Find the Radius of the circle | (Important Geometry and Algebra skills explained) |

[@predator1702]

Fantastic solution and explanation 👍, thank you teacher 🙏.

[@PreMath]

Glad you liked it! ❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@skverma7278]

There is a simple way. Draw a perpendicular from centre on the centre of chord. Angle at the centre will also be 45 degrees. Hence length of perpendicular will be 4 units. Radius will be root of (4*4+12*12)=√160 =12.65 units

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@skverma7278]

@@PreMath thanks sir.

[@flesby]

Quite an elegant solution. For those interested who did not understand your approach immediately: 1. From the middle of the circle, that is point O, draw the perpendicular to the chord CD, you may elongate it even further until it touches the circumference of the circle. 2. Observe that the perpendicular bisects the chord CD in half in the Point we shall call P. 3. Since we know that CE equals 16 and ED equals 8, the chord CD has a length of 24. 4. Therefor CP = 24/2 = 12 5. Since the perpendicular is just that, it hits CD at an angle of 90 degrees 6. You receive the triangle OPE. Since the angle in P is 90 degrees as stated before and the angle in E is given (45 degrees), the remaining angle in O must be 45 degrees as well. (45 + 45 + 90 = 180 degrees) 7. Since CP = 12 and ED = 8 the remainder that is PE must be 4 since 12 + 8 + 4 = 24 = CD. 9. Since the triangle OPE from Step 6 is an equilaterals triangle, we can conclude that PO = PE = 4 10. Draw the line OC. - Observe that OC is the radius. 11. Observe that OCP is a right triangle with a leg length of 12 for CP and 4 for OP, a right angle in P and the hypthenuse of OC which is the radius r. 12. Apply the Pythagorean theorem: 4² + 12² = r² = 16 + 144 = 160. => sqrt (160) = r = 12,649...something. => Rounded to 12,65. 13. DONE! I hope this clarifies the way @skverma7278 took, to those who did not see it at first glance. - I apologize in case anything i wrote sounds clumsy or cumbersome, since English is not my native tongue and I seldom talk/write about mathematical ideas in English.

[@geoninja8971]

This is what I did - only took a minute or two, a nice simple pythagorean solve..... I appreciate the author's different approach though, there is always something to learn, and it is always cool to watch the exact solution flow from completely different methods....

[@tombufford136]

At a quick glance, A perpendicular from the midpoint, m, of a chord on a circle passes through the center of the circle. Then DM and CM = (16 + 8)/2 = 12. Forming a Right angled triangle OME, An isosceles triangle is formed OM = EM = 16 -12 = 4. Forming a right angled triangle, OMC. r^2 = 12^2 + 4^2 =160. Hence r= sqrt(160) = 12.65. The Radius of the circle is 12.65.

[@marioalb9726]

Intersecting chords theorem: (r+x).(r-x) = 8 . 16 r² - x² = 8 . 16 r² - (8 cos45°)² = 128 r² = 128 + (8/√2)² r = 12,65 cm ( Solved √ )

[@santiagoarosam430]

Llamaremos F al punto medio de la cuerda CD》 CE=16=CF+FE=[(16+8)/2]+4=12+4》FO=FE=4》Potencia de F respecto a la circunferencia =12^2=(r-4)(r+4)=r^2 -4^2 》r^2=160》r=4(sqrt10) Gracias y un saludo cordial.

[@phungpham1725]

Drop the perpendicular OH to chord CD ad extend OH to complete the diameter. we have: H is the midpoint of CD so CH = DH =12 The small 45 degree right isosceles so OH=OE=4 ----> chord theorem: (R+4)(R-4) =sq12---->sqR -sq4 =sq12------> sqR= sq12+sq4= 144+16= 160 R=12.65 units

[@michaelkouzmin281]

Justt one more solution: Let x=OE; r=OA=OC. (r+x)*(r-x)=16*8 => r^2-x^2=16*8; (1) r^2=16^2-2*16*x*cos(D45)+x^2 => r^2-x^2=16^2-2*16*x*cos(D45); (2) Left sides of (1) and (2) are equal => 16*8= 16^2-2*16*x*cos(D45) => x=4*sqrt(2); let us put our found x into (1): r^2=128+x^2=128+32=160 => r= sqrt(160)=4sqrt(10).

