Can you calculate the Radius of the Circle? | (Square) |

[@jamestalbott4499]

Thank you!

[@PreMath]

You're welcome! Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@cyruschang1904]

Square side length = √576 = 24 Triangle base = 432/24 = 18 24 : 18 = 4 : 3 => hypotenuse = 30 Trapezoid area = 576 - 216 = 360 = 30r/2 + 6r/2 + 24r/2 + 24(24 - r)/2 = (15 + 3 + 12 - 12)r + (24^2)/2 360 - 288 = 72 = 18r r = 4

[@PreMath]

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[@cyruschang1904]

@ Thank you. May the New Year bring you and your love ones happiness and success.

[@MdShahriarHossain-l1n]

A very easy solution - White region area(it's a trapezoid) = area of the square - Triangle ABC= 360. Draw some lines from the center of the circle( let o be the center) to the 4 corners of the trapezoid. Then there will be 4 Triangles. Then the sum of the area of the Triangles are= 1/2×r(6+24+30)+(24-r)×24×1/2= 360 ⇨30r+288-12r=360 r=4

[@PreMath]

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[@marcelowanderleycorreia8876]

Cool question!! 👍

[@PreMath]

Glad you liked it! 😊 Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@jimlocke9320]

Extend AE and OF down until they meet and label the intersection as point H. AB and OH are parallel.

[@egillandersson1780]

I did exactly the same. Easier and quickier !

[@phungpham1725]

@@egillandersson1780 Another possible solution: 1/Yellow triangle is a 3-4-5 triples. 2/Extend AE and DC down intersecting at point I. --> the triangle ECI is also a 3-4-5 triples --> EC= 6->CI=8 and EI = 10. 3/ Build the inscribed circle (center O’) of the triangle ECI. Note that the circle ( center O) is the ex circle of triangle ECI. We have points I, O’ and O are collinear. And label r and R as radiuses the ins and ex circle We have r = 2 (easy) From O’ drop the perpendicular O’H to CI. By similarity r/R=6/(8+R) -> 2/R=6/(8+R) -> 1/R=3/(8+R) ->R=(8+R)/3 R= 4cm😅😅😅

[@PreMath]

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[@waheisel]

Triangle AEC area=72. Triangle AEC area =6r/2+30r/2. r=4. Thanks for the fun puzzle!

[@PreMath]

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[@santiagoarosam430]

ABCD=576→ AB=√576=24=4*6→ BE=2*216/24=18=3*6→ AE=5*6=30 → AECD=576-216=360 =(30r/2)+(6r/2)+(24r/2)+24[(24-r)/2] → Radio =r=4. Gracias y un saludo cordial.

[@PreMath]

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[@gaylespencer6188]

My approach was from Monty Python - a little bit different. The bottom side is divided into two segments, 18 plus 6. And the 6 segment can be further divided into r plus (6-r). Getting the angles of the triangle was easy enough; so the supplement of the bottom angle was 180-arctan(24/18), or about 126.87 degrees. Divide that by 2, or about 63.44 degrees. Set up equation tangent(63.44) = r/(6-r). r=4.

[@PreMath]

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[@matthieudutriaux]

Mental calculation : 576=24^2 (=AB=BC) 216=24*9=24*18/2 (=AB*BE/2) tan(a)=AB/BE=24/18=4/3 tan(Pi-a)=tan(-a)=-tan(a)=-4/3 OF/EF=tan((Pi-a)/2) tan(Pi-a)=2*x/(1-x^2) with x=tan((Pi-a)/2) -4/3=2*x/(1-x^2) -4*(1-x^2)=6*x 4*x^2-6*x-4=0 x^2-3/2*x-1=0 (x-2)*(x+1/2)=0 but x>0 then x=2 OF/EF=tan((Pi-a)/2)=x=2 r/(6-r)=2 r=12-2*r 3*r=12 r=4

[@PreMath]

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[@raya.pawley3563]

Thank you

[@PreMath]

You are very welcome! ❤ Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@johnryder1713]

Hope you had a great Christmas Premath and everyone here

[@phungpham1725]

@johnryder1713 you too!

