I prefer often an analytical solution which is more straightforward. Give coordinates to the different points: (0,0) to the center of the circle and then (0,h), (2,h), (2,4-h) and (5,h-4) to the points of the blue segments. The points (2,h) and (5,h-4) are on the circle which gives: 2^2+h^2 = 5^2+(h-4)^2=r^2 which gives h=37/8 and r=5.sqrt(65)/8
Let OM be the veryicsl radius of quarter circle O and ON be the horizontal radius, length r for both. Drop a perpendicular from C to E, and from D to F on ON. Let CE = DF = x. As ∠EOA = ∠OAB = ∠ABE = ∠BEO = 90°, then ABEO is a rectangle, OE = AB = 2, and OA = BE = BC+CE = 4+x. As ∠FEC = ∠ECD = ∠CDF = ∠DFE = 90°, then CDFE is a rectangle and CD = FE = 3. Draw radii OB and OD. Triangle ∆BEO: OE² + BE² = OB² 2² + (4+x)² = r² 4 + 16 + 8x + x² = r² x² + 8x + 20 = r² --- [1] Triangle ∆DFO: DF² + OF² = OD² x² + (2+3)² = r² x² + 25 = r² --- [2] x² + 8x + 20 = x² + 25
Une fois de plus la solution proposée escamote le problème. La première question qui vient à l'esprit obligatoirement est la suivante : une telle figure est-elle possible ? Si la réponse est non il faut expliquer pourquoi , si la réponse est oui il faut proposer une construction géométrique. La réponse exige de remarquer que pour construire la figure correctement en vraie grandeur il faut déterminer le centre du quart de cercle qui se trouve être à l'intersection d'une droite de la figure et d'une médiatrice du segment déterminé par les deux points de la ligne brisée qui se trouvent être sur ce quart de cercle. A défaut de commencer par là , il faut si on commence par les calculs, s'assurer que les résultats sont cohérents avec une construction géométrique .
I imagine there are several ways to tackle this: Extend BC down to the diameter line for 4+y. OB is a radius. The base is 2. (4+y)^2 + 2^2 = r^2 20 + 8y + y^2 = r^2 OD is also a radius and extend CD to the left to make another right triangle with sides 5, y, and r. y^2 + 5^2 = r^2 We now have y^2 + 8y + 20 = y^2 + 25 Subtract one equation from the other to zero it: 8y - 5 = 0 8y = 5 y = 5/8 Substitute (5/8)^2 + 5^2 = r^2 25/64 + 25 = r^2 r^2 = 1625/64 Verify with the other triangle: (37/8)^2 + 2^2 = r^2 1369/64 + 256/64 = 1625/64, so verified. Now square root r^2 (sqrt(25)*(sqrt(65))/8 5*sqrt(65)/8, and that seems to be as far as I can go without decimal rounding. 40.31/8, so about 5.04 r approximates to 5.04 Yes, we used mainly the same method, but with slight variations in labelling and calculation.
You can sanity check that your answer isn't correct by noticing that your answer is less than 5, but from the diagram the radius is the hypotenuse of a right triangle one of whose legs is 5. In fact drawing the two right triangles and equating their hypotenuses is a way to solve the problem, since the y-coordinate of C and D will satisfy 2^2+(4+y)^2=r^2=5^2+y^2.
I want to apologize and say that I incorrectly calculated my answer. The radius is (5*sqrt(65))/8. I am deeply sorry. And I correct the calculation that I did.
Intersecting Chords Theorem at C
7*3=4*(4+2a)
a=5/8
I did it that way, too.
I prefer often an analytical solution which is more straightforward. Give coordinates to the different points: (0,0) to the center of the circle and then (0,h), (2,h), (2,4-h) and (5,h-4) to the points of the blue segments. The points (2,h) and (5,h-4) are on the circle which gives: 2^2+h^2 = 5^2+(h-4)^2=r^2 which gives h=37/8 and r=5.sqrt(65)/8
I like this method
x²+(y+a)²=r²
(2,4): 4+(4+a)²=r²
(5,0): 25+a²=r²
20+8a+a²=25+a², a=5/8
r=√[25+25/64]=5√65/8
Potencia de C respecto a la circunferencia =3(2+2+3)=4(h+h+4)---> h=5/8 --->[4+(5/8)]²+2²=r²---> r=(5√65)/8.
Gracias y saludos.
