2 circles sit inside rectangle ABCD, such that

AB is tangent to both circles,

AC and CD are tangent to the larger circle

BD is tangent to the smaller circle.

## 18 thoughts on “Radius Problem #1”

1. Y Scott says:

is there a piece of information missing? As I have tried several strategies and keep getting r = r or 4 = 4

• Are you using the 12? And pythagoras…

2. Y Scott says:

sorry ignore my comment I think I’ve solved it !

• Juan Carlos Saavedra says:

I get 2.1444 cms

• Juan Carlos Saavedra says:

Sorry is r2 = 2.14359360665022975737527008 cms

3. Y Scott says:

Yes

4. Mary Anne says:

I’m not seeing it

• Draw a right angled triangle where the hypotenuse is the 2 radii as shown

5. Sarah says:

Is it 2cm?

• Juan Carlos Saavedra says:

Yes and no, if you consider that (4- X)+(8-X)=4+X

6. Juan Carlos Saavedra says:

2.152636 cm. = (12 x 8) – (16 x pi ) this is the area of the small circle ( 45.734 squared cms ) the squared root of this amount is divided by pi ( 3.14159 ) it give you 2.152636 cm.

• height of the rectangle is not 8…

• nevermind that was dumb but how do you account for the area of the rest of the rectangle?

7. Juan Carlos Saavedra says:

Another method that give me 2.157 cms is to draw a circle half the size of the big circle (2 cms) and place it at the center of the small circle by making an octagono I get that the circumscribe lenght of 2.157 cms.

8. Juan Carlos Saavedra says:

Final answer is a quadratic equation in r2 = (4 – r2) squared + ( 8 – r2 ) squared = ( 4 + r2 ) squared.

• Juan Carlos Saavedra says:

So I get 2.1444 cms