MEDAL AND FAN 1 All circles are similar to each other. True False 2 Find the exact area of a circle having the given circumference. 8 A = 4√2 16 64 3 Find the exact area of a circle having the given circumference. 3 A = 2.25 3 9 4 Find the exact circumference of a circle with an area equal to 36 sq. in. 12 18 324

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MEDAL AND FAN 1 All circles are similar to each other. True False 2 Find the exact area of a circle having the given circumference. 8 A = 4√2 16 64 3 Find the exact area of a circle having the given circumference. 3 A = 2.25 3 9 4 Find the exact circumference of a circle with an area equal to 36 sq. in. 12 18 324

Mathematics
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1. T
2. Use the circumference formula to find the radius. Then use the area formula to find the area.
3. Solve it same way as 2.

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4. Use the area formula to find the radius. Then use the circumference formula to find the circumference.
If you have questions, just ask.
cant remember the area formula for a circle
can you work out number 2 for me?
Here are the formulas you need. \( \Large C_{circle} = 2 \pi r\) \(\Large A_{circle} = \pi r^2\)
|dw:1438148434752:dw|
2 Find the exact area of a circle having the given circumference. 8 A = 4√2 16 64 If the circumference is 8, we set 8 equal to teh circumference formula and find the radius. \(\large C_{circle} = 2 \pi r\) \(\large 8 = 2 \pi r\) \(\large \dfrac{8}{2 \pi} = \dfrac{2 \pi r}{2 \pi} \) \(\large \dfrac{4}{\pi} = r\)
You stared it out correct. You need to divide both sides by pi. |dw:1438148608264:dw|
Ok with the radius?
do i need to divide 4 by pi or is that the final radius?
No. Don't divide. The problem wants an exact area, so we can't divide by pi and give an approximate decimal. We need to keep pi in the radius. Now we find the area.
\(\large A = \pi r^2\) \(\large A = \pi \times \left( \dfrac{4}{\pi} \right) ^2\) Ok so far with the area?
|dw:1438148873842:dw|
|dw:1438148931770:dw|
so the answer is \[16\]
16 pi
thanks
No. Not 16*pi It's 16/pi
Wait. There is no pi at all in problem 2 or in the choices. Was the circumference 8 or 8pi?

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