Let OM be the veryicsl radius of quarter circle O and ON be the horizontal radius, length r for both. Drop a perpendicular from C to E, and from D to F on ON. Let CE = DF = x.
As ∠EOA = ∠OAB = ∠ABE = ∠BEO = 90°, then ABEO is a rectangle, OE = AB = 2, and OA = BE = BC+CE = 4+x. As ∠FEC = ∠ECD = ∠CDF = ∠DFE = 90°, then CDFE is a rectangle and CD = FE = 3.
Draw radii OB and OD.
Triangle ∆BEO:
OE² + BE² = OB²
2² + (4+x)² = r²
4 + 16 + 8x + x² = r²
x² + 8x + 20 = r² --- [1]
Triangle ∆DFO:
DF² + OF² = OD²
x² + (2+3)² = r²
x² + 25 = r² --- [2]
x² + 8x + 20 = x² + 25
(2)^2=4 (3)^2=9 (4)^2=16:={4+9+16}=29 90°ABCDPO/29 =3.3 (ABCDPO ➖ 3ABCDPO+3).
AB = 2; BC = 4; CD = 3; PC = 2; BE = BC + CM + EM = 4 + CM + EM ↔BM = EM →
CE = k → 21 = 4k → k = 21/4 → BM = 37/8 → r = √((37/8)^2 + 4) ) = 5√65/8
r^2=2^2+(4+a)^2....r^2=5^2+a^2...eguagliò le due espressioni risulta a =5/8...r^2=25+25/64..r=5√65/8
Une fois de plus la solution proposée escamote le problème. La première question qui vient à l'esprit obligatoirement est la suivante : une telle figure est-elle possible ? Si la réponse est non il faut expliquer pourquoi , si la réponse est oui il faut proposer une construction géométrique. La réponse exige de remarquer que pour construire la figure correctement en vraie grandeur il faut déterminer le centre du quart de cercle qui se trouve être à l'intersection d'une droite de la figure et d'une médiatrice du segment déterminé par les deux points de la ligne brisée qui se trouvent être sur ce quart de cercle.
A défaut de commencer par là , il faut si on commence par les calculs, s'assurer que les résultats sont cohérents avec une construction géométrique .
Answer = 5*(sqrt(65))/8 ≈ 5.04 units
I imagine there are several ways to tackle this:
Extend BC down to the diameter line for 4+y.
OB is a radius.
The base is 2.
(4+y)^2 + 2^2 = r^2
20 + 8y + y^2 = r^2
OD is also a radius and extend CD to the left to make another right triangle with sides 5, y, and r.
y^2 + 5^2 = r^2
We now have y^2 + 8y + 20 = y^2 + 25
Subtract one equation from the other to zero it:
8y - 5 = 0
8y = 5
y = 5/8
Substitute (5/8)^2 + 5^2 = r^2
25/64 + 25 = r^2
r^2 = 1625/64
Verify with the other triangle:
(37/8)^2 + 2^2 = r^2
1369/64 + 256/64 = 1625/64, so verified.
Now square root r^2
(sqrt(25)*(sqrt(65))/8
5*sqrt(65)/8, and that seems to be as far as I can go without decimal rounding.
40.31/8, so about 5.04
r approximates to 5.04
Yes, we used mainly the same method, but with slight variations in labelling and calculation.
2+3 =5
Acertei !!!!
5((65)^2)/8
I got this one!
The answer is (5*sqrt(15))/4. That is the radius. And boy oh boy I better use that as practice!!! And maybe compare that to Maths Premi!!!
You can sanity check that your answer isn't correct by noticing that your answer is less than 5, but from the diagram the radius is the hypotenuse of a right triangle one of whose legs is 5. In fact drawing the two right triangles and equating their hypotenuses is a way to solve the problem, since the y-coordinate of C and D will satisfy 2^2+(4+y)^2=r^2=5^2+y^2.
I suggest a quick recalculation: 5*sqrt(15) < 20 (because sqrt(16) is 4.), so you have
I want to apologize and say that I incorrectly calculated my answer. The radius is (5*sqrt(65))/8. I am deeply sorry. And I correct the calculation that I did.
@@MrPaulc222 I will look at my calculation. I think that I mis-typed what I calculated or I mis-read what I calculated. I apologize.
@@michaeldoerr5810 No need to apologise. We all get some wrong at times.