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[SOLVED] GRE Circles exam question: Find the shaded area as a fraction

karush

Well-known member
Jan 31, 2012
2,648
Screenshot_20191109-122447_Docs.jpg

Ok this is considered a "hard" GRE geometry question... notice there are no dimensions
How would you solve this in the fewest steps?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
Ok this is considered a "hard" GRE geometry question... notice there are no dimensions
How would you solve this in the fewest steps?
You have three circles inside. Call the radius of the largest inner circle r. Then the radius of the two smaller circles are r/2 and the radius of the big circle on the outside is 2r.

You are supposed to give the answer as a ratio so the r's eventually cancel out.

Can you finish?

-Dan
 

karush

Well-known member
Jan 31, 2012
2,648
You have three circles inside. Call the radius of the largest inner circle r. Then the radius of the two smaller circles are r/2 and the radius of the big circle on the outside is 2r.

You are supposed to give the answer as a ratio so the r's eventually cancel out.

Can you finish?

-Dan
Kinds I know that the area of a circle is
$A=\pi r^2$
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
639
Come on ... you can do this

Big circle - (medium circle + 2 small circles)
 

karush

Well-known member
Jan 31, 2012
2,648
If the radius of the big circle is 2 then the total of the 3 interior radius' is 1 since they are all on the diameter of the big circle.

$\dfrac{\pi(1)^2}{\pi(2)^2}=\dfrac{1}{4}$

Kinda...like..
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
If the radius of the big circle is 2 then the total of the 3 interior radius' is 1 since they are all on the diameter of the big circle.

$\dfrac{\pi(1)^2}{\pi(2)^2}=\dfrac{1}{4}$

Kinda...like..
Check those radii again... Try calling the radius of one of the small circles to be r = 1 and give it another try. Then look at this:
Area of large circle in the center: \(\displaystyle A =\pi r^2\)
Area of the two smaller circles: \(\displaystyle 2a = 2 ( \pi (r/2)^2 ) = (1/2) \pi r^2\)
Area of the whole circle: \(\displaystyle A_{big} = \pi (2r)^2 = 4 \pi r^2\)

Sum of the area of the inner circles: \(\displaystyle \pi r^2 + (1/2) \pi r^2 = (3/2) \pi r^2\)

So what is the ratio?


-Dan
 

karush

Well-known member
Jan 31, 2012
2,648
Check those radii again... Try calling the radius of one of the small circles to be r = 1 and give it another try. Then look at this:
Area of large circle in the center: \(\displaystyle A =\pi r^2\)
Area of the two smaller circles: \(\displaystyle 2a = 2 ( \pi (r/2)^2 ) = (1/2) \pi r^2\)
Area of the whole circle: \(\displaystyle A_{big} = \pi (2r)^2 = 4 \pi r^2\)


Sum of the area of the inner circles: \(\displaystyle \pi r^2 + (1/2) \pi r^2 = (3/2) \pi r^2\)

$\dfrac{4 \pi r^2-(3/2) \pi r^2}{4 \pi r^2}=\dfrac{8 \pi r^2-(3) \pi r^2}{8 \pi r^2}=\dfrac{5}{8}$

maybe
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
639
not maybe ... 5/8 is correct
 

karush

Well-known member
Jan 31, 2012
2,648
ahhh......I can eat lunch now in peace....