I am so happy for making parts of this channel 3 months ago. Thanks to your lessons , I was able to solve this on my own like it was nothing. 👍💯 you're the best.
The point P is the centre of the outer square. So another circle of size equal to the yellow one can be drawn in the diagonally opposite corner. So a measure taken from point T through O to centre O’ and to T’ projected on to a side of the outer square measure r + r/sqrt2 +rsqrt2 + r and so value is r= 4.10
Alternative method using incircle formula A = RS: 1. Join BD to form triangle ABD. BD = diagonal of square ABCD = 14sqrt2 Area of triangle ABD = (1/2)(14)(14) = 98 = A Semiperimeter of triangle ABD = (1/2)(14 + 14 + 14sqrt2) = 7(2 + sqrt2) = S 2. Yellow circle is incircle of triangle ABD. Hence radius R = A/S = 98/[7(2 + sqrt2)] = 7(2 - sqrt2)
@@andryvokubadra2644 "tangent" means that the triangle DOQ is the identical to the triangle DOP, I don't understand why you are saying that DP > DQ. Can you please explain?
Blue square area=49 Side of the square=7 Connect P to Q (Q on AD) So AQ=DQ=7 PQ=DF=7 Connect A to P In ∆ APQ AP^2=7^2+7^2 AP=AO+OP AO=r√2 ; OP=r 2(7^2)=(r√2+r)^2 So r=14-7√2=4.1 units.❤❤❤
Perpendicular distance from O to Line PE = 7 - r Perpendicular distance from O to line PF = 7 - r Then: 2(7 - r) ² = r² solve to find r = 7(2-√2) = 4.1
Here's a super simple solution you can do in your head: Draw BP, which obviously equals 7 sqrt 2 since it's the diagonal of a 7x7 square. BT must equal that because it's a second tangent from the same point to the circle. Now simply subtract that value from 14 and you have your r value. Tada :-)
Possibly the simplest solution as it involves only basic trigonometry. With the additional knowledge of (√2 and) 1/√2, then with a calculator, it is faster to first divide by √2: 7 = r(1 + 1/√2) -> r = 7 / 1.707 = 4.10 Without a calculator, the above is easier because you multiply by (2-1.414 = 0.586) in stead of dividing by 1.707.
Formula for inradius of an inscribed circle in a triangle: Inradius = area/semiperimeter. Simply setup the triangle , get the semi-perimeter and divide that into the area of the triangle. Triangle area is 98. Hyptenuse of triangle is 19.7989. Tangent to circle. Semi-perimiter is 23.8994. I get an inradius of 4.10.
Very simple. Point P is the center of the big square (easy) Then we use an orthonormal center A and first axis(AD) The equation of the circle is (x -R)^2 + (y - R)^2 = R^2 or x^2 + y^2 -2.R.x - 2.R.y =R^2 = 0 P(7; 7) is on the circle, so: 49 + 49 -14.R -14.R +R^2 = 0 or R^2 -28.R +98 = 0. Deltaprime = 14^2 - 98 = 98 =49.2 So R = 14 -7.sqrt(2) or R = 14 +7.sqrt(2) which is rejected as beeing superior to te side length of the big square. Finally: R = 14 - 7.sqrt(2)
Square PFCE: Aₛ = s² 49 = s² s = √49 = 7 Draw AC. As ∠PFC = ∠ADC =90° and ∠FCP = ∠DCA, ∆PFC and ∆ADC are similar triangles. As the diagram is symmetrical about AC, ∆CEP is congruent to ∆PFC and ∆CBA is congruent to ∆ADC. Triangle ∆PFC: PF² + FC² = PC² 7² + 7² = PC² PC² = 49 + 49 = 98 PC = √98 = 7√2 As FC = DC/2, by similarity, PC = AC/2. As AP+PC = AC, AP = PC = 7√2. As BA and AD are tangent to circle O at T and Q respectively, ∠OTA = ∠AQO = 90°. As OT = OQ = r and ∠TAQ = 90°, then ∠QOT = 90° as well and OTAQ is a square with side lengths r. Triangle ∆AQO: AQ² + OQ² = OA² r² + r² = OA² OA² = 2r² OA = √(2r²) = √2r AP = AO + OP 7√2 = √2r + r r(√2+1) = 7√2 r = 7√2/(√2+1) r = 7√2(√2-1)/(√2+1)(√2-1) r = (14-7√2)/(2-1) r = 14 - 7√2 = 7(2-√2) ≈ 4.10 units
