Let the radii of the (quarter/semi/full) circles be _p_ for purple, _b_ for blue and _y_ for yellow. Then we can write the equation system _{p² + (p − b)² = (p + b)², (b − y)² + x² = (b + y)², (p − y)² + (p − x)² = (p + y)², p² = 1296}._ The only solution with only positive numbers is _b_ = 9, _p_ = 36, _x_ = 12 and _y_ = 4. Therefore, the area of the yellow circle is π4² = 16π square units.
Nice problem sir relation between radii of these circles is 1/√r = 1/√a+1/√b where 'r' is the radius of middle circle and 'a' , 'b' for other circles here a = 36 , b= 9 1/√r = 1/6 +1/3 = 1/2 r = 4
At a quick glance: Denoting the Radii, RP, RB and RY for the Purple, Blue and Yellow circles. I took a look a comment below with a Pythagorean formula for the blue and purple circles. Trying this, Project Point E onto DC line, RB from D. Then RP^2 + (RP-RB)^2 = (RP+RB)^2. Then RP^2-2*RB + RB^2 + RP^2 = RP^2+2*RB +RB^2. Then RP^2 - 4*RB = 0. Then RB = 36/4 = 9. Projecting Point O onto lines AB and AD , Points F and G.. Then (RB+RY)^2 = AG^2 + (RB-RY)^2. Then (RP-AG)^2 +(RP-RY)^2 =(RP+RY)^2. Then (9+RY)^2 = AG^2 + (9-RY)^2, Equation (1) and (36-AG)^2 + (36-RY)^2 = (36 + RY)^2, Equation (2). 81 + 18*RY+RY^2 = AG^2 + 81 -18*RY + RY^2. Then 36 * RY = AG^2. Then 1296 -72 AG+AG^2 + 1296 - 72*RY + RY^2 = 1296 +72 * RY + RY^2. Then AG^2 - 72 * AG - 144*RY + 1296 = 0 . RY = AG^2/36 then AG^2 - 72 * AG - 144/36 * AG^2 + 1296 = 0. Then - 3 * AG^2 - 72 * AG + 1296 = 0. Then AG^2 + 24 AG - 432 = 0. Then ( AG + 36) * (AG -12) = 0. Then AG = 12 and RY = 144/36 = 4. The Yellow Radius is 4 and the yellow circle area is PI * 16 = 50.27 area units.
Since BCE, EFO and CHO are all Pythagorean triangles, we can predict that the area of the triangle CEO will be an integer. (the legs of a primitive Pythagorean triangle always include an odd and an even number, so its area is always an integer)
I did it almost the same as you with a little bit different for the last part: sq OH= sq(36+r)-(sq36-r)= (36+r+36-r)(36+r-36+r)=72.2r=144r (1) sq OF = sq(9+r) -sq(9-r) =(9+r+9-r)(9+r-9+r)= 18 .2r = 36r (2) (1)/(2)-----> sq (OH/OF) = 144r/36r= sq12/sq6----> OH/OF= 2 -----> OH/2=OF/1= (OH+OF/(2+1)= 36/3= 12 OF= 12-----> from (2) : r= 144/36= 4 Area of the yellow circle = 16 pi sq units
We find the radius of the blue semi circle just as here: it is 9. Then in an adapted orthonormal of center D we have O(R, u) C(36;0) E(9;36) VectorOC(36-R;-u) Vector OE(9-R;36-u), R beeing the radius of the yellow circle. OC^2 = (R+36)^2, so 1296 -72.R +R^2 +u^2 = R^2 + 72.R +1296, we simplify: 144.R = u^2 and so u = 12.sqrt(R) OE^2 = (R+9)^2, so 81 -18.R +R^2 +1296 -72.u +u^2 = R^2 +18.R +81, we simplify: -36.R+1296 -72.u +u^2 = 0, then we replace u by 12.sqrt(R) and u^2 by 144.R We have then 108.R -864.sqrt(R) +1296 = 0, or R -8.sqrt(R) +12 = 0 (when simplified by 108), this is a second degree equation with deltaprime is (-4)^2 -12 = 4 So: sqrt(R) = 4 +2 = 6 or sqrt(R) = 4 -2 = 2, and R =2 or R =6. We reject R = 6 (if R =6, then u = 12.sqrt(6) is about 29 and O should be inside the blue semi cercle) Finally R = 2 and the area of the yellow circle is 4.Pi
Sorry,, I made a fault. On line 6, please read: sqrt(R) = 6 or 2, then R = 36 or 4, and we reject R = 36 which is the length of the whole square, so R = 2, and the area of the yellow circle is 4.Pi
Yes, it is. If you make the whole picture two times larger, all radiuses would be twice as large => the product of the two smaller would be four times larger and therefore not equal the big radius.
