Challenge Questions

I gave this a "good answer" vote because it shows at least two things graphically:

1. That based on observation of the three circle sets, you can quickly see that the inner circle can be reduced to a point, making the answer obvious.

2. There is an interesting optical illusion which makes the horizontal line between the large circles on the far right appear shorter than the line between the smaller circles on the far left.

Typo error in line 1. Should read

Since r1=OX and r2= OY where O is the centre of the circles

Taking a triangle OXY, where O is the common centre, we get:

sinθ = r₁/r₂

cosθ = 2/r₂

sin²θ + cos²θ = 1

so r₂² = r₁² + 4

Area = (r₂² - r₁²)

so Area = 4π = 12.57 in² = 0.0081 m²

I guess what could be considered unusual is that the annular area is equal to (XY)² · pi, which is valid for any length XY.

PS

If R² - r² is always 2² then the area is constant at (2² x ∏)/4 = 3.141592654 ?

Regards JD.

2² x ∏ = 12.56637061.

Regards JD.

the answer is 4*3.14

R is radius of big circle

r is the radius of small circle

R*R=r*r+2*2

so R²-r²=4

area of annulus 3.14*( R²-r²)
so area of annulus is 4 * 3.14

The answer is 4 in².

Since the area is independent of circle diameters assume the inner circle to be of '0' diameter. Then the area = 4 sq inches

Of course! 4*pi sq inches!

The answer is 12.5663 in².

This is in continuation to the answer of Π×4 which shows that r is zero which does not satisfy the condition given in narrative of the problem that their are two cocentric circles. The answer has to be for different values of r.thanks

22/7

22/7

A case of 2 concentric circles and a line segment with one end tangent to the small circle and the other end touching the big circle. A matter of areas of the 2 circles, A_b= πR² and A_s= πr². Area of big circle as A_b with radius R and area of small circle A_s with radius r. The line segment XY and the radii R and r by construction will form a right triangle with vertices at the center C and the ends of the line segment XY (2 in. long). Thus, the right triangle CXY formed with R as the hypotenuse. Area between the circles as A= πR^2- πr² or π(R^2- r² ). Applying the Pythagorean Theorem, R²= r²+2² or R²- r²=4. By substitution A=4π sq.in., the area of the annular region between the two circles. _{(Is this really a Challenge Question?)}

XY = 2". Full chord = 2XY = 4".

Since the area has to be independent of the inner circle, we can consider the limit of this going to zero. Under this condition the chord, which is a tangent of the inner circle, becomes the diametaer of the outer circle, whose area = 4∏ sq.in.

The anuular area = 4∏ sq.in.

A good chance it's 4*pi in² , without going to the trouble of working it out.

From the way the question is put it looks like area only depends on the tangent length, not the radii of the 2 circles. So make the inner circle zero radius.

Codey

4∏

Certainly the answer is 4π. But is very important to take into account we need some conditions:

but certainly this works if r+m>2 but r>0

r>2-m>0

finally it works for every value of r if r>2>m

You seem to be making something simple rather complicated!

If you insist on solving a quadratic I make it m = √(r²+4) - r. If r = 0, m = 2 as you'd expect. Less obvious that m = 0 when r = ∞, but if it's written

m = r*√(1 + 4/r²) - r =

r*(1 + 4/r²)^1/2 - r =

(by the binomial theorem, when r is large) r*(1 + 2/r²) - r =

r - r = 0 as r → ∞

Cheers.......Codey

I would like to add this result for the discution

this means if the basis of the triangle is 3 then A=9*pi

and for any value of b it is constant and is not dependant of the diameters of the circles.

Given no limits on the radii, lets assume that r1 is zero. the area between the point and the outer ring would be Pi*r^2. Substituting 2 for r the area is 4*Pi.

The general formula for a sector of a annulus is 1/2*h(s1+s2)

where: h=r2-r1 and s1 and s2 are the lengths of the arcs defined by the angle of the sector theta. In the case of a full circle instead of a sector, s1 and s2 are the circumferences of the two circles. Again with r1=0 the result is 1/2*2(0+2*Pi*2)=4*Pi

For a non-zero r1, assume it is 1. Then the hypotenuse for the right triangle with height 1 and base 2 is sqrt(1+4)=2.236. substituting into the general formula:

1/2*(2.236-1)*(2Pi*1+2pi*2.236)=1/2*1.236*(2pi+4.472Pi)=0.618*6.472Pi=4Pi

On the far end of the possibilities assume that r1 = 100.

then r2 = sqrt(10000+4)=100.02. Substituting again

1/2*(100.02-100)*(2Pi*100+2Pi*100.02)=1/2(.02)(400.02Pi)=4.0002Pi

(small error from rounding)

Assuming that the 2 inch line terminates on the outer circle:

A line 0X from the center of the inner circle to the line tangent to the inner circle is the radius of the inner circle Rs. The line OY represents the radius of the outer circle, Ro. The area between the circles is A=Pi * (Ro^2 - Rs^2). Ro^2 -Rs^2 =xy^2 =4^2=16 sq-in. Therefore A=16*Pi sq in.

Forgive the brain slip... I squared 2 in my head and wrote down 4 and then turned around and squared the four. It should have been 4*Pi sq in.

A = pi/4*(D^2 - d^2) = pi/4*{ (2*R)^2 - (2*r)^2 }

= pi/4*{ 4*(R^2 - r^2) } = pi/4*{ 4*2 } = 2*pi

This is a correction:

A = pi*(R^2 - r^2)

From the figure, R^2 - r^2 = 2^2 = 4

A = 4*pi

If R is radius of large circle and r that of small circle then (R*R-r*r)=4

so (R+r)*(R-r)=4

Now i am going to take the easy route

By examination if R=2.5 and r=1.5 the above equation is satisfied.

so the diameter of large is 5 and small is 3.

so 0.785398(25-9)=12.566371 Sq inches

The area is 12.5664 square inches. If you reduce the center circle to a radius of zero then you are left only to calculate the area of the outer circle. The 2 inch line becomes the radius of the circle. The formula for the area of a circle is pi r squared or approximately 3.14*2in*2in = 12.5664 square inches