Can you find the Radius of the circle? | (Chord) |

[@sardarstationers2165]

Nice Explanation.

[@PreMath]

Glad it was helpful! Thanks for the feedback ❤️

[@uwelinzbauer3973]

Nice geometry question- I wish you a happy Sunday 😊

[@terryshell9045]

Good problem. Excellent explanation.

[@MrPaulc222]

Extend DC to the top of the circumference and call that point E. DE = 23 (when comparing to AB's 13). DE's midpoint is M. OME is a right triangle with sides x, 23/2. and r. N is the midpoint of AB> Right triangle ANO has sides x+6, 13/2, and r. Find x: x^2 + (23/2)^2 = (x + 6)^2 + (13/2)^2, because both = r^2. x^2 + 529/4 = x^2 + 12x + 36 + 169/4 529/4 = 12x + 36 + 169/4 529 = 48x + 144 + 169 529 = 48x + 313 48x = 216 Reduce: 12x = 54 Again: 2x = 9 Again: x = 4.5 or 9/2 if preferred. You now have 2 right triangles, each containing r. One to calculate and another to verify. First: x, 23/2, and r (9/2)^2 + (23/2)^2 = r^2 81/4 + 529/4 = 610/4 = r^2 Try the second before square rooting, in case it's wrong: x + 6 = 10.5 or, if preferred, 21/2. (21/2)^2 = 441/4.+ 169/4 = 610/4 = r^2 Both calculations give r^2 = 610/4 r = sqrt(610)/2 It's not immediately clear if any square numbers (other than 1), are factors, so I will look up sqrt(610) as it stands. 24.7 looks to be a reasonable approximation, so I will try my luck with r = 12.35. It looks thereabouts, so I will look at the video now. Well, I got it right, but did cause myself a couple of extra calculations - though we broadly used the same method.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santiagoarosam430]

Prolongamos DC y AC y obtenemos las cuerdas DE=DC+CE y AF=AC+CF → CE=CD-AB=18-13=5 → Potencia de C respecto a la circunferencia =5*18=6*CF→ CF=15→ BF²=AB²+AF² → (2r)²=13²+(6+15)²→ r=(√610)/2 =12,34908...ud. Gracias y un saludo cordial.

[@Emerson_Brasil]

Você também pode usar esse teorema quando as cordas são perpendiculares. Note que o ponto C é onde ocorre perpendicularismo, logo: AC²+CF² + DC² + CE² = (2R)²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

Thank you! I did it the same😅

[@scottdort7197]

Alternatively. 13/2=6.5. 18-6.5=11.5. 11.5*2=23. 23-18=5. Use chord product theorem 5*18=6*X. X must equal 15. Use the RMS technique to solve for r. r=(5^2+6^2+15^2+18^2)^.5/2=610^•5/2 or approximately 12.3491 or 12.35 as Premath stated.

[@Emerson_Brasil]

5×18=6×X→ X=16??? 😢😢😢😢, X=15. Redo your calculations again.

[@Emerson_Brasil]

Chord theorem: If two chords CD and AB are perpendicular at point O, the following property holds: AO² + OB² + OD² + OC² = (2R)², where R is the radius of the circle.

[@PreMath]

Thanks for the feedback ❤️

[@scottdort7197]

@@Emerson_Brasil sorry. Typo. Corrected.

[@Emerson_Brasil]

​@@scottdort7197OK, no problem!

[@marioalb9726]

Intersecting chords theorem: 6 . x = 18 . 5 --> x = 15cm c = x + 6 = 21cm Intersecting chords theorem again: (R+13/2)(R-13/2)= (c/2)² R² - (13/2)² = (21/2)² R² = 10,5² + 6,5² R = 12,349 cm ( Solved √ )

[@PreMath]

Thanks for sharing ❤️

[@quigonkenny]

