At a quick glance: AB * 2 + 16 * 2 = 80. AB = 24. Rb and Rr are the Radii of the blue and Red circles. Drawing a perpendicular from O to F and finding the intersection, M with a horizontal line from P. EQ(1): OP ^2 = OM ^2 + FQ ^2 and OF = Rb-OM, EQ(2):OM = 8- Rr. OP= Rb+Rr = 8+Rr. FQ = 24- AF - QB. EQ(3):FQ = 24- 8 - Rr = 16 - Rr. Substituting EQ's (2) and (3) in EQ(1): OP^2 = (8-Rr)^2 + (16-Rr) ^2. OP = Rr + 8. EQ(4): (Rr+8)^2 = (8-Rr)^2 + (16-Rr)^2. Expanding this equation. Rr^2 + 2*8*Rr + 64 = 64 -16Rr + Rr^2 + 256 -32Rr +Rr^2. This reduces to 16Rr+Rr^2 = -48Rr+ 2Rr^2 +256 and 64Rr-Rr^2 - 256 =0. (-Rr+16)(Rr-16 ) + 64Rr = 0, Using quadratic formula -b +-/sqrt( b^2-4ac)/2a. a=-1, b=64 and c= -256. This gives Rr = 4.287
Very nice example! I did it as follows: Radius of big circle = R = 8. Heigth of rectangle = 2R = 16. Draw a vertical line in O and a horizontal line in P. They meet at T. Then consider the right angled triange TP² + TO² = OP² Since [80 - (2·16)]/2 = 24 is the width of the outer rectangle, we have TP = 24 - R - r = (16 - r), TO = (8 - r) and OP = R + r = (8 + r). This gives the condition (16 - r)² + (8 - r)² = (8 + r)² Now we calculate 256 - 32r + r² + 64 - 16r + r² = 64 + 16r + r² 2r² - 48r + 256 = r² + 16r r² - 64r + 256 = 0 The p/q-Formula then leads to r = 32 ± √(32² - 256) = 32 ± √768 Since 768 = 3·256 = 3·2⁸ one can write 32 ± √(3·2⁸) = 32 ± 2⁴·√3 = 32 ± 16√3 The 32 + 16√3 makes no sense here, so the solution is r = 32 - 16√3 = 16(2 - √3)
My solution was nearly identical, I feel that in the last step using the so called pq-forumla (x²+px+q = 0 => x = - p/2 +- sqrt((-p/2)² - q) to solve for r is faster, but maybe it is just a matter of preference
4.28719 Answer Using 16 - r , and 8 - r, as the base and 8 + r as the hypotenuse since perimeter = 80, then 2 ( L + W) = 80, L + W = 40 Since blue circle with radius 8 covers the entire width, then Width= diameter = 16, Hence length = 24 Draw a line from the center of the blue to p = hypotenuse 8 + radius of red Draw a parallel line from the center of blue to F ( O F), and one from the center of red to F : label the intersecting line , T. Then the distance P T = 16 - radius of red The center of blue to T = 8- radius of red, Using Pythagorean theorem (16- radius of red)^2 + (8-radius of red)^2 = (8 + radius of red)^2 let the radius of red = r , then (16-r)^2 + (8- r)^2 = (8+ r)^2 256 + r^2 -32 r + 64 + r^2 -16 r = 64 + 16r + r^2 r^2 = 64 r + 256 = 0 r = 4.28719 (quadratic formulae calculator)
Let :a=length ; b=width b=8+8=16 units 2(a+b)=80 a+b=40 a=40-16=24 units Connect OE; OF ; OG; OP; PS; PQ r is radius of small circle DE=AF=8 units CE=BF=a-8 BS=BQ=r OP=r+8 OT right PT PT=24-8-r=16-r OT=8-r (OT)^2+(PT)^2=(OP)^2 (8-r)^2+(16-r)^2=(8+r)^2 64-16r+r^2+256-32r+r^2=64+16r+r^2 2r^2-48r+256=r^2+16r r^2-64r+256=0 So: r=32-16√3=4.29units❤❤❤Thanks
If R = 4.29 units the D should be 2R and if that was true the red circle should be bigger? Because the red circle should be taller than the center of the blue circle?
But what if we just make a line from C to AB and create a trapezoid. Then we later with formula of circle inside of a trapezoid we say: 16+c(it means c to ab line)=24+24-x?
Why dont just 80-16-16 since diameter 16 so we get 28 and then we divide by 2 for one side =24 after this since it diameter 16 we can substrate 24 with 16 an get ref diameter which is 8.Then 8/2 to get red radius?
beatiful
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At a quick glance: AB * 2 + 16 * 2 = 80. AB = 24. Rb and Rr are the Radii of the blue and Red circles. Drawing a perpendicular from O to F and finding the intersection, M with a horizontal line from P. EQ(1): OP ^2 = OM ^2 + FQ ^2 and OF = Rb-OM, EQ(2):OM = 8- Rr. OP= Rb+Rr = 8+Rr. FQ = 24- AF - QB. EQ(3):FQ = 24- 8 - Rr = 16 - Rr. Substituting EQ's (2) and (3) in EQ(1): OP^2 = (8-Rr)^2 + (16-Rr) ^2. OP = Rr + 8. EQ(4): (Rr+8)^2 = (8-Rr)^2 + (16-Rr)^2. Expanding this equation. Rr^2 + 2*8*Rr + 64 = 64 -16Rr + Rr^2 + 256 -32Rr +Rr^2. This reduces to 16Rr+Rr^2 = -48Rr+ 2Rr^2 +256 and 64Rr-Rr^2 - 256 =0. (-Rr+16)(Rr-16 ) + 64Rr = 0, Using quadratic formula -b +-/sqrt( b^2-4ac)/2a. a=-1, b=64 and c= -256. This gives Rr = 4.287
Very nice example! I did it as follows:
Radius of big circle = R = 8. Heigth of rectangle = 2R = 16.
