Можно найти радиус по формуле R=a*b*c/(4*S) для ∆DCB. Стороны находим по теореме Пифагора, S=0.5 * DC*PB=78, CB=4√10, DB=15, R=14*4√10*15/(4*78)=2,5√10.
AP = 3 (in triangle APC) Then 3.(x + 5) = 4.(2.x -5) (chords th.), giving that x = 7. So AP = 3, BP = 12, DP = 9, CP = 4 We use then an orthonormal center P and first axis (PB) We have A(-3; 0) B(12, 0) C( 0; 4) D(0; -9) The equation of the circle is x^2 + y^2 + a;x + b.y + c = 0 A and B are on the circle, so 9 -3.a + c = 0 and 144 + 12.a +c = 0, giving c = - 36 and a = -9. C is on the circle, so 16 + 4.b - 36 = 0, giving that b = 5. The equation of the circle is x^2 + y^2 -9.x +5.y -36 = 0, or (x -9/2)^2 + (y +5/2)^2 = 250/4 So the radius of the circle is sqrt(250/4) = (5/2).sqrt(10).
1. Use intersecting chords theorem to find x = 7. 2. Use Pythagoras theorem to find BC = 4sqrt(10) and BD = 15 while BP = 12. 3. Use area formula (area = 1/2 base x height) to find area of triangle BCD = (1/2)(13)(12) = 78. 4. Use circumradius formula for triangle BCD = abc/4xarea = (13)(15)(4sqrt10)/(4)(78) = (5sqrt10)/2.
El triángulo APC es rectangular→ Si AC=5 y PC=4→ AP=3→ Potencia de P respecto a la circunferencia =3(X+5)=4(2X-5)→ X=7→ AB=3+12=15 ; CD=4+9=13→ Si Q es la proyección ortogonal de P sobre el diámetro horizontal→ En el triángulo OQC: OQ²+QC²=OC²=r²→ (9/2)²+(13/2)²=r²→ r=(5√10)/2 ud. Gracias y saludos.
After finding the value of x, I solved the problem by rotating the figure so that AB coincides with the X-axis and CD with the Y-axis and point P at the origin. Now Boeing the 4 points are at the circumference of a circle, I proceeded to solve for the equation of that circle.
In ∆APC AP^2+CP^2=AC^2 AP^2+4^2=5^2 AP=√5^2-4^2=3 (AP)(BP)=(CP)(DP) 3(x+5)=4(2x-5) So x=7 BP=7+5=12 DP=2(7)-5=14-5=9 Connect O to E (E middle AB and OE right AB ) AE=BE=15/2 Connect O to F (F middle CD and right CD) CF=DF=13/2 PF=9-6.5=2.5 Connect O to B In ∆ OBE OE^2+BE^2=OB^2 (2.5)^2+(7.5)^2=r^2 So r=5√10/2 units=7.91.units.❤❤❤
Teorema delle corde..4:3=(x+5):((2x-5)...8x-20=3x+15..x=7..il triangolo inscritto AOB ha lati 15,√90,15..r=abc/4A=225*√90/4*67,5=225√90/270=5√90/6=5√10/2
PA = 3, due to 3,4,5. 3(x+5) = 4(2x-5) 3x + 15 = 8x - 20 3x = 8x - 35 x = 7 Make the midpoint of AB, M. AB = 7+5+3 = 15 Move AB down to lie across the diameter. Either side of the other chord, it now has lengths of 12+a and 3+a. It bisects the other chord (DC) which is 14 - 5 + 4 long, so 13/2 on each side of the newly positioned AB. Although AB is 15+2a long, is is also 2r. We now need to find the length of a. (a+12)(a+3) = (13/2)(13/2) will find length a. a^2 + 15a + 36 = 169/4 4a^2 + 60a + 144 = 169 4a^2 + 60a - 25 = 0 (-60+or-sqrt(3600 - 4*4*-25))/8 = a (-60+or-sqrt(4000))/8 = a (-60+or-4*sqrt(250))/8 = a Discard the minus result as it will be a negative value. (-15+sqrt(250))/2 = a (-15+5*sqrt(10))/2 = a a = (5*sqrt(10) - 15)/2 = 0.4057 (rounded). the diameter is 15+2a, so r is 7.5+a. r = 7.9057 (rounded) Just checked the video. Different paths, same conclusion.
