Question

The circumference, in inches, of a certain circle is C, where Absolute value (C minus 20.5) less than or equal to 0.5. Which of the following is a possible area of the circle, in square inches?

(A) 64 over pi
(B) 81 over pi
(C) 100 over pi
(D) 121 over pi
(E) 144 over pi

Answer 1

\(\huge |C-20.5|\le0.5\)

Answer 2

know how to solve absolute value equations ?

Answer 3

I think so.

Answer 4

make the 0.5 either + or -?

Answer 5

and then solve as a regular equation?

Answer 6

to eliminate |...|, yes.
if |a| < b
then,
a<b , a>-b
yes, then solve as regular equation...

Answer 7

Note: circumference cannot be negative.
so you only get one inequality for C.

Answer 8

Right.

Answer 9

Let me try now...

Answer 10

I was on the same path, just was a bit doubtful :/

Answer 11

okay, so what u get for C ?

Answer 12

Alright so C=21 , simple.

Answer 13

Then we use 21=2pir to find r

Answer 14

\(C\le21\)

Answer 15

thats correct.

Answer 16

Right....

Answer 17

we'd use 21 right?

Answer 18

yes. find r.

Answer 19

r= 21/2pi

Answer 20

now area = ?

Answer 21

plug that into pir^2

Answer 22

441 / 4 pi isnt the answer!!!

Answer 23

\[\large -0.5 \le C - 20.5 \le 0.5 \\ \large 20 \le C \le 21\]

Answer 24

you only get one inequality for C.<----i was incorrect here..

Answer 25

@sirm3d hmm. So we'd use 20.5 as C?

Answer 26

yes you may, if it is among the choices

Answer 27

if |a| < b
then,
a<b , a>-b

so u also have
C-20.5 >= -0.5
add 20.5.

Answer 28

Right.

Answer 29

I get it fine up until here, now what? my area is coming all funky

Answer 30

\[\large 20 \le C \ 21 \\ \large 20 \le 2 \pi r \le 21 \\ \large 10 \pi \le r \le (21/2\pi)\]

Answer 31

C>=20
so, r >= 10/pi
so AREA >= pi * 100/pi^2
>=100/pi

Answer 32

Riiight!!! awesome. Ok now we have r. Which we had before...

Answer 33

uhm, \[\large (10/\pi) \le r \le (21/2\pi) \\ \large 100/\pi^2 \le r^2 \le 21^2/(4^2\pi^2) \\ \large 100/\pi \le \pi r^2 \le 21^2/(16\pi)\]

Answer 34

square everything then divide by pi?

Answer 35

for pi r ^2

Answer 36

square, then mult by pi

Answer 37

you got A<= 441/4pi <= 110.25 /pi
and A>=100/pi
only one option satisfies both these condition.

Answer 38

\(\large (10/\pi) \le r \le (21/2\pi) \\ \large 100/\pi^2 \le r^2 \le 21^2/(4\pi^2) \\ \large 100/\pi \le \pi r^2 \le 21^2/(4\pi)\)

Answer 39

the numeric must be between 100 and 441/16

Answer 40

441/4 @sirm3d

Answer 41

i stand corrected =)

Answer 42

yeah i reached the 441/16 part! didnt know how to do both together though!!

Answer 43

sorry /4

Answer 44

A<= 441/4pi <= 110.25 /pi
and A>=100/pi

Answer 45

\[\large \frac{ 100 }{ \pi } \le \pi r^2 \le \frac{ 441/4 }{ \pi }\]\[\large \frac{ 100 }{ \pi } \le \pi r^2 \le \frac{ 110.25 }{ \pi }\] as @hartnn gave

Answer 46

Gotchaaaa!!! The answer would be C! Thank you both! @sirm3d @hartnn

Answer 47

welcome ^_^