[@majidsetoudeh1654]

If you drop a perpendicular from O to the chord CD it will divide the chord in two equal segments and form a small right triangle OEH. The length of CH will be (16+8)/2 =12 and so HE=4. OH will also be 4 unit because OHE is a right angle triangle with two angles being equal to 45. The hypotheses OE will be 4 root 2. So AE=r+4root2 and BE=r-4root2, the rest of the solution will be as the rule of intersecting two chords like you explained resulting r=4root10.

[@farzad1343]

I'm completely agree with you. This is my first solution also, much simpler

[@giuseppemalaguti435]

cosECO=12/r....16cos45=rcos(45-ECO)...r=√160.,.r=√928...credo sia corretta r=√160

[@zdrastvutye]

the largest angle this solves without changing l1 and l2 ist 58 degrees 10 l1=16:l2=8:w=45*pi/180:y3=l1*sin(w):y4=-l2*sin(w):sw=.01:r=sw+y3 20 lx=(l1+l2)*cos(w):goto 50 30 x3=-sqr(r^2-y3^2)+r:dgu1=y4^2/r^2:dgu2=(x3+lx-r)^2/r^2:dg=dgu1+dgu2-1 40 return 50 gosub 30 60 dg1=dg:r1=r:r=r+sw:if r>100*l1 then stop 70 r2=r:gosub 30:if dg1*dg>0 then 60 80 r=(r1+r2)/2:gosub 30:if dg1*dg>0 then r1=r else r2=r 90 if abs(dg)>1E-10 then 80 100 print r:x4=x3+lx:mass=200/r:goto 120 110 xb=x*mass:yb=(y-r)*mass:return 120 x=r:y=0:gosub 110:circle xb,yb,r*mass:x=x3:y=y3:gosub 110:xba=xb:yba=yb 130 x=x4:y=y4:gosub 110:xbn=xb:ybn=yb:line xba,yba,xbn,ybn 140 x=0:y=0:gosub 110:xba=xb:yba=yb:x=2*r:y=0:gosub 110:xbn=xb:ybn=yb:line xba,yba,xbn,ybn 13.6006298 > run in bbcbasic sdl and hit ctrl tab to copy

[@safwanmfarij3023]

I solved it in another way I drew two lines from C to O and from D to O as a Radius. I named OB as X Then i used the law of cosines In the two triangles CEO & DEO to calculate the radius through the variable X Then i equaled the two equations to solve for X then i put the value in one of the two equations to solve for the radius

[@AmirgabYT2185]

At a quick glance as diametre and chord are equal to each other diametre will be 16+8=24 so the radius seems to be 12😅

[@HappyFamilyOnline]

Great explanation 👍 Thanks for sharing 😊

[@soli9mana-soli4953]

Setting OE = X and radius = R and H the proiection of C point on the diameter we can write 2 equations first one with the intersecting chords theorem EB : ED = EC : EA => (R-X) : 8 = 16 : (R+X) the secons with Pythagorean theorem: CH² + HO² = OC² => (8√ 2)² + (8√ 2 - X)² = R² (being HEC an isosceles right triangle with hypotenuse 16)

[@Geometri_asiklari]

Merhaba, CD kirişine dik indirip, dikme kirişi 2 ye böler, dikten E ye kadar 4 olur, 45,45 ikizkenar üçgenden dikme 4 olur, sonra pisagor dan 4 ün karesi + 12 nin karesi eşittir r nin karesi, r kare 160 kök 160 olur🤓

[@MrArcan10]

to find OE=x I used the theorem of cosines for triangles COE and DOE (got radius via x from both triangles to build the equation)

[@timeonly1401]

Know OA = r. Let OE = a. Then AE = r+a and BE = r-a. By Intersecting Chords Thm, (AE)(BE)=(CE)(DE) (r+a)(r-a)=16*8 r² - a² = 128, r² = a² + 128 -- (1) Draw radius CO = r. Using Law of Cosines: r² = 16² + a² - 2(16)a cos(45). Subbing (1) into LHS & simplifying: a² + 128 = 256 + a² - 32a (√2 / 2) 0 = 128 - 16√2 a So, a = 128/(16√2) = 4√2. From (1) r² = (4V2)² +128 = 32 + 128 = 160 So, r = √160 = 4√10 . Done!

