D%5E%7B2%7D%20)
Answer:
11
Step-by-step explanation:
See Work Below
I hope this helps. If you have any more questions, please feel free to post them and someone will be able to help you, whether it's myself or others. Please leave a like, rating, and if possible, Brainliest. Have a great day!
A number (we will call X) times 8 Is 8x