@ 13:41 ...Cuckoo for Cocoa Puffs! If that "Tiny little yellow Circle" was a Hydrogen Atom, bam! right off the top of my head @ 15:13 r=0.037nanometers. 🙂
Let's find the area of the yellow circle: . .. ... .... ..... Since the blue semicircle and the purple quarter circle have exactly one point of intersection, this point (P) is located on the line connecting the centers of the circles (E and C). The same argument can be used to show that Q is located on CO and that T is located on EO. Now we apply the Pythagorean theorem to the right triangle BCE. With p being the radius of the pink quarter circle (and the side length of the square as well) and b being the radius of the blue semicircle we obtain: EC² = EB² + BC² (EP + CP)² = (AB − AE)² + BC² (b + p)² = (p − b)² + p² b² + 2*p*b + p² = p² − 2*p*b + b² + p² 4*p*b = p² ⇒ b = p/4 = √A(ABCD)/4 = √1296/4 = 36/4 = 9 A line throught point O, which is parallel to AD, may intersect AB at point R and CD at point S. Now we apply the Pythagorean theorem to the right triangles ERO and CSO. With y being the radius of the yellow circle we obtain: EO² = ER² + RO² (ET + OT)² = (AE − AR)² + RO² (b + y)² = (b − y)² + RO² b² + 2*b*y + y² = b² − 2*b*y + y² + RO² RO² = 4*b*y = 4*9*y = 36*y ⇒ RO = 6*√y CO² = CS² + SO² (CQ + OQ)² = (CD − DS)² + SO² (p + y)² = (p − y)² + SO² p² + 2*p*y + y² = p² − 2*p*y + y² + SO² SO² = 4*p*y = 4*36*y = 144*y ⇒ SO = 12*√y Since RO+SO=RS=AD=p, we can conclude: 6*√y + 12*√y = 18*√y = p = 36 ⇒ √y = 2 ⇒ y = 4 ⇒ A(yellow) = πy² = 16π Best regards from Germany
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IMHO, Premath provides the most clearly explained solutions to geometry problems.
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Where does 72R come from sir? Is it 2 sides added?
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Let the radii of the (quarter/semi/full) circles be _p_ for purple, _b_ for blue and _y_ for yellow.
Then we can write the equation system _{p² + (p − b)² = (p + b)², (b − y)² + x² = (b + y)², (p − y)² + (p − x)² = (p + y)², p² = 1296}._
The only solution with only positive numbers is _b_ = 9, _p_ = 36, _x_ = 12 and _y_ = 4.
Therefore, the area of the yellow circle is π4² = 16π square units.
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Nice problem sir
relation between radii of these circles is
1/√r = 1/√a+1/√b
where 'r' is the radius of middle circle and 'a' , 'b' for other circles
here a = 36 , b= 9
1/√r = 1/6 +1/3 = 1/2
r = 4
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At a quick glance: Denoting the Radii, RP, RB and RY for the Purple, Blue and Yellow circles. I took a look a comment below with a Pythagorean formula for the blue and purple circles. Trying this, Project Point E onto DC line, RB from D. Then RP^2 + (RP-RB)^2 = (RP+RB)^2. Then RP^2-2*RB + RB^2 + RP^2 = RP^2+2*RB +RB^2. Then RP^2 - 4*RB = 0. Then RB = 36/4 = 9. Projecting Point O onto lines AB and AD , Points F and G.. Then (RB+RY)^2 = AG^2 + (RB-RY)^2. Then (RP-AG)^2 +(RP-RY)^2 =(RP+RY)^2. Then (9+RY)^2 = AG^2 + (9-RY)^2, Equation (1) and (36-AG)^2 + (36-RY)^2 = (36 + RY)^2, Equation (2). 81 + 18*RY+RY^2 = AG^2 + 81 -18*RY + RY^2. Then 36 * RY = AG^2. Then 1296 -72 AG+AG^2 + 1296 - 72*RY + RY^2 = 1296 +72 * RY + RY^2. Then AG^2 - 72 * AG - 144*RY + 1296 = 0 . RY = AG^2/36 then AG^2 - 72 * AG - 144/36 * AG^2 + 1296 = 0. Then - 3 * AG^2 - 72 * AG + 1296 = 0. Then AG^2 + 24 AG - 432 = 0. Then ( AG + 36) * (AG -12) = 0. Then AG = 12 and RY = 144/36 = 4. The Yellow Radius is 4 and the yellow circle area is PI * 16 = 50.27 area units.
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So beautiful! And the icing on the cake is that BCE, EFO and CHO are all Pythagorean triangles 🤩
Yes they are!
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Since BCE, EFO and CHO are all Pythagorean triangles, we can predict that the area of the triangle CEO will be an integer.
