1/ By using chord theorem: (7x-5).(x+2) = x.(5x+1) --> we have the equation Sqx + 4x -5 = 0 Or (sq x + 2.2 x + 4)-4 -5 = 0 --> sq(x+2) = 9 x+2 = +/- (3) x=1. ( negative result rejected) 2/ Chord AB= 1+6=7 and chord CD= 2+3 = 5 Distance from center O to AB= 1/2 and distance from O to CD= 5/2 Just draw the diameter perpendicular to CD and by using chord theorem again: ( r-5/2).(r+5/2)= sq5/2--> sq r-sq5/2= sq5/2 --> sqr= sq5/2 + sq5/2= 25/4 + 25/4 sqr = 50/4 r = 5sqrt2/2😅
Let's find the radius: . .. ... .... ..... First of all we calculate x by applying the intersecting chords theorem: AE*BE = CE*DE x*(5*x + 1) = (7*x − 5)*(x + 2) 5*x² + x = 7*x² + 14*x − 5*x − 10 0 = 2*x² + 8*x − 10 0 = x² + 4*x − 5 0 = (x − 1)*(x + 5) Since AE=x>0, the only useful solution is: x = 1 ⇒ AE = 1 ∧ BE = 6 ∧ CE = 2 ∧ DE = 3 ⇒ AB = 7 ∧ CD = 5 Now let's assume that A'B' is parallel to AB, that C'D' is parallel to CD and that A'B' and C'D' are both diameters of the circle. For reasons of symmetrie we can conclude: A'E' = AE + a B'E' = BE + a C'E' = CE + b D'E' = DE + b By applying the intersecting chords theorem we obtain: A'E'*B'E' = CE'*DE' (AE + a)*(BE + a) = (CD/2)*(CD/2) (1 + a)*(6 + a) = (5/2)*(5/2) a² + 7*a + 6 = 25/4 a² + 7*a − 1/4 = 0 a = −(7/2) ± √[(7/2)² + 1/4] = −(7/2) ± √(49/4 + 1/4) = −(7/2) ± √(50/4) = (−7 ± 5√2)/2 AE'*BE' = C'E'*D'E' (AB/2)*(AB/2) = (CE + b)*(DE + b) (7/2)*(7/2) = (2 + b)*(3 + b) 49/4 = b² + 5*b + 6 0 = b² + 5*b − 25/4 b = −(5/2) ± √[(5/2)² + 25/4] = −(5/2) ± √(25/4 + 25/4) = −(5/2) ± √(50/4) = (−5 ± 5√2)/2 Since a>0 and b>0, there is only one useful solution for a and only one useful solution for b. For the first case (shifting AB to A'B') we obtain: 2*R = A'B' = A'E' + B'E' = AE + a + BE + a = AB + 2*a = 7 + (−7 + 5√2) = 5√2 ⇒ R = (5/2)√2 For the second case (shifting CD to C'D') we obtain: 2*R = C'D' = C'E' + D'E' = CE + b + DE + b = CD + 2*b = 5 + (−5 + 5√2) = 5√2 ⇒ R = (5/2)√2 ✓ Best regards from Germany
Intersecting chords theorem x(5x+1)=(x+2)(7x-5) 5x^2+x=7x^2-5x+14x-10 0=2x^2+8x-10 0=x^2+4x-5 0= (x-1)(x+5) x-1 =0 or x+5=0 x=1 or x=-5 x=1 since x= |x| let's use coordinate geometry (E;EB;EC) E(0;0) ; B(6;0) ; C(0;2) ; D(0;-3) ; A(-1;0) let O(h;k) Equation of a circle : (x-h)^2+(y-k)^2=r^2 let's derive equations (6-h)^2+(-k)^2=r^2 (1) (-h)^2+(2-k)^2=r^2 (2) (-h)^2+(-3-k)^2=r^2 (3) (-1-h)^2+(-k)^2= r^2 (4) subtract equation 1 and 4 to get (6-h)^2-(-1-h)^2=0 (6-h)^2-(1+h)^2=0 (6-h+1+h)(6-h-1-h)=0 7*(5-2h)=0 5-2h=0 h=5/2 subtract equations (2) and (3) to get (2-k)^2-(-3-k)^2=0 (2-k)^2-(3+k)^2=0 (2-k+3+k)(2-k-3-k)=0 5(-1-2k)=0 -5(1+2k)=0 1+2k=0 k=-1/2 so coordinates of O are (5/2;-1/2) let's substitute h and k in one of the equations (-1-5/2)^2+(1/2)^2= r^2 (-7/2)^2+1/4=r^2 49/4+1/4=r^2 50/4=r^2 so r= sqrt (50/4)= 5*sqrt2/2
( 7x -5)× ( x +2)=x ×( 5x +1) Quadratic 2 roots...testing Forming square Get the value of side Diagonal = root 2 × side D^2= a ^2 + b^2 + c^2 + d^2 R =D/2
By Intersecting chords theorem, AE•EB = CE•ED. AE•EB = CE•ED x(5x+1) = (x+2)(7x-5) 5x² + x = 7x² + 14x - 5x - 10 2x² + 8x - 10 = 0 x² + 4x - 5 = 0 (x+5)(x-1) = 0 x = -5 ❌ | x = 1 AE = 1 EB = 5(1) + 1 = 6 CE = 7(1) - 5 = 2 ED = (1) + 2 = 3 As AB and CD are chords, any radii that perpendicularly intersect the chords will do so at their midpoints. Draw radii OM and ON perpendicular to AB and CD respectively. The intersection points will be P and Q respectively. As AB = 1+6 = 7, AP = PB = 7/2. As CD = 2+3 = 5, CQ = QD = 5/2. As AE = 1, EP = OQ = 7/2-1 = 5/2, and as CE = 2, EQ = OP = 5/2-2 = 1/2. Triangle ∆OPB: OP² + PB² = OB² (1/2)² + (7/2)² = r² r² = 1/4 + 49/4 = 50/4 = 25/2 r = √25/2 = 5/√2 = (5√2)/2
STEP-BY-STEP RESOLUTION PROPOSAL : 1) Intersecting Chords Theorem : 2) (7X - 5) * (X + 2) = X * (5X + 1) 3) X = - 5 or X = 1 4) Positive Solution Check : 2 * 3 = 1 * 6 ; 6 = 6 5) AB = (1 + 6) = 7 lin un 6) CD = (2 + 3) = 5 lin un 7) AB / 2 = (7 / 2) = 3,5 lin un 8) CD / 2 = (5 / 2) = 2,5 lin un 9) Draw two Diameters; (passing through Point O); one parallel to Blue Chord and the other parallel to Red Chord. 10) Horizontal Distance from Diameter to Red Chord equal to : 3,5 - 1 = 2,5 lin un. 11) Vertical Distance from Diameter to Blue Chord equal to : 2,5 - 2 = 0,5 lin un. 12) Using Pythagoras Theorem : 13) R^2 = 0,5^2 + 3,5^2 14) R^2 = 0,25 + 12,25 15) R^2 = 12,5 16) R = sqrt(12,5) 17) R ~ 3,355 lin un 18) Radius equal to sqrt(25/2) Linear Units or Radius approx. equal to 3.355 Linear Units.
(7X-5)(X+2)=X(5X+1) ; X=1 ; Longitud de la cuerda vertical =2v; v=(2+3)/2=5/2; su distancia al centro es b=[(1+6)/2]-1=5/2=v; Radio =r =(5/2)√2=5√2/2. Gracias y un saludo cordial.