1/ Note that the two triangles APC and DPB are 3/4/5 triplets. 2/ x=7-> PB=12 ->PD=9 and BD=15 Because AB= 15 too, so the triangle ABD is an isosceles one, so: Just draw the diameter BB’, -> BB’ is the perpendicular bisector of the chord AD. We have: sqAD=sqAP+sqBD=9+81=90 AD=3sqrt10 Label the angle ABB’=alpha We have: Angle ADP=angleABB’=alpha( having sides perpendicular) -> cos alpha=PD/AD=3/sqrt10 Focus on the triangle B’AB AB/BB’= cos alpha-> BB’=AB/cos alpha 2R=15/(3sqrt10)-> R=5sqrt10/2= 7.91 untd😅😅😅
7.91 AP = 3 (3-4-5) right triangle 3(x+5) = 4(2x-5) intersection chord theorem 3x + 15 = 8x -20 35 = 5x x =7 PB = 12 (7+5) PD = 9 AB=15, hence the center = 7.5, but since AP = =3, its center is from CD , let label it RS = 4.5 (7.5 -3) CD =13, hence the center = 6.5 Draw a line from the center of the CD to the circle's center; the distance is 4.5 Draw a line from C to O , this is the radius r^2 = 4.5^2 + 6.5^2 r =7.91
Alternative solution: Construct OD, which is also a radius and is the hypotenuse of ΔOFD. Length DF = 13/2 and length OF = 15/2 - 3 = 9/2. Applying the Pythagorean theorem, r² = (13/2)² + (9/2)² = 169/4 + 81/4 = 250/4 and, after simplifying, r = (5√(10))/2, as PreMath found from ΔOEB. Note that ΔOFC can also be used, but is congruent to ΔOFD. Similarly, ΔOEA is congruent to PreMath's ΔOEB.
Start with the intersecting chord theorum. Do the algebra. X = 7. Thus AP is 3, PB is 12, CP is 4 , PD is 9. Next, square each element, then add them together. Comes out to 250. Then, divide by 4, get square root of quotient. Makes r = 7.0956. Can't remember the name of the formula or who discovered it.
Let's find the radius: . .. ... .... ..... The triangle ACP is a right triangle, so we can apply the Pythagorean theorem: AP² + CP² = AC² ⇒ AP² = AC² − CP² = 5² − 4² = 25 − 16 = 9 ⇒ AP = 3 By using the intersecting chords theorem we obtain: AP*BP = CP*DP 3*(x + 5) = 4*(2*x − 5) 3*x + 15 = 8*x − 20 35 = 5*x ⇒ x = 7 Now let's assume that O is the center of the coordinate system and that AB is parallel to the xaxis. Since CD is perpendicular to AB, CD must then be parallel to the y-axis. Therefore we can conclude: xB = AB/2 = (AP + BP)/2 = [3 + (x + 5)]/2 = [3 + (7 + 5)]/2 = 15/2 yC = CD/2 = (CP + DP)/2 = [4 + (2*x − 5)]/2 = [4 + (2*7 − 5)]/2 = 13/2 yB = yC − CP = 13/2 − 4 = 13/2 − 8/2 = 5/2 xC = AP − xB = 3 − 15/2 = 6/2 − 15/2 = −9/2 Now we are able to calculate the radius R and to check the result: R² = xB² + yB² = (15/2)² + (5/2)² = 225/4 + 25/4 = 250/4 ⇒ R = (5/2)√10 R² = xC² + yC² = (−9/2)² + (13/2)² = 81/4 + 169/4 = 250/4 ⇒ R = (5/2)√10 ✓ Best regards from Germany
The radius is 1/2[5*sqrt(10)]. I am glad that I did my own sanity check on my work. I think that you should create of platliat that make use of intersecting chirds theorem as well perpendicular busector theorems. Or playlists that involve a combination of at least two or three theorems.
