Circle MANIA Part-2

Geometry Level 2

In the Figure ,

when all the outer circle have radii r,

then the radius of the inner circle is....

root 2* r 1/root2 *r 2/(root 2 +1)r (Square root 2 -1)r

4 solutions

Vi Le
Apr 23, 2014

Let the centre points of the outer cirles be A, B, C, D. Let the centre point of the smaller one be O ABCD is a square of which AB = 2R Therefore AC = square root 2 * R So r = AC/2 - R = (square root 2 - 1)*R

Hypotenuse 'x' from center of any big circle to center of small circle=Sqrt(r^2+r^2) and this Hypotenuse=Radius of Big circle+Radius of Small circle, So radius of small circle=[Sqrt(r^2+r^2)]-r=r(Sqrt2 -1)

Devendra Marghade - 7 years, 1 month ago

Basically what ViLe said.

Treat the centers of the circles as point A, B, C and D.

Draw a square by connecting those points, draw another line from any point to the point across from it to form a right triangle.

Calculate the length of the hypotenuse of that right triangle.

Then since the edge of the center circle is exactly 'r' away from the corners of the box, subtract 2r from the length of the hypotenuse.

This gives the circumference, now divide by two to get the radius.

Dorian Thiessen - 7 years, 1 month ago

Mostafa Algammal
May 22, 2014

sin45=(r+x)/(2r);
sin(45)={1/sqroot(2)}
x(r)=r[sqroot(2)-1]

Asad Ullah
Apr 30, 2014

so simple just concentrate on the figure.

Shreejeet Praveen
Apr 27, 2014

Join the center points of 3 circle to form a triangle whose hypotaneous is passing through small circle and then solve it by pythogoras th.

Circle MANIA Part-2

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