-gre.ge.3 Circles Find the shaded area as a fraction

[karush]

View attachment 9352

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

 

[topsquark]

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]

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]

Come on ... you can do this

Big circle - (medium circle + 2 small circles)

 

[karush]

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]

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:

Spoiler

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]

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:

Spoiler

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: \(\displaystyle A_{big} = \pi (2r)^2 = 4 \pi r^2\)

Spoiler

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

 

[skeeter]

not maybe ... 5/8 is correct

 

[karush]

mahalo sorry my likes were so late