My way of solution ▶ We can find [AP] by using the Pythagoream theorem: CA²= AP²+PC² [PC]= 4 [CA]= 5 5²= AP²+4² ⇒ [AP]= 3 According to the intersecting chords theorem : [AP]*[PB]= [CP]*[PD] [AP]= 3 [PB]= x+5 [CP]= 4 [PD]= 2x-5 ⇒ 3*(x+5)= 4*(2x-5) 3x+15= 8x-20 5x= 35 x= 7 [AB]= [AP]+[PB] [AB]= 3+7+5 [AB]= 15 length units [CD]= [CP]+[PD] [CD]= 4+(2*7-5) [CD]= 13 length units I) Let's consider the triangle ΔAOB: [AO]= [OB]= r the height of this isosceles triangle is y and would divide the base [AB] in two equal parts: [EO]= y [AE]= [EB] = 15/2 E ∈ [AB] if we apply the Pythagorean theorem for the ΔOBE: [EO]²+[BE]²= [OB]² [OB]= r [EO]= y [BE]= 15/2 ⇒ y²+(15/2)²= r² II) if we consider the second isosceles triangle ΔCDO: [OC]= [DO]= r F ∈ [CD] [CF]= [FD]= 13/2 [PF]=[EO] [PF]= y ⇒ 4+y= [FD] 4+y= 13/2 y= 13/2 - 4 y= 5/2 if we replace the y value in the Pythagorean theorem above, we get: y²+(15/2)²= r² (5/2)²+225/4= r² r²= 250/4 r= √250/2 r= √5²*10/2 r= 5√10/2 r≈ 7,9057 length units
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After finding x = 7, you can use formula for perpendicular chords
r2=(x2+y2+z2+w2)/4
So
r2=(9+16+81+144)/4
r2=250/4
r=5√10/2
Можно найти радиус по формуле R=a*b*c/(4*S) для ∆DCB. Стороны находим по теореме Пифагора, S=0.5 * DC*PB=78, CB=4√10, DB=15, R=14*4√10*15/(4*78)=2,5√10.
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merci beaucoup sire excellente exercice
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AP = 3 (in triangle APC)
Then 3.(x + 5) = 4.(2.x -5) (chords th.), giving that x = 7. So AP = 3, BP = 12, DP = 9, CP = 4
We use then an orthonormal center P and first axis (PB)
We have A(-3; 0) B(12, 0) C( 0; 4) D(0; -9)
The equation of the circle is x^2 + y^2 + a;x + b.y + c = 0
A and B are on the circle, so 9 -3.a + c = 0 and 144 + 12.a +c = 0, giving c = - 36 and a = -9.
C is on the circle, so 16 + 4.b - 36 = 0, giving that b = 5.
The equation of the circle is x^2 + y^2 -9.x +5.y -36 = 0, or (x -9/2)^2 + (y +5/2)^2 = 250/4
So the radius of the circle is sqrt(250/4) =
(5/2).sqrt(10).
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(x+5)3=(2x-5)4
x= 7; PB= 12; PD= 9;
BD= 15; AD= 3√9.
Let's connect points A and B with points D and O.
1. Use intersecting chords theorem to find x = 7.
2. Use Pythagoras theorem to find BC = 4sqrt(10) and BD = 15 while BP = 12.
3. Use area formula (area = 1/2 base x height) to find area of triangle BCD = (1/2)(13)(12) = 78.
4. Use circumradius formula for triangle BCD = abc/4xarea = (13)(15)(4sqrt10)/(4)(78) = (5sqrt10)/2.
El triángulo APC es rectangular→ Si AC=5 y PC=4→ AP=3→ Potencia de P respecto a la circunferencia =3(X+5)=4(2X-5)→ X=7→ AB=3+12=15 ; CD=4+9=13→ Si Q es la proyección ortogonal de P sobre el diámetro horizontal→ En el triángulo OQC: OQ²+QC²=OC²=r²→ (9/2)²+(13/2)²=r²→ r=(5√10)/2 ud.
Gracias y saludos.
After finding the value of x, I solved the problem by rotating the figure so that AB coincides with the X-axis and CD with the Y-axis and point P at the origin. Now Boeing the 4 points are at the circumference of a circle, I proceeded to solve for the equation of that circle.
In ∆APC
AP^2+CP^2=AC^2
AP^2+4^2=5^2
AP=√5^2-4^2=3
(AP)(BP)=(CP)(DP)
3(x+5)=4(2x-5)
So x=7
BP=7+5=12
DP=2(7)-5=14-5=9
Connect O to E (E middle AB and OE right AB )
AE=BE=15/2
Connect O to F (F middle CD and right CD)
CF=DF=13/2
PF=9-6.5=2.5
Connect O to B
In ∆ OBE
OE^2+BE^2=OB^2
(2.5)^2+(7.5)^2=r^2
So r=5√10/2 units=7.91.units.❤❤❤
Teorema delle corde..4:3=(x+5):((2x-5)...8x-20=3x+15..x=7..il triangolo inscritto AOB ha lati 15,√90,15..r=abc/4A=225*√90/4*67,5=225√90/270=5√90/6=5√10/2
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PA = 3, due to 3,4,5.