1. Use intersecting chords theorem to find x = 1. 2. AE =1, CE = 2, DE = 3 3. For triangle ACD, sides AC = sqrt 5 and AD = sqrt 10 by Pythagoras theorem and CD = 2 + 3 = 5 4. Area of triangle ACD = area of triangle ACE + area of triangle ADE = 2.5 5. Radius of circumcircle of triangle ACE = (AC)(AD)(CD)/4 x area = sqrt 5 x sqrt 10 x 5/ 4 x 2.5 = 5 x sqrt 2/2
Naturally applying the theorem of intersecting chords, x(5x+1)=(7x-5)(x+2), x^2+4x-5=0, x=1 or -5, rejected, then we have two perpendicular chords with length 7=2×7/2 and 5=2×5/2 with middle rectangle 1/2×5/2, therefore r^2=(7/2)^2+(1/2)^2=(5/2)^2+(5/2)^2=50/4, r=(5/2)sqrt(2).😊 The puzzles figures are well designed.🎉🎉🎉
Thanks Sir
That’s very nice
We are learn very much from like these exercises .
With my respects
❤❤❤
It's my pleasure!
Thanks for the feedback ❤️🌹
1/ By using chord theorem:
(7x-5).(x+2) = x.(5x+1)
--> we have the equation
Sqx + 4x -5 = 0
Or
(sq x + 2.2 x + 4)-4 -5 = 0
--> sq(x+2) = 9
x+2 = +/- (3)
x=1. ( negative result rejected)
2/ Chord AB= 1+6=7 and chord CD= 2+3 = 5
Distance from center O to AB= 1/2 and distance from O to CD= 5/2
Just draw the diameter perpendicular to CD and by using chord theorem again:
( r-5/2).(r+5/2)= sq5/2--> sq r-sq5/2= sq5/2
--> sqr= sq5/2 + sq5/2= 25/4 + 25/4
sqr = 50/4
r = 5sqrt2/2😅
Let's find the radius:
.
..
...
....
.....
First of all we calculate x by applying the intersecting chords theorem:
AE*BE = CE*DE
x*(5*x + 1) = (7*x − 5)*(x + 2)
5*x² + x = 7*x² + 14*x − 5*x − 10
0 = 2*x² + 8*x − 10
0 = x² + 4*x − 5
0 = (x − 1)*(x + 5)
Since AE=x>0, the only useful solution is:
x = 1
⇒ AE = 1 ∧ BE = 6 ∧ CE = 2 ∧ DE = 3
⇒ AB = 7 ∧ CD = 5
Now let's assume that A'B' is parallel to AB, that C'D' is parallel to CD and that A'B' and C'D' are both diameters of the circle. For reasons of symmetrie we can conclude:
A'E' = AE + a
B'E' = BE + a
C'E' = CE + b
D'E' = DE + b
By applying the intersecting chords theorem we obtain:
A'E'*B'E' = CE'*DE'
(AE + a)*(BE + a) = (CD/2)*(CD/2)
(1 + a)*(6 + a) = (5/2)*(5/2)
a² + 7*a + 6 = 25/4
a² + 7*a − 1/4 = 0
a = −(7/2) ± √[(7/2)² + 1/4] = −(7/2) ± √(49/4 + 1/4) = −(7/2) ± √(50/4) = (−7 ± 5√2)/2
AE'*BE' = C'E'*D'E'
(AB/2)*(AB/2) = (CE + b)*(DE + b)
(7/2)*(7/2) = (2 + b)*(3 + b)
49/4 = b² + 5*b + 6
0 = b² + 5*b − 25/4