One could use the following path : Area of Triangle ABD = 196 / 2 = 98 01) (14R + 14R + 14Rsqrt(2)) = 196 02) 14R * (1 + 1 + sqrt(2)) = 196 03) R * (2 + sqrt(2)) = 14 04) R = 14 / (2 + sqrt(2)) 05) R ~ 4,100505
First a rough estimate to insure against a maths gremlin: PA is 7*sqrt(2) and r is a bit less than half of that, so around 3*sqrt(2)ish or 4 ish.. Make a point in the circle called H such that PHO is a right triangle with sides r (the hypotenuse), 7-r. and 7-r. (7-r)^2 + (7-r)^2 = r^2 49 - 14r + r^2 + 49 - 14r + r^2 = r^2 2r^2 - 28r + 98 = r^2 r^2 - 28r + 98 = 0 (28+or-sqrt(784 - 4*98))/2 = r (28+or--sqrt(392)/2 = r (28+or-2*sqrt(98))/2 = r 14+or-sqrt(98) = r 14+or-7*sqrt(2) = r Unusually, the positive answer must be discarded, as it would be larger than the side length of the original square. 14 - 7*sqrt(2) 14-9.9=4.1 r=4.1 (rounded) which fits with the original rough estimate.
My reasonning is based on 2 circles inscribed in a square: 2 circles of radius R, tangent to each other and each tangent to the sides of the square. Consecutively, there is a FORMULA to get R from the square side length (n): R = n - n√2/2. Then R = 14 - 14√2/2 = 4.100505 cm
@@andryvokubadra2644 Finding the radius of a circle inscribed in a right triangle. The Formula is the following : Let BD = a Let AD = b Let BD = c R = (a + b - c) / 2 In the present case a = 14 b = 14 c = 14*sqrt(2) a + b = 28 R = [28 - 14*sqrt(2)] / 2 R = 14*(2 - sqrt(2) / 2 R = 14*(0,586) / 2 R = 7*(0,586) R = 4,1005,,,
@@andryvokubadra2644 Формула выводится для прямоугольного треугольгика. Для других треугольнтков формула другая, радиус равен площадь разделить на периметр. Здесь сумма плошадей трёх треугольников, образованных сторонами треугольника и биссектрисами углов, сходящихся в центре окружности. высотами, в этих треугольниках. являются радиусы вписанной окружности. проведенные из её центра, перпендикулярно сторонам.
considering the square as a grid which begins at (0,0) and ends at (14,14), we can initialize the circle's equation at (x-r)^2+(y+r-14)^2=r^2. it gives us 3 points along the circle: (0,14-r), (r,14) and (7,7) first 2 points plugged in the equation give us r^2=r^2. the third will give us eventually r^2-28r+98=0, which gives us two solutions, of which we only care about r
Let's find the radius: . .. ... .... ..... Since AD and AB are tangents to the circle, we know that ∠AQO=90° and ∠ATO=90°. Additionally we know that ∠QAT=90° and OQ=OT=R (with R being the radius of the yellow circle). Therefore AQOT is a square with a side length being identical to the radius of the circle and we can conclude that O is located on the diagonal AC. Since we have a ratio of √2 between the length of the diagonal of a square and its side length, we obtain: AC = √2*AD CP + OP + AO = √2*AD √2*CE + R + √2*AQ = √2*AD √2*√A(CEPF) + R + √2*R = √2*AD √2*√49 + R + √2*R = 14√2 7√2 + R + √2*R = 14√2 R + √2*R = 7√2 R*(√2 + 1) = 7√2 R*(√2 + 1)*(√2 − 1) = 7√2*(√2 − 1) R*(2 − 1) = 14 − 7√2 ⇒ R = 14 − 7√2 ≈ 4.101 Best regards from Germany