3(x+5) = 4(2x-5)
3x + 15 = 8x - 20
3x = 8x - 35
x = 7
Make the midpoint of AB, M.
AB = 7+5+3 = 15
Move AB down to lie across the diameter. Either side of the other chord, it now has lengths of 12+a and 3+a.
It bisects the other chord (DC) which is 14 - 5 + 4 long, so 13/2 on each side of the newly positioned AB. Although AB is 15+2a long, is is also 2r. We now need to find the length of a.
(a+12)(a+3) = (13/2)(13/2) will find length a.
a^2 + 15a + 36 = 169/4
4a^2 + 60a + 144 = 169
4a^2 + 60a - 25 = 0
(-60+or-sqrt(3600 - 4*4*-25))/8 = a
(-60+or-sqrt(4000))/8 = a
(-60+or-4*sqrt(250))/8 = a
Discard the minus result as it will be a negative value.
(-15+sqrt(250))/2 = a
(-15+5*sqrt(10))/2 = a
a = (5*sqrt(10) - 15)/2 = 0.4057 (rounded).
the diameter is 15+2a, so r is 7.5+a.
r = 7.9057 (rounded)
Just checked the video. Different paths, same conclusion.
Basta usar esse teorema quando as cordas são perpendiculares:
CP²+DP²+AP²+PB²=4R², onde R é o raio do círculo.
1/ Note that the two triangles APC and DPB are 3/4/5 triplets.
2/ x=7-> PB=12 ->PD=9 and BD=15
Because AB= 15 too, so the triangle ABD is an isosceles one, so:
Just draw the diameter BB’, -> BB’ is the perpendicular bisector of the chord AD.
We have:
sqAD=sqAP+sqBD=9+81=90
AD=3sqrt10
Label the angle ABB’=alpha
We have: Angle ADP=angleABB’=alpha( having sides perpendicular)
-> cos alpha=PD/AD=3/sqrt10
Focus on the triangle B’AB
AB/BB’= cos alpha-> BB’=AB/cos alpha
2R=15/(3sqrt10)-> R=5sqrt10/2= 7.91 untd😅😅😅
7.91
AP = 3 (3-4-5) right triangle
3(x+5) = 4(2x-5) intersection chord theorem
3x + 15 = 8x -20
35 = 5x
x =7
PB = 12 (7+5)
PD = 9
AB=15, hence the center = 7.5, but since AP = =3, its center is from CD , let label it RS = 4.5 (7.5 -3)
CD =13, hence the center = 6.5
Draw a line from the center of the CD to the circle's center; the distance is 4.5
Draw a line from C to O , this is the radius
r^2 = 4.5^2 + 6.5^2
r =7.91
I like these videos. They expose my Hubris that cost many exams 🙂
If in the circle we have 2 perpendicular across each other , the formula is 4r^2=the sum of Square of each side.i.e= 4^2+9^2+12^2+3^2==>r=7.90
Teorema de Faure 4R²=a²+b²+c²+d²
Alternative solution: Construct OD, which is also a radius and is the hypotenuse of ΔOFD. Length DF = 13/2 and length OF = 15/2 - 3 = 9/2. Applying the Pythagorean theorem, r² = (13/2)² + (9/2)² = 169/4 + 81/4 = 250/4 and, after simplifying, r = (5√(10))/2, as PreMath found from ΔOEB. Note that ΔOFC can also be used, but is congruent to ΔOFD. Similarly, ΔOEA is congruent to PreMath's ΔOEB.
Good evening sir
Start with the intersecting chord theorum. Do the algebra. X = 7. Thus AP is 3, PB is 12, CP is 4 , PD is 9. Next, square each element, then add them together. Comes out to 250. Then, divide by 4, get square root of quotient. Makes r = 7.0956. Can't remember the name of the formula or who discovered it.
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3(x + 5) = 4(2x - 5) → x = 7 → r = √((13/2)^2 + (9/2)^2) = 5√10/2
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Let's find the radius:
.
..
...
....