Draw OM, where M is the point on chord AB where OM and AB are perpendicular. As AB is a chord and O is the center of the circle, OM bisects AB and AM = MB = 13/2. Let OM = x. Draw BE, where E is the point on DC where BE and DC are perpendicular. As ∠BEC = ∠ECA = ∠CAB = 90°, then ∠ABE = 360°-3(90°) = 90° as well and ABEC is a rectangle. As AB = EC = 13, DE = DC-EC = 18-13 = 5. Extend DC to intersect circumference at F and let the point where OM intersects chord DF be N. As ∠ONF = 90°, ON is a perpendicular bisector of both ABEC and DF, and thus by symmetry CF = DE = 5. DF = DC+CF = 18+5 = 23. DN = NF = 23/2. ON = OM-NM = x-6. Draw radii OB and OD, length r. Triangle ∆OMB: OM² + MB² = OB² x² + (13/2)² = r² r² = x² + 169/4 --- [1] Triangle ∆OND: ON² + DN² = OD² (x-6)² + (23/2)² = r² r² = x² - 12x + 36 + 529/4 r² = x² - 12x + 673/4 --- [2] x² + 169/4 = x² - 12x + 673/4

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@南雲誠一]

Extend DC to the top of the circumference and call that point E. Extend AC to the right of the circumference and call that point F. AC x CF = DC x CE -> 6 x CF = 18 x 5 then CF = 15. AF = 21 Triangle ABF is right triangle in circle. so BF is diameter. BF = √(21²+13²) Radius is BF/2 ...

[@marcgriselhubert3915]

We choose an orthonormal center A and first axis (AC). Then A(0; 0), B(0; -13), and D(6; -18) are on the circle. The equation of the circle is x^2 + y^2 + a.x + b.y + c = 0. A is on the circle so c = 0, B is on the circle, so 169 -13.b = 0 and so b = 13, D is on the circle, so 36 + 324 +6.a -18.13 = 0 and so a = -21 The equation of the circle is then x^2 + y^2 +13.x - 21.y = 0 or (x +(13/2)^2 + (y -(21/2)^2 = (13/2)^2 + (21/2)^2. If R is the radius of the circle, then R^2 = (13/2)^2 = (21/2)^2 = 610/4, and R = sqrt(610)/2

[@giuseppemalaguti435]

R^2=6,5^2+(6+a)^2...a^2=R^2-11,5^2...calcolo a=4,5...R^2=152,5

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Emerson_Brasil]

*_Outra maneira de fazer essa questão:_* Se eu conseguir formar um triângulo numa circunferência, eu posso usar a fórmula: S=abc/4R, onde S é a área do triângulo, a, b e c são os lados e, por fim, R é o raio da circunferência. É possível formar o triângulo ABD, onde podemos encontrar a área S, da seguinte maneira: S=AC×AB/2=6×13/2= *3×13* Seja P o pé da perpendicular em relação ao lado CD, logo BP=6 e PD=18-13=5. Assim, BD²=BP²+PD²→BD²=5²+6²=61→ *BD=√61.* Por outro lado, AD²=AC²+CD²=6²+18²=360 *AD=6√10* Assim, S=AD×BD×AB/4R 3×13=(6×√10×√61 ×13)/4R 1=(2×√10×√61)/4R 1=(√10×√61)/2R 1=(√610)/2R *R=(√610)/2*

[@toninhorosa4849]

Dear teacher. I liked the way you solved this problem. But I find another way. First I extended the line CD, from point C until it reached the top line of the circumference and marked the point "E". The chord ED was formed. I extended the line AC until it reached the other side of the circle and marked the point "F". The chord AF was formed. From point "B" I drew a parallel line to line AC until I met line CD and marked point "G". GD = CD - AB GD = 18 - 13 = 5 EC = GD = 5 Then the relationship below is valid: EC*CD = AC*CE 5*18 = 6*CE CE = 90/6 CE = 15 And: AE = 6+15 = 21 The midpoint of AE is "M" so that AM = ME = 21/2. The midpoint of AB is "N" só that AN = NB = 13/2. Then applying Pythagoras in ∆ANO AO = R (radius) AN = 13/2 NO = AM = 21/2 R^2 = (13/2)^2 + (21/2)^2 R^2 = 169/4 + 441/4 R^2 = 610/4 R = (√610)/2 units^2.