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@johnryder1713]

@@PreMath And Please God a good one enjoying your fine content

[@MrPaulc222]

Draw a vertical line up from E. The rectangle formed has an area of 432. As the height is 24, the horizontal is 432/24 = 18. Therefore, EC = 6. The diagonal, AE is the sqrt of 24^2 + 18^2 so 576 + 324 = 900, whose sqrt = 30. Now I will turn ADCE into a series of triangles: AOE has an area of 15r. EOC area of 3r. COD area of 12r. AOD is an odd one as its base is 24 and its height is 24-r. This has an area of 12*(24-r) = 288-12r. Add all those up for 30r + (288-12r) = 18r + 288 18r + 288 = 576 - 216. 18r + 288 = 360. 18r = 72. r = 4. This one took me a while as I went down a couple of dead ends trying to figure out a viable method.EDIT: Looking further, I could have simplified it to 15r + 3r = (6*24)/2

[@PreMath]

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[@quigonkenny]

Square ABCD: A = s² 576 = s² s = √576 = 24 Triangle ∆ABE: A = bh/2 216 = BE(24)/2 = 12BE BE = 216/12 = 18 AB² + BE² = AE² 24² + 18² = AE² AE² = 576 + 324 = 900 AE = √900 = 30 As BC = 24 and BE = 18, EC = 6. As FC = r, EF = 6-r. As FE and PE are tangents to circle O that intersect at E, PE = FE = 6-r. AP = AE-PE = 30-(6-r) = 24+r. Draw AC. As AC is the diagonal of square ABCD, AC = 24√2. As OFCG is a square with side length r and shares C as a corresponding vertex, then O is collinear with AC. As OC is the diagonal of OFCG then OC = √2r. AO = AC-OC = 24√2-√2r = √2(24-r) Triangle ∆APO: AP² + OP² = AO² (24+r)² + r² = (√2(24-r))² 576 + 48r + r² + r² = 2(576-48r+r²) 576 + 48r + 2r² = 1152 - 96r + 2r² 144r = 576 [ r = 576/144 = 4 cm ]

[@PreMath]

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[@phungpham1725]

1/The yellow triangle ABE is a 3-4-5 triples AB= 24-> BE= 18-> AO= 30 2/ Label angle EAC= alpha and BAE = beta We have tan (beta) = 3/4 So tan( alpha)=tan( 45 degrees-beta) = (1-3/4)/(1+3/4)=(1/4)/(7/4)=1/7 3/ AP= 24+r -> r/(24+r)= 1/7 -> 7r=24+r -> r=4cm😅😅😅

[@PreMath]

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[@wackojacko3962]

@ 5:03 one truly enters the realm of mathematics and allows for observations to solve the problem. Throw a slice of bread in a toaster and whadd'ya get? ...ya get toast! 😊

[@PreMath]

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[@marcgriselhubert3915]

The side length of the square is sqrt(576) = 24, the area of the yellow triangle is (1/2).BA.BE = (1/2).24.BE = 216, so BE = 216/12 = 18 We now use an orthonormal center B and first axis (BC). We have A(0; 24), E(18; 0), VectorAC(18; -24) is colinear to VectorU(-3; 4) The equation of (AE) is (x - 18).(4) - (y).(-3) = 0 or 4.x + 3.y - 72 = 0. If R is the radius of the circle, we have O(24 - R; R) Distance from O to (AC): abs(4.(24 - R) + 3.(R) - 72)/sqrt(4^2 + 3^2) = (24 - R)/5, this distance is also equal to R, so we have (24 - R)/5 = R, and so R = 4.

[@PreMath]

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[@georgebliss964]

Square side length = 24. BE = 18. EC = 6. Extending lines AE & DC downwards to intersect at point H. Then triangle ECH is similar to triangle ABE. So 24 /18 = CH / 6. CH = 8. Tan BAE = 18 / 24 = 0.75. Angle BAE = 36.870 degrees. Joining centre point O to H. This will bisect angle AHD. Thus angle OHG = 18.435 degrees. Joining OG which = radius r. In triangle OHG, tan 18.435 = r / (r + 8). 0.33333 = r / (r + 8). 1 / 3 = r / (r + 8) 3r = r + 8. 2r = 8 r = 4.

[@PreMath]

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[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Blue Square Side (BSS) = sqrt(576) ; BSS = 24 02) BE = (216 * 2) / 24 ; BE = 432 / 24 ; BE = 18 lin un 03) EC = (24 - 18) ; EC = 6 lin un 04) R = Radius 05) EF = PE = (6 - R) lin un 06) AE^2 = 324 + 576 ; AE^2 = 900 ; AE = 30 07) OP = R 08) AP = 30 - (6 - R) ; AP = (24 + R) lin un 09) AC = 24*sqrt(2) lin un 10) AO = AC - OC 11) OC = R*sqrt(2) 12) AO = (24*sqrt(2) - R*sqrt(2)) 13) AP^2 + OP^2 = AO^2 14) (24 + R)^2 + R^2 = (24*sqrt(2) - R*sqrt(2))^2 15) Solution : R = 4 Therefore, Beyond any Resonable Doubt, Radius equal to 4 Linear Units.