My way of solution ▶ Ablue= 49 cm² a= 7 cm the length of the diagonal would be: d= 7√2 cm the diagonal of the big square is equal to: D= 14√2 cm D= d+2r+x 14√2= 7√2+2r+x 7√2= 2r+x..........(1) the radius of the circle is r if we draw a square in this cirlcle with a= r d= √2 r ⇒ √2 r= r+x............(2) √2 r= r+x √2 r-r = x r(√2-1)= x if we put this in equation (1) we get: 7√2= 2r+x 7√2= 2r+r(√2-1) 7√2= r(2+√2-1) r= 7√2/(1+√2) r= 7√2*(1-√2)/(1-2) r= 7√2*(√2-1) cm r= 14- 7√2 r= 7(2-√2) cm r ≈ 4,1 cm
I did it in a different way. First I realised that the edge of the small square is (1+0.707) times the radius of the circle = 7 Then radius = r = 7/1.707 = 4.1 units.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) OT = OQ = R 02) OA^2 = OT^2 + OQ^2 03) OA^2 = R^2 + R^2 04) OA^2 = 2R^2 05) OA = R*sqrt(2) 06) AP = 7*sqrt(2) 07) AP = OA + OP 08) 7*sqrt(2) = R + R*sqrt(2) 09) 7*sqrt(2) = R*(1 + sqrt(2)) 10) R = 7*sqrt(2) / (1 + sqrt(2)) 11) R ~ 4,1 ln un 12) ANSWER : Radius equal to approx. 4,1 Linear Units. . Best wishes from The International Islamic Institute for Studying Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate - Al Andalus
Waaaaaaaaaay too complicated approach. Extend the line segment FP to intersect AB (call that intersect M). Extend the line segment QO to intersect the FM segment, call that intersect N. Now, QO is a radius length, and OP is also a radius length (r). The triangle OPN is a right triangle (intersection of perpendicular line segments), specifically a 45, 45, 90 triangle, so the hypotenuse OP = r = sqrt(2) ON. Since we are told P is the center of the square we can say AB (or side length a) halved is the same as QO + ON r + r/sqrt(2) = a/2 [2+sqrt(2)] r = a r = a / [2+sqrt(2)] r = 14/3.4142 = 4.1 u
I am so happy for making parts of this channel 3 months ago. Thanks to your lessons , I was able to solve this on my own like it was nothing. 👍💯 you're the best.
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The point P is the centre of the outer square. So another circle of size equal to the yellow one can be drawn in the diagonally opposite corner.
So a measure taken from point T through O to centre O’ and to T’ projected on to a side of the outer square measure r + r/sqrt2 +rsqrt2 + r and so value is r= 4.10
Should be r+ r/sqrt2 + r/sqrt2 + r = 14 r= 14/(2+sqrt2) = 4.100
R + R cos45° = (14 - √49)
R (1+cos45°) = 7
R = 7 / (1 + 1/√2)
R = 4,1 cm ( Solved √ )
Alternative method using incircle formula A = RS:
1. Join BD to form triangle ABD.
BD = diagonal of square ABCD = 14sqrt2
Area of triangle ABD = (1/2)(14)(14) = 98 = A
Semiperimeter of triangle ABD = (1/2)(14 + 14 + 14sqrt2) = 7(2 + sqrt2) = S
2. Yellow circle is incircle of triangle ABD.
Hence radius R = A/S = 98/[7(2 + sqrt2)] = 7(2 - sqrt2)
Thanks for the feedback ❤️
DQ = 14-r , DB = SQRT(14^2 + 14^2) = 14*SQRT(2) , so DP = 7*SQRT(2), DP=DQ=14-r, so 7*SQRT(2) = 14-r ==> r= 14-7*SQRT(2) = 7*(2-SQRT(2)) = 4.1
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wait. . . DP = DQ, really?
