A Very Nice Geometry Problem | 3 Different Methods
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- Опубліковано 23 чер 2024
- A Very Nice Geometry Problem | 3 Different Methods
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My method:
Let radius of big quarter circle be R, and radius of small circle r.
By Pythagorean theorem:
(2r)²+8²=R²
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Shaded area: πR²/4-πr²=?
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4r²-R²=-64
R²-4r²=64
R²/4-r²=16
πR²/4-πr²=π(R²/4-r²)=16π≈50,29
My method: Big circle radius = R, little circle radius = r. The desired answer is pi (R^2/4 - r^2), so we're on the lookout for R^2/4 - r^2 in order to potentially get a shortcut. Connect O to D, then OC^2 + 64 = R^2, but OC = 2r, so 4r^2 + 64 = R^2. But that's just a little algebra from what we want: R^2/4 - r^2 = 16. So we're done, we get 16 pi.
I see that's the same as your method 1.
Your method 2 is a cute power of a point setup, but in the end you're still just hunting for something equivalent to R^2/4 - r^2.
Your method 3 is essentially just expanding the proof of the theorem behind method 2.
We note that length OC is not given, which implies that the solution is the same over a range of values for OC. When presented with this type of problem, I look for special cases. The most obvious special case here is length OC = 0. The small circle's radius becomes 0 and CD and OB become equal in length, so OB = 8. The quarter circle has area (1/4)(π)(8)² = 16π. If this were a multiple choice test, or we were not required to show our work, we could select 16π as the correct answer and move on. However, to check our work, or if the first special case were not apparent, another special case may be considered. Construct OD. As in method #1, we see right triangle ΔOCD. One side, CD, has length 8. We take the familiar Pythagorean triple 3 - 4 - 5 and double each side, so we have a 6 - 8 - 10 right triangle, so assign 8 to CD. OC = 6 and OB = 10. The radius of the quarter circle is OD, or 10, so its area = (1/4)(π)(10)² = 25π. The radius of the circle is OC/2 = 6/2 = 3. Its area is π(3)² = 9π. The difference in area is 25π - 9π = 16π, same answer as for the other special case.
The second special case does lead us to the general case solution given in method #1.
A more challenging problem: what are the radii for the circle and quarter circle when the circle is tangent to OA, OB and AB? (These should be the maximum radii.)
Exactly. I looked at it and shrank the small circle to 0 in my head, giving a quarter circle of r=8. A couple of seconds later you know the answer. No pencil needed.
My answer to the challenging problem: R = 8 / (root(8 * root(2) - 11)) (aprox. 14.28) and r = R * (root(2) - 1) (aprox. 5.91)
S=16π≈50,29
A = π (8+8)
A = 16π cm² ( Solved √ )
Potencia del punto C respecto a la circunferencia de redio "R" =8² =(R-2r)(R+2r)→ R²=64+4r² → Área sombreada =(πR²/4)-(πr²)=[π(64+4r²)/4]-(πr²) =π(16+r²-r²) =16π.
Gracias y un saludo.
10:55-13:05 Right Triangle Altitude Theorem:
CD²=AC•CE => (R-2r)(R+2r)=64 😁
This problem is similar to one three method problem that had similar first and second methods. The third method is clearly different. I shall use these problems for practice!!!
I want issue a correction: the third method is EXACTLY like the third method as the LAST three method problem a week ago.
Parabéns pelos seus vídeos! São bastantes instrutivos!
(8CD)^2 =64CD 90°/64CD 1.26 1^1.2^13 2^13^1 2^1^1 2^1 (CD ➖ 2CD+1)
4o metodo.
No triângulo ADE, usando as relações métricas no triângulo retângulo, 8^2=m.n
etc
❤❤🎉🎉😊
Can it have OC/OA>2*(sqrt(2)-1) ?
Blue area=16π
Dalctonic?