Draw a vertical line in O and a horizontal line in P. They meet at T.
Then consider the right angled triange TP² + TO² = OP²
Since [80 - (2·16)]/2 = 24 is the width of the outer rectangle, we have
TP = 24 - R - r = (16 - r),
TO = (8 - r) and
OP = R + r = (8 + r).
This gives the condition
(16 - r)² + (8 - r)² = (8 + r)²
Now we calculate
256 - 32r + r² + 64 - 16r + r² = 64 + 16r + r²
2r² - 48r + 256 = r² + 16r
r² - 64r + 256 = 0
The p/q-Formula then leads to
r = 32 ± √(32² - 256) = 32 ± √768
Since 768 = 3·256 = 3·2⁸ one can write
32 ± √(3·2⁸) = 32 ± 2⁴·√3 = 32 ± 16√3
The 32 + 16√3 makes no sense here, so the solution is r = 32 - 16√3 = 16(2 - √3)
Awesome!
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L = 8 + (8 + r) cos(α) + r = (8 + r) + (8 + r) cos(α) = y + y cos(α) => cos(α) = (L - y) / y, where 8 + r = y
W = 8 + (8 + r) sin(α) + r = y + y sin(α) => sin(α) = (W - y) / y
(cos(α))^2 + (sin(α))^2 = 1 => ((L - y) / y)^2 + ((W - y) / y)^2 = 1
Resolved the y-based quadratic equation, r = y - 8
My solution was nearly identical, I feel that in the last step using the so called pq-forumla (x²+px+q = 0 => x = - p/2 +- sqrt((-p/2)² - q) to solve for r is faster, but maybe it is just a matter of preference
4.28719 Answer
Using 16 - r , and 8 - r, as the base and 8 + r as the hypotenuse
since perimeter = 80, then 2 ( L + W) = 80, L + W = 40
Since blue circle with radius 8 covers the entire width, then Width= diameter
= 16, Hence length = 24
Draw a line from the center of the blue to p = hypotenuse 8 + radius of red
Draw a parallel line from the center of blue to F ( O F), and one from the center
of red to F : label the intersecting line , T. Then the distance P T = 16 - radius of red
The center of blue to T = 8- radius of red,
Using Pythagorean theorem
(16- radius of red)^2 + (8-radius of red)^2 = (8 + radius of red)^2
let the radius of red = r , then
(16-r)^2 + (8- r)^2 = (8+ r)^2
256 + r^2 -32 r + 64 + r^2 -16 r = 64 + 16r + r^2
r^2 = 64 r + 256 = 0
r = 4.28719 (quadratic formulae calculator)
Thanks dear ❤️🌹
Let :a=length ; b=width
b=8+8=16 units
2(a+b)=80
a+b=40
a=40-16=24 units
Connect OE; OF ; OG; OP; PS; PQ
r is radius of small circle
DE=AF=8 units
CE=BF=a-8
BS=BQ=r
OP=r+8
OT right PT
PT=24-8-r=16-r
OT=8-r
(OT)^2+(PT)^2=(OP)^2
(8-r)^2+(16-r)^2=(8+r)^2
64-16r+r^2+256-32r+r^2=64+16r+r^2
2r^2-48r+256=r^2+16r
r^2-64r+256=0
So: r=32-16√3=4.29units❤❤❤Thanks
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If R = 4.29 units the D should be 2R and if that was true the red circle should be bigger? Because the red circle should be taller than the center of the blue circle?
Bro he literally said that the diagram may not be 100% to scale 0:45
@@sofianehawati3587 I know "BUT" that is really messes things up because it might be solved a different way that is all i was pointing out.
Please make some more videos on algebra
I'm working on them...
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Yes, I got that one, but it took me a while to figure out where the triangle goes :)
Again, thank you.
Awesome!
Thanks❤️
Rectángulo b×h》h=2×8=16》b=[80-(2×16)]/2=24》(8+r)^2 -(16-r)^2 =(8-r)^2》r=16(2-sqrt3) =4.2872
Gracias y saludos.
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But what if we just make a line from C to AB and create a trapezoid. Then we later with formula of circle inside of a trapezoid we say: 16+c(it means c to ab line)=24+24-x?
r=16(2-√3)≈4,32 un
Why dont just 80-16-16 since diameter 16 so we get 28 and then we divide by 2 for one side =24 after this since it diameter 16 we can substrate 24 with 16 an get ref diameter which is 8.Then 8/2 to get red radius?
Love it!!!!!!!!!!!!!
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Very nice .......I thought it was going to be positive
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Dai dati risulta b=24,h=16...l'equazione risolutiva è 8+√((8+r)^2-(8-r)^2)+r=24....r=32-16√3
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Nice! sin(δ) = (8 - r)/(8 + r) = (√3/13)(4 - √3)
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16 = sqrt(32r)+r and r = 4.32
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First comment very nice video❤❤❤❤
Thanks you Ibrahim❤
I got it without looking at the answer, word for word.
Super!
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I used plank length for the units. Then it becomes a small problem… 🙄🎶🎵
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