Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 600.
Let r1= radius of circle dia. AB, r2= radius of circle dia. BC and r3=radius of circle dia. CDApply Pythagoras theorem to circles with radius r1 and r2 we get(r1+r2)^2-(r1-r2)^2= 81 or 4.r1.r2=81Similarly with circles r2 and r3 (r2+r3)^2-(r2-r3)^2=1 or 4.r2.r3=1With circles r1 and r3(r1+2.r2+r3)^2-(r1-r3)^2=100 or 4.r2^2+4.r1.r3=18Solve system equations above we get r1=3/2 , r2= 27/2 , r3=1/6Replace values of r1, r2 and r3 in following formula we will get the resultArea S= ½( pi.(r1+r2+r3)^2 – pi.r1^2-pi.r2^2-pi.r3^2))Peter Tran
Minor typo correctionsolution of system of equations give:r1=27/2 ; r2=3/2 and r3=1/6Peter Tran
Pravin,Should the final answer not be 91π/4 ?Ajit
Applying Problem 812. Take a=9, b=1. Then S = π/4 (a^2 + b^2 + ab)= π/4 (81 + 1 + 9)= 91π/4
r1 --> Big circler2 --> medium circler3 --> small circler4 --> Biggest circleUsing formula of tangent of two circles we will have 3 equation:1. (r1 + r2)^2 = (r1 - r2)^2 + 812. (r1 + r2 + r3)^2 = (r1 - r3)^2 + 1003. (r2 + r3)^2 = (r2 - r3)^2 + 1from the 1st equation: r1 . r2 = 81/4from the 3rd equation: r2 . r3 = 1/4r2 = 1/4r3With the similarity of right angle triangle:4. (r3 + r2)/(r1 + 2r2 + r3) = 1/105. (r2 - r3)/(r1 - r2) = 1/10from the 4th equation and 5th equation:9r3 + 8r2 = 10r2 - 9r39r3 = r29r3 = 1/4r3r3 = 1/6r2 = 3/2r1 = 27/2r4 = 27Area of S = 91π/4Thank u