[@PreMath]

COOL! Thanks for sharing ❤️

[@tontonbeber4555]

Brute force Circle equation x²+y²+ax+by+c = 0 3 points on the circle : (0,0),(0,-13),(6,-18) => find a,b,c : c=0 169-13b = 0 => b = 13 36+324+6a-234 = 0 => a=-21 x²-21x + y²+13y = 0 (x-10.5)² + (y-6.5)² = 10.5²+6.5² radius is V(10.5²+6.5²) = 12.349

[@marioalb9726]

Easiest solution Intersecting chords theorem: 6 . x = 18 . 5 --> x = 15 cm Pytagorean theorem: (2R)² = c₁² + c₂² (2R)² = 13² + (6+x)² R = 12,349 cm ( Solved √ )

[@marioalb9726]

Intersecting chords theorem: 6 . x = 18 . 5 --> x = 15 cm Quadratic mean, of the chords: (2R)² = c₁² + c₂² (2R)² = (18²+6²)+(5²+15²) R = 12,349 cm ( Solved √ )

[@PreMath]

Thanks for sharing ❤️

[@rabotaakk-nw9nm]

6:08-7:52 169/4, 529/4, 144/4, 216/4 ? 😮 12x=(23/2)²-(13/2)²-36= =(23/2+13/2)(23/2-13/2)-36= =(36/2)(10/2)-36= =18•5-36=90-36=54 😁

[@joeschmo622]

✨Magic!✨ But they can never make the answers something simple, like, *3.* 🙄

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@wackojacko3962]

I wrote a Compare and Contrast essay on a problem similar to this one just because the English teacher wasn't a mathematician. She didn't understand a word of it. 🙂

[@PreMath]

Wow!😀 Thanks for sharing ❤️

[@Birol731]

My way of solution ▶ E ∈ [AB] for the right triangle ΔEOA we can write the Pythagoren theorem [AE]²+[EO]²= [OA]² [AE]= 13-z [EO]= 6+y [OA]= r ⇒ (13-z)²+(6+y)²= r² for the right triangle ΔBOE we can write the Pythagoren theorem: [OE]²+[EB]²= [BO]² [OE]= 6+y [EB]= z [OA]= r ⇒ (6+y)²+z²= r² ⇒ (13-z)²+(6+y)²= (6+y)²+z² (13-z)²= z² 169-26z+z²= z² 169= 26z z= 13/2 II) Let's take a point between C and D : F ∈ [CD] and [FD] ⊥ [BF] if [AB]= [CF]= 13 [FD]= 18-13 [FD]= 5 Also, G ∈ [CD] and [CG] ⊥ [EG] [OG]= y By applying the Pythagorean theorem for the ΔDOG, we get [OG]²+[GD]²= [DO]² [OG]=y [GD]= z+5 [GD]= 13/2+ 5 [DO]= r ⇒ y²+(13/2+5)²= r² we also know that: (6+y)²+z²= r² (6+y)²+(13/2)²= r² ⇒ y²+(13/2+5)²= (6+y)²+(13/2)² y²+169/4+65+25= 36+12y+y²+169/4 12y= 90-36 y= 54/12 y= 9/2 ⇒ r²= y²+(13/2+5)² r²= (9/2)²+(13/2+5)² r²= 81/4 + 169/4+ 65+25 r²= 250/4+ 90 r²= 610/4 r= √(610/4) r= √610/2 r ≈ 12,349 length units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Let's call the Line from Point O to the Big Chord, X 02) X^2 + (23/2)^2 = R^2 03) (X + 6)^2 + (13/2)^2 = R^2 04) X^2 + 529/4 = (X + 6)^2 + 169/4 05) X^2 + 529/4 = X^2 + 12X + 36 + 169/4 06) 12X = 529/4 - 169/4 - 36 07) 12X = 360/4 - 144/4 08) X = 216/48 ; X = 9/2 09) Using Formula on Item 2) : R^2 = 81/4 + 529/4 ; R^2 = 610/4 ; R^2 = 305/2 ; R = sqrt(305/2) 10) R ~ sqrt(152,5) ; R ~ 12,35 Therefore, OUR BEST ANSWER : The Radius is equal to sqrt(305/2) Linear Units or approx. equal to 12,35 Linear Inits.

[@PreMath]

Great! Thanks for sharing ❤️

[@marioalb9726]

Cyclic quadrilateral: a = 13 cm b = 18 + (18-13) = 23 cm c=d = √(6²+5²) = √61cm p = a+b+2c = 51.62 cm A² = (½p-a).(½p-b).(½p-c)² A² = 12,81x2,81x18² A = 108 cm² Circumference: (4R.A)²=(ab+cd)(ac+bd)² (4R.A)²= 360 x 79056 R = 12,349 cm ( Solved √ ) First is Brahmagupta's formula for cyclic quadrilateral areas Second is Parameshvara's formula for circumradius !! Both Indian mathematicians !!! Just as a point of interest, it's not the most practical solution

[@marioalb9726]