I'm really sure DQ > DP 🍵🍵🍵
@@andryvokubadra2644 the circle is tangent to both lines DA and DB
@@franzgorg I see but the problem when DP > DQ, an elementary school's student must agree DP > DQ 😁😁😁
@@andryvokubadra2644 "tangent" means that the triangle DOQ is the identical to the triangle DOP, I don't understand why you are saying that DP > DQ. Can you please explain?
14-7√2=7(2-√2)≈4,095≈4,1
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Blue square area=49
Side of the square=7
Connect P to Q (Q on AD)
So AQ=DQ=7
PQ=DF=7
Connect A to P
In ∆ APQ
AP^2=7^2+7^2
AP=AO+OP
AO=r√2 ; OP=r
2(7^2)=(r√2+r)^2
So r=14-7√2=4.1 units.❤❤❤
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You confused me, with your construction of PQ as a perpendicular on AD, since Q already exists and is not in a spot where PQ is perpendicular...
AQ = DQ
PQ = DF, Really? 🤔🤔🤔
An elementary school's student must agree if your assume is totally wrong 🍵🍵🍵
Perpendicular distance from O to Line PE = 7 - r
Perpendicular distance from O to line PF = 7 - r
Then: 2(7 - r) ² = r² solve to find r = 7(2-√2) = 4.1
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Here's a super simple solution you can do in your head: Draw BP, which obviously equals 7 sqrt 2 since it's the diagonal of a 7x7 square. BT must equal that because it's a second tangent from the same point to the circle. Now simply subtract that value from 14 and you have your r value. Tada :-)
Thanks for the feedback ❤️
AC = 14√2; PC = 7√2 → AP = 14√2 - 7√2 = 7√2 = r(√2 + 1) → r = 7(2 - √2)
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Possibly the simplest solution as it involves only basic trigonometry.
With the additional knowledge of (√2 and) 1/√2, then with a calculator, it is faster to first divide by √2:
7 = r(1 + 1/√2) -> r = 7 / 1.707 = 4.10
Without a calculator, the above is easier because you multiply by (2-1.414 = 0.586) in stead of dividing by 1.707.
7sqrt(2)=(1+sqrt(2))r, r=7
sqrt(2)/(sqrt(2)+1)=7(2-sqrt(2)).😊
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Formula for inradius of an inscribed circle in a triangle: Inradius = area/semiperimeter. Simply setup the triangle , get the semi-perimeter and divide that into the area of the triangle. Triangle area is 98. Hyptenuse of triangle is 19.7989. Tangent to circle. Semi-perimiter is 23.8994. I get an inradius of 4.10.
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Very simple. Point P is the center of the big square (easy)
Then we use an orthonormal center A and first axis(AD)
The equation of the circle is (x -R)^2 + (y - R)^2 = R^2
or x^2 + y^2 -2.R.x - 2.R.y =R^2 = 0
P(7; 7) is on the circle, so: 49 + 49 -14.R -14.R +R^2 = 0
or R^2 -28.R +98 = 0. Deltaprime = 14^2 - 98 = 98 =49.2
So R = 14 -7.sqrt(2) or R = 14 +7.sqrt(2) which is rejected as beeing superior to te side length of the big square.
Finally: R = 14 - 7.sqrt(2)
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Square PFCE:
Aₛ = s²
49 = s²
s = √49 = 7
Draw AC. As ∠PFC = ∠ADC =90° and ∠FCP = ∠DCA, ∆PFC and ∆ADC are similar triangles. As the diagram is symmetrical about AC, ∆CEP is congruent to ∆PFC and ∆CBA is congruent to ∆ADC.
Triangle ∆PFC:
PF² + FC² = PC²
7² + 7² = PC²
PC² = 49 + 49 = 98
PC = √98 = 7√2
As FC = DC/2, by similarity, PC = AC/2. As AP+PC = AC, AP = PC = 7√2.
As BA and AD are tangent to circle O at T and Q respectively, ∠OTA = ∠AQO = 90°. As OT = OQ = r and ∠TAQ = 90°, then ∠QOT = 90° as well and OTAQ is a square with side lengths r.