[@janwendlandt3126]

√((16-12)²+12²) = 12.65 Explanation: If you plumb from the chord perpendicular to the center of the circle, so you hit it with 45 degrees to the horizontal diameter line. An isosceles triangle is created. The plumb point is exactly at the center of the chord, divides this as (16+8)/2 = 12. So the two legs are each 16-12 = 4 This is also the distance of the chord from the center of the circle. On the upper side of the chord you now have a right-angled triangle with legs 12 and 4, the hypotenuse of this triangle is the wanted radius: √(12²+4²) = 12.65

[@dainiusb1114]

Perpendicular from O to CD. CO^2=R^2=4^2+{(16+8)/2}^2.

[@じーちゃんねる-v4n]

OE=4√2 (r+4√2)(r-4√2)=r^2-32=8*16=128 ∴r^2=160 ∴r=4√10

[@venelinarnaudov7416]

Nice problem and nice solution! I had it in another way. I drew a chord perpendicular to the diameter (AB) and passing through point E. Let the new chord crosses the arc CB in point M. The symetric point on arc AD should be N. It is obvious that EM=EN and that EM^2=16*8 or EM=8*sqrt(2) The angle CEM=45°. We apply cosin formula => EC^2+EM^2-2*EC*EM*cos(45°)=CM^2 => CM^2= 16^2+16*8-2*16*8*sqrt(2)*sqrt(2)/2 CM^2 = 16^2 + 16*8 - 2*16*8 CM^2 = 16*8 CM = EM => the triangle CME is orthogonal and equaliteral, where angle CME=90°, MCE=45° => CM||AB => EB = (AB-CM)/2 & OE=CM/2 Let OB=R and OE=x (=8*sqrt(2)/2 = 4*sqrt(2)) We apply the chord formula AE*EB=EM^2 (R-x)*(R+x) = EM^2 (R-CM)*(R+CM)=(2*CM)^2 R^2 - CM^2 = 4*CM^2 R^2= 5*CM^2 R = CM*sqrt(5) R= 4*sqrt(2)*sqrt(5) R = 4*sqrt(10)

[@WaiWai-qv4wv]

Very thanks

[@tombufford136]

I am learning fast. Thales theorem, 2 tangent theorem, interior and external angle theorems .......

[@じーちゃんねる-v4n]

OE=4√2 distance O and CD is 4 OC=√(12^2+4^2)=4√(9+1)=4√10

[@mathbynisharsir5586]

Wonderful sir

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@jakalasi7540]

Hi sir, May you please show us how to calculate the probability of winning a lottery. Thank you 🙏

[@eoinrobson2725]

Genius!

[@johnwindisch1956]

Did it using the radius chord properly

[@Nothingx303]

❤❤❤❤❤❤❤❤❤❤❤veryyyy informative 👌 sir

[@PreMath]

Thanks and welcome ❤️ You are awesome. Keep it up 👍

[@misterenter-iz7rz]

Let OE be x, OA be r, then 16×8=(r+x)(r-x)=r^2-x^2, r^2=x^2+128, r^2=x^2+128=x^2+(16-sqrt(2)x)^2+sqrt(2)x(16-sqrt(2)x), 128=(16-sqrt(2)x)^2+sqrt(2)x(16-sqrt(2)x), 128=256+2x^2-32sqrt(2)x+16sqrt(2)x-2x^2, 16sqrt(2)x=128, x=128/(16sqrt(2))=8/sqrt(2)=4sqrt(2), r^2=32+128=160, r=4sqrt(10).😅

[@JSSTyger]

I think r=sqrt(4^2+12^2)=sqrt(160)=4sqrt(10).

[@jhill4874]

Gack! I had this up to the next to last step, but I subtracted 32 instead of adding it! Dang calculation glitch!

[@ИванФортунов-й6о]

Why CF is equal 8?

[@parthtomar6987]

Because the sectors are congruent

[@parthtomar6987]

You can easily compare both of these

[@hadialfuraiji6531]

Proving that two sectors match is not clear !

[@misterenter-iz7rz]

This is difficult. 😅

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