Intersecting chords theorem: 6 . x = 18 . 5 x = 15 cm c = x + 6 = 21 cm Intersecting chords theorem again: (R+13/2)(R-13/2)= (c/2)² R² - (13/2)² = (21/2)² R² = 10,5² + 6,5² R = 12,349 cm ( Solved √ )

[@marioalb9726]

Intersecting chords theorem: 6 . x = 18 . 5 --> x = 15 cm Quadratic mean, of chords: (2R)² = c₁² + c₂² (2R)² = (18²+6²)+(5²+15²) R = 12,349 cm ( Solved √ )

[@marioalb9726]

Easiest solution: Intersecting chords theorem: 6 . x = 18 . 5 --> x = 15 cm Pytagorean theorem: (2R)² = c₁² + c₂² (2R)² = 13² + (6+15)² R = 12,349 cm ( Solved √ )

[@PreMath]

Great! Thanks for sharing ❤️

[@JSSTyger]

I'll say r = sqrt(305/2)

[@unknownidentity2846]

Let's find the radius: . .. ... .... ..... Let's assume that O is the center of the coordinate system and that AB and CD are parallel to the y-axis. In this case AC must be parallel to the x-axis. Since A and D are located on the circle, with R being the radius of the circle we can conclude: xA² + yA² = R² xD² + yD² = R² The x-axis is perpendicular to AB, so it intersects this chord exactly at its midpoint. Therefore we know that yA=AB/2=13/2. From the known lengths of AC and CD we obtain: xD = xC = xA + 6 yD = yC − CD = yA − CD = 13/2 − 18 = 13/2 − 36/2 = −23/2 xA² + (13/2)² = R² (xA + 6)² + (−23/2)² = R² xA² + (13/2)² = (xA + 6)² + (−23/2)² xA² + 169/4 = xA² + 12*xA + 36 + 529/4 −504/4 = 12*xA −126 = 12*xA ⇒ xA = −126/12 = −21/2 Now we are able to calculate the radius of the circle: R² = xA² + yA² = (−21/2)² + (13/2)² = 441/4 + 169/4 = 610/4 = 305/2 ⇒ R = √(305/2) Let's check this result: xD = xA + 6 = −21/2 + 6 = −21/2 + 12/2 = −9/2 yD = −23/2 R² = xD² + yD² = (−9/2)² + (−23/2)² = 81/4 + 529/4 = 610/4 = 305/2 ✓ Best regards from Germany

[@PreMath]

Thanks for sharing ❤️

[@michaeldoerr5810]

The radius is [sqrt(610)]/2. Looks like I shall use that as intermediate level geometry practice!!!

[@PreMath]

Good idea! Thanks for the feedback ❤️

[@Waldlaeufer70]

Intersecting Chord Theorem: 6x = 13 (18 - 13) 6x = 13 * 5 = 65 x = 65/6 Pythagors: r² = (13/2)² + [(65/6 + 6)/2]² r² = 169/4 + [(65/6 + 36/6)/2]² r² = 169/4 + [101/12]² r² = 169/4 + 10201/144 = 6084/144 + 10201/144 = 16285/144 r = 10.63 units So, my result is different. Where's the mistake?

[@AllmondISP]

CD isn't a chord, neither AC. Both are part of a chord, but not a chord, so you can't use the Chord Theorem.

[@marioalb9726]

You wrote wrongly "13" instead of "18" !!! at the beginning Intersecting chords theorem: 6 . x = 18 . (18-13) x = 15 cm c = x + 6 = 21 cm Intersecting chords theorem again: (R+13/2)(R-13/2)= (c/2)² R² - (13/2)² = (21/2)² R² = 10,5² + 6,5² R = 12,349 cm ( Solved √ )

[@Waldlaeufer70]

@@AllmondISP Thanks, I always wondered where my mistake was. I guess I was too tired when I tried to solve the problem.

[@Waldlaeufer70]

@@marioalb9726 Thank you very much. So, my way was not completely wrong, just wrong enough. ;)

[@marioalb9726]

@@Waldlaeufer70 You was right choosing "intersecting chord theorem" method.. You could even do it twice, as my answer shows

[@wasimahmad-t6c]

This math you riten see

[@peterkrauliz5400]

Use Analytic Geometry for God's sake, man, and cross 2 perpendicular lines to get the center of the circle! This teacher doesn't like Analytic Geometry !