(the legs of a primitive Pythagorean triangle always include an odd and an even number, so its area is always an integer)
Rp=36...arctg36/(36-R)=arccos(36-R)/(36+R).. applico tg..R=9...inoltre arccos(36-r)/(36+r)=arctg(36-2√Rr)/(36-r)..applico tg..risulta √144r=36-2√Rr...(R=9)..18√r=36...r=4..Ay=16π
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I did it almost the same as you with a little bit different for the last part:
sq OH= sq(36+r)-(sq36-r)= (36+r+36-r)(36+r-36+r)=72.2r=144r (1)
sq OF = sq(9+r) -sq(9-r) =(9+r+9-r)(9+r-9+r)= 18 .2r = 36r (2)
(1)/(2)-----> sq (OH/OF) = 144r/36r= sq12/sq6----> OH/OF= 2
-----> OH/2=OF/1= (OH+OF/(2+1)= 36/3= 12
OF= 12-----> from (2) : r= 144/36= 4
Area of the yellow circle = 16 pi sq units
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We find the radius of the blue semi circle just as here: it is 9.
Then in an adapted orthonormal of center D we have O(R, u) C(36;0) E(9;36) VectorOC(36-R;-u) Vector OE(9-R;36-u), R beeing the radius of the yellow circle.
OC^2 = (R+36)^2, so 1296 -72.R +R^2 +u^2 = R^2 + 72.R +1296, we simplify: 144.R = u^2 and so u = 12.sqrt(R)
OE^2 = (R+9)^2, so 81 -18.R +R^2 +1296 -72.u +u^2 = R^2 +18.R +81, we simplify: -36.R+1296 -72.u +u^2 = 0, then we replace u by 12.sqrt(R) and u^2 by 144.R
We have then 108.R -864.sqrt(R) +1296 = 0, or R -8.sqrt(R) +12 = 0 (when simplified by 108), this is a second degree equation with deltaprime is (-4)^2 -12 = 4
So: sqrt(R) = 4 +2 = 6 or sqrt(R) = 4 -2 = 2, and R =2 or R =6. We reject R = 6 (if R =6, then u = 12.sqrt(6) is about 29 and O should be inside the blue semi cercle)
Finally R = 2 and the area of the yellow circle is 4.Pi
Sorry,, I made a fault.
On line 6, please read: sqrt(R) = 6 or 2, then R = 36 or 4, and we reject R = 36 which is the length of the whole square, so R = 2, and the area of the yellow circle is 4.Pi
(270°+90°)=360° (36)^2=1296° (36)^2=1296° ,(1296°+1296°)=2592° (2592°-360°)=√2232 2^√11 2^√16 2^1 2^√4^√4 √2^1 √2^√2^√2^√2^2 √1^√1 √1^√1^√1 1^2 (x+1x-2)
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Is it just coincidence that the big radius is the product of the 2 smaller radio?
Yes, it is. If you make the whole picture two times larger, all radiuses would be twice as large => the product of the two smaller would be four times larger and therefore not equal the big radius.
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Clear and crispy don't thanks❤
16 pi with pythagorean tool 😁
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@ 13:41 ...Cuckoo for Cocoa Puffs! If that "Tiny little yellow Circle" was a Hydrogen Atom, bam! right off the top of my head @ 15:13 r=0.037nanometers. 🙂
😀
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😆
R=9
r=4
36=9*4😂
Let's find the area of the yellow circle:
.
..
...
....
.....
Since the blue semicircle and the purple quarter circle have exactly one point of intersection, this point (P) is located on the line connecting the centers of the circles (E and C). The same argument can be used to show that Q is located on CO and that T is located on EO. Now we apply the Pythagorean theorem to the right triangle BCE. With p being the radius of the pink quarter circle (and the side length of the square as well) and b being the radius of the blue semicircle we obtain:
EC² = EB² + BC²
(EP + CP)² = (AB − AE)² + BC²
(b + p)² = (p − b)² + p²
b² + 2*p*b + p² = p² − 2*p*b + b² + p²
4*p*b = p²
⇒ b = p/4 = √A(ABCD)/4 = √1296/4 = 36/4 = 9
A line throught point O, which is parallel to AD, may intersect AB at point R and CD at point S. Now we apply the Pythagorean theorem to the right triangles ERO and CSO. With y being the radius of the yellow circle we obtain:
EO² = ER² + RO²
(ET + OT)² = (AE − AR)² + RO²
(b + y)² = (b − y)² + RO²
b² + 2*b*y + y² = b² − 2*b*y + y² + RO²
RO² = 4*b*y = 4*9*y = 36*y
⇒ RO = 6*√y
CO² = CS² + SO²
(CQ + OQ)² = (CD − DS)² + SO²
(p + y)² = (p − y)² + SO²
p² + 2*p*y + y² = p² − 2*p*y + y² + SO²
SO² = 4*p*y = 4*36*y = 144*y
⇒ SO = 12*√y
Since RO+SO=RS=AD=p, we can conclude:
6*√y + 12*√y = 18*√y = p = 36
⇒ √y = 2
⇒ y = 4
⇒ A(yellow) = πy² = 16π
Best regards from Germany
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