b = −(5/2) ± √[(5/2)² + 25/4] = −(5/2) ± √(25/4 + 25/4) = −(5/2) ± √(50/4) = (−5 ± 5√2)/2
Since a>0 and b>0, there is only one useful solution for a and only one useful solution for b. For the first case (shifting AB to A'B') we obtain:
2*R = A'B' = A'E' + B'E' = AE + a + BE + a = AB + 2*a = 7 + (−7 + 5√2) = 5√2 ⇒ R = (5/2)√2
For the second case (shifting CD to C'D') we obtain:
2*R = C'D' = C'E' + D'E' = CE + b + DE + b = CD + 2*b = 5 + (−5 + 5√2) = 5√2 ⇒ R = (5/2)√2 ✓
Best regards from Germany
Intersecting chords theorem
x(5x+1)=(x+2)(7x-5)
5x^2+x=7x^2-5x+14x-10
0=2x^2+8x-10
0=x^2+4x-5
0= (x-1)(x+5)
x-1 =0 or x+5=0
x=1 or x=-5
x=1 since x= |x|
let's use coordinate geometry (E;EB;EC)
E(0;0) ; B(6;0) ; C(0;2) ; D(0;-3) ; A(-1;0)
let O(h;k)
Equation of a circle :
(x-h)^2+(y-k)^2=r^2
let's derive equations
(6-h)^2+(-k)^2=r^2 (1)
(-h)^2+(2-k)^2=r^2 (2)
(-h)^2+(-3-k)^2=r^2 (3)
(-1-h)^2+(-k)^2= r^2 (4)
subtract equation 1 and 4 to get
(6-h)^2-(-1-h)^2=0
(6-h)^2-(1+h)^2=0
(6-h+1+h)(6-h-1-h)=0
7*(5-2h)=0
5-2h=0
h=5/2
subtract equations (2) and (3) to get
(2-k)^2-(-3-k)^2=0
(2-k)^2-(3+k)^2=0
(2-k+3+k)(2-k-3-k)=0
5(-1-2k)=0
-5(1+2k)=0
1+2k=0
k=-1/2
so coordinates of O are (5/2;-1/2)
let's substitute h and k in one of the equations
(-1-5/2)^2+(1/2)^2= r^2
(-7/2)^2+1/4=r^2
49/4+1/4=r^2
50/4=r^2
so r= sqrt (50/4)= 5*sqrt2/2
( 7x -5)× ( x +2)=x ×( 5x +1)
Quadratic 2 roots...testing
Forming square
Get the value of side
Diagonal = root 2 × side
D^2= a ^2 + b^2 + c^2 + d^2
R =D/2
By Intersecting chords theorem, AE•EB = CE•ED.
AE•EB = CE•ED
x(5x+1) = (x+2)(7x-5)
5x² + x = 7x² + 14x - 5x - 10
2x² + 8x - 10 = 0
x² + 4x - 5 = 0
(x+5)(x-1) = 0
x = -5 ❌ | x = 1
AE = 1
EB = 5(1) + 1 = 6
CE = 7(1) - 5 = 2
ED = (1) + 2 = 3
As AB and CD are chords, any radii that perpendicularly intersect the chords will do so at their midpoints. Draw radii OM and ON perpendicular to AB and CD respectively. The intersection points will be P and Q respectively. As AB = 1+6 = 7, AP = PB = 7/2. As CD = 2+3 = 5, CQ = QD = 5/2. As AE = 1, EP = OQ = 7/2-1 = 5/2, and as CE = 2, EQ = OP = 5/2-2 = 1/2.
Triangle ∆OPB:
OP² + PB² = OB²
(1/2)² + (7/2)² = r²
r² = 1/4 + 49/4 = 50/4 = 25/2
r = √25/2 = 5/√2 = (5√2)/2
R=5√2/2≈3,54
Another Method
Draw perpendiculars OX on AB & OY on CD.
Join OE & OB
++ OB is the radius.
