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

Intersecting Chords Theorem Proof link:

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

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[predator]

Fantastic solution and explanation 👍, thank you teacher 🙏.

[skverma]

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

[tombufford]

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.

[marioalb]

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 √ )

[santiagoarosam]

Llamaremos F al punto medio de la cuerda CD》 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.

[majidsetoudeh]

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.

[safwanmfarij]

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

[phungpham]

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 sqR= sq12+sq4= 144+16= 160
R=12.65 units

[michaelkouzmin]

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).

[giuseppemalaguti]

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

[solimana-soli]

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🤓

[AmirgabYT]

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😅

[venelinarnaudov]

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 => =>
CM^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)

[janwendlandt]

√((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

[HappyFamilyOnline]

Great explanation 👍
Thanks for sharing 😊

[MrArcan]

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)

[zdrastvutye]

the largest angle this solves without changing l1 and l2 ist 58 degrees
10
20 lx=(l1+l2)*cos(w):goto 50
30
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
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:xbn=xb:ybn=yb:line xba, yba, xbn, ybn
13.6006298
>
run in bbcbasic sdl and hit ctrl tab to copy

[timeonly]

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!

[dainiusb]

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