Triangle ∆AQO:
AQ² + OQ² = OA²
r² + r² = OA²
OA² = 2r²
OA = √(2r²) = √2r
AP = AO + OP
7√2 = √2r + r
r(√2+1) = 7√2
r = 7√2/(√2+1)
r = 7√2(√2-1)/(√2+1)(√2-1)
r = (14-7√2)/(2-1)
r = 14 - 7√2 = 7(2-√2) ≈ 4.10 units
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Great 🎉🎉🎉
One could use the following path :
Area of Triangle ABD = 196 / 2 = 98
01) (14R + 14R + 14Rsqrt(2)) = 196
02) 14R * (1 + 1 + sqrt(2)) = 196
03) R * (2 + sqrt(2)) = 14
04) R = 14 / (2 + sqrt(2))
05) R ~ 4,100505
Thanks for the feedback ❤️
First a rough estimate to insure against a maths gremlin: PA is 7*sqrt(2) and r is a bit less than half of that, so around 3*sqrt(2)ish or 4 ish..
Make a point in the circle called H such that PHO is a right triangle with sides r (the hypotenuse), 7-r. and 7-r.
(7-r)^2 + (7-r)^2 = r^2
49 - 14r + r^2 + 49 - 14r + r^2 = r^2
2r^2 - 28r + 98 = r^2
r^2 - 28r + 98 = 0
(28+or-sqrt(784 - 4*98))/2 = r
(28+or--sqrt(392)/2 = r
(28+or-2*sqrt(98))/2 = r
14+or-sqrt(98) = r
14+or-7*sqrt(2) = r
Unusually, the positive answer must be discarded, as it would be larger than the side length of the original square.
14 - 7*sqrt(2)
14-9.9=4.1
r=4.1 (rounded) which fits with the original rough estimate.
Excellent!
Thanks for the feedback ❤️
My reasonning is based on 2 circles inscribed in a square: 2 circles of radius R, tangent to each other and each tangent to the sides of the square. Consecutively, there is a FORMULA to get R from the square side length (n): R = n - n√2/2. Then R = 14 - 14√2/2 = 4.100505 cm
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Thanks for the feedback ❤️
Acircle is inscribed in right triangle ABD, R= (AB+AD-BD)/2=(28-14\/2)/2=7(2-\/2)=4,1.
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Yes, no need to go thru other methods if we can assume the solver should know this basic formula.
What do formula you use?
@@andryvokubadra2644
Finding the radius of a circle inscribed in a right triangle.
The Formula is the following :
Let BD = a
Let AD = b
Let BD = c
R = (a + b - c) / 2
In the present case
a = 14
b = 14
c = 14*sqrt(2)
a + b = 28
R = [28 - 14*sqrt(2)] / 2
R = 14*(2 - sqrt(2) / 2
R = 14*(0,586) / 2
R = 7*(0,586)
R = 4,1005,,,
@@andryvokubadra2644 Формула выводится для прямоугольного треугольгика. Для других треугольнтков формула другая, радиус равен площадь разделить на периметр. Здесь сумма плошадей трёх треугольников, образованных сторонами треугольника и биссектрисами углов, сходящихся в центре окружности. высотами, в этих треугольниках. являются радиусы вписанной окружности. проведенные из её центра, перпендикулярно сторонам.
considering the square as a grid which begins at (0,0) and ends at (14,14), we can initialize the circle's equation at (x-r)^2+(y+r-14)^2=r^2.
it gives us 3 points along the circle: (0,14-r), (r,14) and (7,7)
first 2 points plugged in the equation give us r^2=r^2. the third will give us eventually r^2-28r+98=0, which gives us two solutions, of which we only care about r
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@ 7:59 we form the conjugate that the Vulcan Spock from Star Trek used too mind meld a telepathic link with any organism. ...I mean same process. 🙂
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Let's find the radius:
.
..
...
....
.....