AE*BE=OB^2 - EO^2---(1)
[I will prove it below ]
In 🔺 EXO
EO^2 =OX^2 + EX^2
= (1/2)^2 +(5/2)^2
[ note
AE*BE=(BX - EX) (BX + EX)
= BX^2 - EX^2
= (BX^2+ OX^2)- (OX^2 +EX^2)
=.OB^2 -EO^2
=13/2----(2)
From (1) & (2)
AE *BE=OB^2 -13/2
>1*6 + 13/2 =OB ^2
Then OB = 5/√2=5√2/2
[ note
AE*BE =(BX - EX)(BX+EX){as BX = AX}
= BX^2 - EX^2
=(BX^2 + OX^2) -(OX^2+ EX^2) = OB^2 - EO^2 ]
x(5x+1)=(x+2)(7x-5)
5x^2+x=7x^2-5x+14x-10
2x^2+8x-10=0
x^2+4x-5=0
So x=1 ; x=-5 rejected
AE=1; BE=6; CE=2; DE=3
4R^2=1^2+2^2+3^2+6^2
R=5√2/2 units.❤❤❤
STEP-BY-STEP RESOLUTION PROPOSAL :
1) Intersecting Chords Theorem :
2) (7X - 5) * (X + 2) = X * (5X + 1)
3) X = - 5 or X = 1
4) Positive Solution Check : 2 * 3 = 1 * 6 ; 6 = 6
5) AB = (1 + 6) = 7 lin un
6) CD = (2 + 3) = 5 lin un
7) AB / 2 = (7 / 2) = 3,5 lin un
8) CD / 2 = (5 / 2) = 2,5 lin un
9) Draw two Diameters; (passing through Point O); one parallel to Blue Chord and the other parallel to Red Chord.
10) Horizontal Distance from Diameter to Red Chord equal to : 3,5 - 1 = 2,5 lin un.
11) Vertical Distance from Diameter to Blue Chord equal to : 2,5 - 2 = 0,5 lin un.
12) Using Pythagoras Theorem :
13) R^2 = 0,5^2 + 3,5^2
14) R^2 = 0,25 + 12,25
15) R^2 = 12,5
16) R = sqrt(12,5)
17) R ~ 3,355 lin un
18) Radius equal to sqrt(25/2) Linear Units or Radius approx. equal to 3.355 Linear Units.
Method 2 is faster
(7x-5)*(x+2)=x(x+1), x^2+4x-5=0, x=1. Letёs draw perpendicular OF to CD, OF=AB/2-x =3,5-1 = 2,5. CF=CD/2=2,5,
OC = r = \/2,5^2+2,5^2=2,5\/2=3,53553390592...!
(7X-5)(X+2)=X(5X+1) ; X=1 ; Longitud de la cuerda vertical =2v; v=(2+3)/2=5/2; su distancia al centro es b=[(1+6)/2]-1=5/2=v; Radio =r =(5/2)√2=5√2/2.
Gracias y un saludo cordial.
1. Use intersecting chords theorem to find x = 1.
2. AE =1, CE = 2, DE = 3
3. For triangle ACD, sides AC = sqrt 5 and AD = sqrt 10 by Pythagoras theorem and CD = 2 + 3 = 5
4. Area of triangle ACD = area of triangle ACE + area of triangle ADE = 2.5
5. Radius of circumcircle of triangle ACE = (AC)(AD)(CD)/4 x area
= sqrt 5 x sqrt 10 x 5/ 4 x 2.5
= 5 x sqrt 2/2
Naturally applying the theorem of intersecting chords, x(5x+1)=(7x-5)(x+2), x^2+4x-5=0, x=1 or -5, rejected, then we have two perpendicular chords with length 7=2×7/2 and 5=2×5/2 with middle rectangle 1/2×5/2, therefore r^2=(7/2)^2+(1/2)^2=(5/2)^2+(5/2)^2=50/4, r=(5/2)sqrt(2).😊
The puzzles figures are well designed.🎉🎉🎉
Thanks sir, I wanted to ask if there were any limitations to the formula you showed
Thank you
arctg x/(7x-5)+arctg(5x+1/7x-5)+arctg x/x+2+arctg5x+1/x+2=180..x=-1/6,-15/4(no)...x=1,x=3/8(no).. l'unico accettabile è x=1..r=√(3,5^2+0,5^2)=√12,5=5/√2
I used the first method as I had not the second in mind.