.....
The triangle ACP is a right triangle, so we can apply the Pythagorean theorem:
AP² + CP² = AC²
⇒ AP² = AC² − CP² = 5² − 4² = 25 − 16 = 9 ⇒ AP = 3
By using the intersecting chords theorem we obtain:
AP*BP = CP*DP
3*(x + 5) = 4*(2*x − 5)
3*x + 15 = 8*x − 20
35 = 5*x
⇒ x = 7
Now let's assume that O is the center of the coordinate system and that AB is parallel to the xaxis. Since CD is perpendicular to AB, CD must then be parallel to the y-axis. Therefore we can conclude:
xB = AB/2 = (AP + BP)/2 = [3 + (x + 5)]/2 = [3 + (7 + 5)]/2 = 15/2
yC = CD/2 = (CP + DP)/2 = [4 + (2*x − 5)]/2 = [4 + (2*7 − 5)]/2 = 13/2
yB = yC − CP = 13/2 − 4 = 13/2 − 8/2 = 5/2
xC = AP − xB = 3 − 15/2 = 6/2 − 15/2 = −9/2
Now we are able to calculate the radius R and to check the result:
R² = xB² + yB² = (15/2)² + (5/2)² = 225/4 + 25/4 = 250/4 ⇒ R = (5/2)√10
R² = xC² + yC² = (−9/2)² + (13/2)² = 81/4 + 169/4 = 250/4 ⇒ R = (5/2)√10 ✓
Best regards from Germany
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Thanks for sharing ❤️
The radius is 1/2[5*sqrt(10)]. I am glad that I did my own sanity check on my work. I think that you should create of platliat that make use of intersecting chirds theorem as well perpendicular busector theorems. Or playlists that involve a combination of at least two or three theorems.
Nice suggestion
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I am trying to be a mathematician. And your channel and but one of DOZENS that I am making use of!!!
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I need help here.
My way of solution ▶
We can find [AP] by using the Pythagoream theorem:
CA²= AP²+PC²
[PC]= 4
[CA]= 5
5²= AP²+4²
⇒
[AP]= 3
According to the intersecting chords theorem :
[AP]*[PB]= [CP]*[PD]
[AP]= 3
[PB]= x+5
[CP]= 4
[PD]= 2x-5
⇒
3*(x+5)= 4*(2x-5)
3x+15= 8x-20
5x= 35
x= 7
[AB]= [AP]+[PB]
[AB]= 3+7+5
[AB]= 15 length units
[CD]= [CP]+[PD]
[CD]= 4+(2*7-5)
[CD]= 13 length units
I) Let's consider the triangle ΔAOB:
[AO]= [OB]= r
the height of this isosceles triangle is y and would divide the base [AB] in two equal parts:
[EO]= y
[AE]= [EB] = 15/2
E ∈ [AB]
if we apply the Pythagorean theorem for the ΔOBE:
[EO]²+[BE]²= [OB]²
[OB]= r
[EO]= y
[BE]= 15/2
⇒
y²+(15/2)²= r²
II) if we consider the second isosceles triangle ΔCDO:
[OC]= [DO]= r
F ∈ [CD]
[CF]= [FD]= 13/2
[PF]=[EO]
[PF]= y
⇒
4+y= [FD]
4+y= 13/2
y= 13/2 - 4
y= 5/2
if we replace the y value in the Pythagorean theorem above, we get:
y²+(15/2)²= r²
(5/2)²+225/4= r²
r²= 250/4
r= √250/2
r= √5²*10/2
r= 5√10/2
r≈ 7,9057 length units
2 circle in a square
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AP^2 = 5^2 - 4^2 ; AP^2 = 25 - 16 ; AP^2 = 9 ; AP = 3
02) 3 * (x + 5) = 4 * (2x - 5) ; 3x + 15 = 8x - 20 ; 5x = 35 ; x = 35 / 5 ; x = 7
03) AB = 15
04) CD = 13
05) OD^2 = 4,5^2 + 6,5^2 ; OD^2 = 20,25 + 42,25 ; OD^2 = 62,5
06) OD = R
07) R = sqrt(62,5)
08) R ~ 7,905 lin un
OUR ANSWER : Radius equal sqrt(62,5) Linear Units or ~ 7,905 Linear Units.
X=7
Thanks ❤️
The figure is absolutely unclear.
New prescription maybe?