Since AD and AB are tangents to the circle, we know that ∠AQO=90° and ∠ATO=90°. Additionally we know that ∠QAT=90° and OQ=OT=R (with R being the radius of the yellow circle). Therefore AQOT is a square with a side length being identical to the radius of the circle and we can conclude that O is located on the diagonal AC. Since we have a ratio of √2 between the length of the diagonal of a square and its side length, we obtain:
AC = √2*AD
CP + OP + AO = √2*AD
√2*CE + R + √2*AQ = √2*AD
√2*√A(CEPF) + R + √2*R = √2*AD
√2*√49 + R + √2*R = 14√2
7√2 + R + √2*R = 14√2
R + √2*R = 7√2
R*(√2 + 1) = 7√2
R*(√2 + 1)*(√2 − 1) = 7√2*(√2 − 1)
R*(2 − 1) = 14 − 7√2
⇒ R = 14 − 7√2 ≈ 4.101
Best regards from Germany
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My way of solution ▶
Ablue= 49 cm²
a= 7 cm
the length of the diagonal would be:
d= 7√2 cm
the diagonal of the big square is equal to:
D= 14√2 cm
D= d+2r+x
14√2= 7√2+2r+x
7√2= 2r+x..........(1)
the radius of the circle is r
if we draw a square in this cirlcle with a= r
d= √2 r
⇒
√2 r= r+x............(2)
√2 r= r+x
√2 r-r = x
r(√2-1)= x
if we put this in equation (1) we get:
7√2= 2r+x
7√2= 2r+r(√2-1)
7√2= r(2+√2-1)
r= 7√2/(1+√2)
r= 7√2*(1-√2)/(1-2)
r= 7√2*(√2-1) cm
r= 14- 7√2
r= 7(2-√2) cm
r ≈ 4,1 cm
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I did it in a different way. First I realised that the edge of the small square is (1+0.707) times the radius of the circle = 7
Then radius = r = 7/1.707 = 4.1 units.
DP = CP
DP = 7sqrt2
DP = DQ (tangents
r = 14 - 7sqrt
14√2-√49√2-r√2=r → r=14-7√2 =4,10050.... Gracias y un saludo cordial.
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4.1 units
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Thanks Sir
Wonderful method for solve
I’m very enjoy
With glades
❤❤❤❤❤❤
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14√2/2=r+√2r...r=7√2/(1+√2)=14-7√2
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Thank you
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Very well explained
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Я решил вообще не используя сторону большого квадрата 🤷🏻♂️
Ответ: 14-7sqrt(2)
большое спасибо🌹
r = 14 - 7√2
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(14✓2-7✓2)÷ 2=7√2÷2=7/√2
r= (14√2-7√2) ÷2= 7✓2÷ 2= 7/√2
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OT = OQ = R
02) OA^2 = OT^2 + OQ^2
03) OA^2 = R^2 + R^2
04) OA^2 = 2R^2
05) OA = R*sqrt(2)
06) AP = 7*sqrt(2)
07) AP = OA + OP
08) 7*sqrt(2) = R + R*sqrt(2)
09) 7*sqrt(2) = R*(1 + sqrt(2))
10) R = 7*sqrt(2) / (1 + sqrt(2))
11) R ~ 4,1 ln un
12) ANSWER : Radius equal to approx. 4,1 Linear Units.
.
Best wishes from The International Islamic Institute for Studying Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate - Al Andalus
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Thank you!
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14-r=7+(r/sqrt(2))
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7/✓2
這個 圓 並不存在於 正方形 的範圍內 這可能是一個錯誤的圖形
Waaaaaaaaaay too complicated approach.
Extend the line segment FP to intersect AB (call that intersect M). Extend the line segment QO to intersect the FM segment, call that intersect N.
Now, QO is a radius length, and OP is also a radius length (r). The triangle OPN is a right triangle (intersection of perpendicular line segments), specifically a 45, 45, 90 triangle, so the hypotenuse OP = r = sqrt(2) ON.
Since we are told P is the center of the square we can say AB (or side length a) halved is the same as QO + ON
r + r/sqrt(2) = a/2
[2+sqrt(2)] r = a
r = a / [2+sqrt(2)]
r = 14/3.4142
= 4.1 u
Thanks for the feedback ❤️