Geometry Problem

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## Saturday, May 7, 2011

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## Saturday, May 7, 2011

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Geometry Problem 600: Circle, Area, Tangent, Semicircle, Tangency Point

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Labels:
area,
center,
circle,
semicircle,
tangency point,
tangent

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Let r1= radius of circle dia. AB, r2= radius of circle dia. BC and r3=radius of circle dia. CD

ReplyDeleteApply Pythagoras theorem to circles with radius r1 and r2 we get

(r1+r2)^2-(r1-r2)^2= 81 or 4.r1.r2=81

Similarly with circles r2 and r3

(r2+r3)^2-(r2-r3)^2=1 or 4.r2.r3=1

With circles r1 and r3

(r1+2.r2+r3)^2-(r1-r3)^2=100 or 4.r2^2+4.r1.r3=18

Solve system equations above we get r1=3/2 , r2= 27/2 , r3=1/6

Replace values of r1, r2 and r3 in following formula we will get the result

Area S= ½( pi.(r1+r2+r3)^2 – pi.r1^2-pi.r2^2-pi.r3^2))

Peter Tran

Minor typo correction

ReplyDeletesolution of system of equations give:

r1=27/2 ; r2=3/2 and r3=1/6

Peter Tran

Comment:

ReplyDelete2S+π(r₁²+r₂²+r₃²)=π(r₁+r₂+r₃)²

S=π(r₁r₂+r₂r₃+r₃r₁)=π(81+1+100)/4=91π/2

Pravin,

ReplyDeleteShould the final answer not be 91π/4 ?

Ajit

Applying Problem 812.

ReplyDeleteTake a=9, b=1.

Then

S = π/4 (a^2 + b^2 + ab)

= π/4 (81 + 1 + 9)

= 91π/4

r1 --> Big circle

ReplyDeleter2 --> medium circle

r3 --> small circle

r4 --> Biggest circle

Using formula of tangent of two circles we will have 3 equation:

1. (r1 + r2)^2 = (r1 - r2)^2 + 81

2. (r1 + r2 + r3)^2 = (r1 - r3)^2 + 100

3. (r2 + r3)^2 = (r2 - r3)^2 + 1

from the 1st equation: r1 . r2 = 81/4

from the 3rd equation: r2 . r3 = 1/4

r2 = 1/4r3

With the similarity of right angle triangle:

4. (r3 + r2)/(r1 + 2r2 + r3) = 1/10

5. (r2 - r3)/(r1 - r2) = 1/10

from the 4th equation and 5th equation:

9r3 + 8r2 = 10r2 - 9r3

9r3 = r2

9r3 = 1/4r3

r3 = 1/6

r2 = 3/2

r1 = 27/2

r4 = 27

Area of S = 91π/4

Thank u

y = radius of the smallest circle (circle on the right)

ReplyDeletex = radius of the circle next to the smallest circle (circle on the middle)

r = radius of the circle on the left

By pythagorean theorem,

(r+x)² = (r-x)² + 81

rx = 81/4 or r = 81/4x (1)

(x+y)² = (x-y)² + 1

xy = 1/4 or y = 1/4x (2)

By similar triangles,

(x-y)/(r-x) = 1/9

9x - 9y = r-x (3)

Substitute (1) and (2) on (3), we get x = 3/2

then solve for r and y

the next is piece of cake