Can you find the Radius of the Yellow circle? | (Squares) |
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8 күн бұрынLearn how to find the Radius of the Yellow circle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; circle theorem; area of the square formula. Step-by-step tutorial by PreMath.com
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Can you find the Radius of the Yellow circle? | (Squares) | #math #maths | #geometry
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@MateusMuila 7 күн бұрын
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.
@PreMath 6 күн бұрын
Fantastic!🌹 Glad to hear that! You are very welcome! Thanks for the feedback ❤️
@davidreid7293 6 күн бұрын
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
@davidreid7293 6 күн бұрын
Should be r+ r/sqrt2 + r/sqrt2 + r = 14 r= 14/(2+sqrt2) = 4.100
@hongningsuen1348 7 күн бұрын
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)
@PreMath 7 күн бұрын
Thanks for the feedback ❤️
@unknownidentity2846 7 күн бұрын
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
@PreMath 7 күн бұрын
Super!👍 Thanks for sharing ❤️
@marioalb9726 3 күн бұрын
R + R cos45° = (14 - √49) R (1+cos45°) = 7 R = 7 / (1 + 1/√2) R = 4,1 cm ( Solved √ )
@thewolfdoctor761 6 күн бұрын
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
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@andryvokubadra2644 6 күн бұрын
wait. . . DP = DQ, really? I'm really sure DQ > DP 🍵🍵🍵
@franzgorg 6 күн бұрын
@@andryvokubadra2644 the circle is tangent to both lines DA and DB
@andryvokubadra2644 6 күн бұрын
@@franzgorg I see but the problem when DP > DQ, an elementary school's student must agree DP > DQ 😁😁😁
@franzgorg 6 күн бұрын
@@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?
@andreasproteus1465 6 күн бұрын
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
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@ravikrpranavam 6 күн бұрын
Very well explained
@PreMath 6 күн бұрын
Glad to hear that! 🌹 Thanks for the feedback ❤️
@jamestalbott4499 6 күн бұрын
Thank you!
@PreMath 6 күн бұрын
You are very welcome!🌹 Thanks ❤️
@raya.pawley3563 6 күн бұрын
Thank you
@PreMath 6 күн бұрын
You are very welcome! Thanks ❤️
@AmirgabYT2185 7 күн бұрын
14-7√2=7(2-√2)≈4,095≈4,1
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@marcgriselhubert3915 6 күн бұрын
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)
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@yalchingedikgedik8007 6 күн бұрын
Thanks Sir Wonderful method for solve I’m very enjoy With glades ❤❤❤❤❤❤
@PreMath 6 күн бұрын
Most welcome, dear 🌹 Thanks for the feedback ❤️
@GillesF31 6 күн бұрын
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
@PreMath 6 күн бұрын
Excellent! Thanks for the feedback ❤️
@wackojacko3962 7 күн бұрын
@ 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. 🙂
@PreMath 7 күн бұрын
Excellent!😀 Thanks for the feedback ❤️
@LuisdeBritoCamacho 6 күн бұрын
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
@PreMath 6 күн бұрын
Thanks for the feedback ❤️
@prossvay8744 7 күн бұрын
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.❤❤❤
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@quigonkenny 7 күн бұрын
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...
@andryvokubadra2644 6 күн бұрын
AQ = DQ PQ = DF, Really? 🤔🤔🤔 An elementary school's student must agree if your assume is totally wrong 🍵🍵🍵
@misterenter-iz7rz 7 күн бұрын
7sqrt(2)=(1+sqrt(2))r, r=7 sqrt(2)/(sqrt(2)+1)=7(2-sqrt(2)).😊
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@murdock5537 7 күн бұрын
AC = 14√2; PC = 7√2 → AP = 14√2 - 7√2 = 7√2 = r(√2 + 1) → r = 7(2 - √2)
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@rjbmarchiac8693 5 күн бұрын
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.
@LucasBritoBJJ Күн бұрын
DP = CP DP = 7sqrt2 DP = DQ (tangents r = 14 - 7sqrt
@timmcguire2869 6 күн бұрын
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 :-)
@PreMath 6 күн бұрын
Thanks for the feedback ❤️
@MrPaulc222 6 күн бұрын
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.
@PreMath 6 күн бұрын
Excellent! Thanks for the feedback ❤️
@ABhaim 6 күн бұрын
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
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@lasalleman6792 6 күн бұрын
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.
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@gauravroy8528 6 күн бұрын
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.
@quigonkenny 7 күн бұрын
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
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@andryvokubadra2644 6 күн бұрын
Great 🎉🎉🎉
@sergeyvinns931 7 күн бұрын
Acircle is inscribed in right triangle ABD, R= (AB+AD-BD)/2=(28-14\/2)/2=7(2-\/2)=4,1.
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@qc1okay 6 күн бұрын
Yes, no need to go thru other methods if we can assume the solver should know this basic formula.
@andryvokubadra2644 6 күн бұрын
What do formula you use?
@LuisdeBritoCamacho 6 күн бұрын
@@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,,,
@sergeyvinns931 6 күн бұрын
@@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.
@kalavenkataraman4445 7 күн бұрын
4.1 units
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@abrahammg9814 Күн бұрын
(14✓2-7✓2)÷ 2=7√2÷2=7/√2
@santiagoarosam430 6 күн бұрын
14√2-√49√2-r√2=r → r=14-7√2 =4,10050.... Gracias y un saludo cordial.
@PreMath 6 күн бұрын
Excellent!👍 Thanks for sharing ❤️
@Birol731 6 күн бұрын
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
@PreMath 6 күн бұрын
Excellent! Thanks for sharing ❤️
@abrahammg9814 Күн бұрын
r= (14√2-7√2) ÷2= 7✓2÷ 2= 7/√2
@giuseppemalaguti435 7 күн бұрын
14√2/2=r+√2r...r=7√2/(1+√2)=14-7√2
@PreMath 7 күн бұрын
Excellent!👍 Thanks for sharing ❤️
@abrahammg9814 Күн бұрын
7/✓2
@dainiusb1114 6 күн бұрын
14-r=7+(r/sqrt(2))
@PreMath 6 күн бұрын
Thanks for sharing ❤️
@nomad7966 6 күн бұрын
Я решил вообще не используя сторону большого квадрата 🤷🏻♂️ Ответ: 14-7sqrt(2)
@PreMath 6 күн бұрын
большое спасибо🌹
@erwinkurniadi1850 7 күн бұрын
r = 14 - 7√2
@PreMath 6 күн бұрын
Excellent Thanks for sharing ❤️
@user-rj9jy3oj7b 5 күн бұрын
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
@SpokeSeadog 6 күн бұрын
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
@PreMath 6 күн бұрын
Thanks for the feedback ❤️
@LuisdeBritoCamacho 6 күн бұрын
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
@PreMath 6 күн бұрын
Bravo!👍 Thanks for sharing ❤️🙏
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