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The easiest way to solve is to assume the problem has an answer. If it does, you can move the length 12 line lower until the inscribed circle shrinks to an infinitely small area, then solve the special case of finding the area of a quartercircle of radius 12, which is 36 pi.
That's Way more elegant solution.
Hi, can you explain it to me? I cant fully understand your method but im very curious.
@@gonzalolivetti6893 Im assuming that the green area stays the same if I change the size of the white circle, if it didn't stay the same, this problem wouldnt have a solution, because then the green area could be a bunch of different answers.
Exactly what I did.
Exactly. This reminded me of the "painting the circular hallway" problem: A contractor is to paint a circular hallway, and the only measurement he has is the length of the chord that is tangent to the interior offices. If this is all that is needed, then he (or she!) realizes that the specs do not define a particular radius or width, and so the interior offices can be shrunk to a circle of zero size, and the length of the chord then becomes the diameter, and the square footage (or meterage) can be calculated easily. You can also solve this hallway problem using algebra and triangles.
Small circle is floating and can be moved to the point tangent both x and y axis. Thus r=6 and R=12.sqrt2.
Great question!
There are a lot of ways of solving this, here's another.
[1.1] (2𝒔 + 𝒌)² = 𝒕² (2𝒔)² + 12²
[1.2] 4𝒔² + 4𝒔𝒌 + 𝒌² 4𝒔² + 144 … rearrange
[1.3] 4𝒔𝒌 + 𝒌² = 144
This - by itself - is kind of sterile. Well … then Léts work on the Area problem.
[2.1] Area = quarter circle minus lil' circle
[2.2] Area = ¼π(2𝒔 + 𝒌)² - π𝒔² … move common π
[2.3] Area = π(¼(2𝒔 + 𝒌)² - 𝒔²) … expand
[2.4] Area = π(¼(4𝒔² + 4𝒔𝒌 + 𝒌²) - 𝒔²) … involve ¼
[2.5] Area = π(𝒔² + 𝒔𝒌 + ¼𝒌² - 𝒔²) … cancel 𝒔²
[2.6] Area = π(𝒔𝒌 + ¼𝒌²)
Notice (to myself) that [2.6] looks like [1.3] scaled by ¼ OK.
[1.4] ¼(4𝒔𝒌 + 𝒌² = 144)
[1.5] 𝒔𝒌 + ¼𝒌² = 36
Yep, that's what we need as a substitute
[2.7] Area = 36π
And that's the end of the story, with 113.097 being the decimal result.
Cool
36*pi
Since they don't specify where the 12-long line is vertically, it must be the case that the green area is independent of that. So push it all the way down to the bottom, making the white circle go away completely. Then you just have a quarter circle with radius 12 - that has area 36*pi.
Solution:
r = radius of the white full circle,
R = Radius of the quarter circle.
Pythagoras: R²-(2r)² = R²-4r² = 12² = 144
Green area = area of the quarter circle - area of the white full circle
= π*R²/4-π*r² = π/4*(R²-4r²) = π/4*144 = 36π ≈ 113.0973
c = 12 x 2 = 24 cm
A = ¼ (¼ π c²)
A = π 24² / 16
A = 113,1 cm² ( Solved √ )
1) x:12=12:(2R-x)
2.) A=pi*R^2/4-pi*((R-x)/2)^2
2Rx-x^2=144
A= pi*R^2/4-pi*((R^2+x^2-2Rx)/4)
A= pi*R^2/4-pi*((R^2+144)/4)
A= pi*R^2/4-pi*R^2)/4+pi(144)/4)=36pi
Green area= 36π
My way of solution is ▶
Let O be the center of the circle and P be the point on the line x = 12 representing the radius OP. Then the radius OP, along with the line x = 12 and the segment OQ, forms a right triangle, where Q is the foot of the perpendicular from P to the line x = 12.
OQ= 2r
QP= 12
OP= R
Following from the Pythagorean theorem:
OQ²+QP²= OP²
(2r)²+12²= R²
4r²+144= R²
⇒
R²-4r²= 144
the greem area, Agreen
Agreen= πR²/4 - πr²
If we divide this equation above by 4 and multiply by π, we obtain this area:
R²-4r²= 144
multiplied by π:
πR²-4πr²= 144π
divided by 4:
πR²/4 -πr²= 144π/4
Agreen= 36 π
Agreen ≈ 113,1 square units
This is a nice problem which I am pleased to have solved in almost the same way you did. I was wondering if there is a way to find the values of R and r. I can’t immediately see that there is. Maybe someone can suggest a way. :)
No, R and r are related but can have an infinite number of values. The area, however, remains constant.