- #1
chwala
Gold Member
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- Homework Statement
- see attached
- Relevant Equations
- pythagoras theorem
Find the question here and the solution i.e number 10 indicated as ##6-4\sqrt{2}##,
I am getting a different solution, my approach is as follows. I made use of pythagoras theorem for the three right angle triangles as follows,
Let radius of the smaller circle be equal to ##c## and distance from mid point of smaller circle to circle with radius ##1=x## then it follows that,
##x^2+(1-c)^2=(1+c)^2##
##x^2=4c##
let us also have the distance from the smaller circle to circle with radius ##2=y##, then it follows that,
##y^2+(2-c)^2=(2+c)^2##
##y^2=8c##
Let ##y^2+x^2=AB##, then it follows that, ##(AB)^2 + (2-1)^2=3^2##
##AB=2\sqrt{2}##
then it follows that, ##\sqrt{8}=2\sqrt{c}+2\sqrt{2c}##
Am i missing something here!
i got it!
##\sqrt{c}##=##\dfrac {\sqrt {8}}{2+2\sqrt{2}}##
⇒##c=0.34314575## Bingo guys! Phew i took time on this men!
I am getting a different solution, my approach is as follows. I made use of pythagoras theorem for the three right angle triangles as follows,
Let radius of the smaller circle be equal to ##c## and distance from mid point of smaller circle to circle with radius ##1=x## then it follows that,
##x^2+(1-c)^2=(1+c)^2##
##x^2=4c##
let us also have the distance from the smaller circle to circle with radius ##2=y##, then it follows that,
##y^2+(2-c)^2=(2+c)^2##
##y^2=8c##
Let ##y^2+x^2=AB##, then it follows that, ##(AB)^2 + (2-1)^2=3^2##
##AB=2\sqrt{2}##
then it follows that, ##\sqrt{8}=2\sqrt{c}+2\sqrt{2c}##
Am i missing something here!
i got it!
##\sqrt{c}##=##\dfrac {\sqrt {8}}{2+2\sqrt{2}}##
⇒##c=0.34314575## Bingo guys! Phew i took time on this men!
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