[@PreMath]

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[@wasimahmad-t6c]

3×3×3.14159268=28.27433

[@AmirgabYT2185]

r=4 cm

[@PreMath]

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[@AmirgabYT2185]

@@PreMath Thanks a lot 🥰 Wish you the same ❤️🎉 Happy New Year 🎊🎊🎊

[@unknownidentity2846]

Let's find the radius: . .. ... .... ..... First of all we calculate the side length s of the blue square: A(ABCD) = s² 576cm² = s² ⇒ s = √(576cm²) = 24cm Now we can calculate the side lengths AB and BE of the right triangle ABE: AB = s = 24cm A(ABE) = (1/2)*AB*BE ⇒ BE = 2*A(ABE)/AB = 2*216cm²/(24cm) = 18cm Now let's assume that B is the center of the coordinate system and that BC is located on the x-axis. With r being the radius of the circle we obtain the following coordinates: A: ( 0cm ; 24cm ) B: ( 0cm ; 0cm ) E: ( 18cm ; 0cm ) O: ( 24cm−r ; r ) P: ( xP ; yP ) The line AE can be represented by the following function: y = (yA − yE)*(x − xE)/(xA − xE) + yE = (24 − 0)*(x − 18cm)/(0 − 18) + 0cm = 4(x − 18cm)/(−3) = 24cm − 4x/3 Since AE is a tangent to the circle, AE is perpendicular to PO. Therefore the product of the slopes of AE and PO is −1 and we can conclude: (yO − yP)/(xO − xP) = −1/(−4/3) = 3/4 OP² = (xO − xP)² + (yO − yP)² r² = (xO − xP)² + (3/4)²(xO − xP)² r² = (xO − xP)² + (9/16)(xO − xP)² r² = (16/16)(xO − xP)² + (9/16)(xO − xP)² r² = (25/16)(xO − xP)² ⇒ r = (5/4)(xO − xP) = (5/3)(yO − yP) (5/4)(xO − xP) = (5/3)(yO − yP) 3(xO − xP) = 4(yO − yP) 3[(24cm − r) − xP] = 4[r − (24cm − 4xP/3)] 3(24cm − r − xP) = 4(r − 24cm + 4xP/3) 72cm − 3r − 3xP = 4r − 96cm + 16xP/3 168cm − 7r = 3xP + 16xP/3 = 9xP/3 + 16xP/3 = 25xP/3 ⇒ xP = 3(168cm − 7r)/25 r = (5/4)(xO − xP) r = (5/4)[(24cm − r) − 3(168cm − 7r)/25] r = 30cm − 5r/4 − 3(168cm − 7r)/20 r = 30cm − 5r/4 − 504cm/20 + 21r/20 20r = 600cm − 25r − 504cm + 21r 24r = 96cm ⇒ r = 4cm Best regards from Germany

[@PreMath]

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[@giuseppemalaguti435]

l=24...AP+PE=AE...√((24-r)^2+(24-r)^2-r^2)+(6-r)=30...calcoli ..r=4

[@PreMath]

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[@nenetstree914]

r=4

[@PreMath]

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[@LuisdeBritoCamacho]

ANSWER : Radius equal to 4 Linear Units. Later I will come with my Reasoning!! Bye.

[@PreMath]

Thanks ❤️🙏

[@sergioaiex3966]

Solution: Blue Square Area = 576 576 = s² s = 24 Yellow Triangle Area = ½ b h 216 = ½ b h 216 = ½ 24 h h = 216/12 h = 18 Applying Pythagorean Theorem in Triangle ABE (18)² + (24)² = AE² AE² = 900 AE = 30 White Area = Square Area - Triangle Área White Area = 576 - 216 White Area = 360 White Area = ∆AEO Area + ∆CEO Area + ∆CDO Area + ∆ADO Area ... ¹ ∆AEO Area = ½ b h ∆AEO Area = ½ 30 r ∆AEO Area = 15r ∆CEO Area = ½ b h ∆CEO Area = ½ 6 r ∆CEO Area = 3r ∆CDO Area = ½ b h ∆CDO Area = ½ 24 r ∆CDO Area = 12r ∆ADO Area = ½ b h ∆ADO Area = ½ 24 (24 - r) ∆ADO Area = 288 - 12r Replacing in Equation ¹ 360 = 15r + 3r + 12r + 288 - 12r 360 - 288 = 18r 18r = 72 r = 4 cm Thus Radius = 4 cm ✅

[@Abdelfattah-hr8tt]

Good job

[@PreMath]

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