We use an orthonormal center O and first axis (OQ). The radius of the yellow semi circle is 5 (see triangle OQP), so its equation is x^2 + y^2 = 25. The equaton of the red circle is (x -3)^2 + (y -4)^2 = 16 or x^2 +y^2 - 6.x -8.y +9 = 0. We surch the intersection. By difference we have 6.x + 8.y -34 = 0 or y = (-3.x +17)/4, then we replace y by this value in x^2 + y^2 = 25 and obtain 25.x^2 -102.x -111 = 0. Deltaprime = 5376 = (2^8).21 So x = (51 -16.sqrt(21))/25 which is the abscissa of A, or x = (51 + 16.sqrt(21))/25 which is the abscissa of B We obtain the ordinates of these two points with y = (-3.x +17)/25. Finally: A((51 -16.sqrt(21))/25; (68 +12.sqrt(21))/25) and B((51 + 16.sqrt(21))/25; (68 - 12.sqrt(21))/25) Then VectorAB((32.sqrt(21))/25; (-24.sqrt(21))/25) = (8.sqrt(21))/25.Vector U with VectorU(4; -3) and norm(VectorU) = 5 Then AB = [(8.sqrt(21))/25].5 = (8/5).sqrt(21).
I am glad that I have learned of a geometry problem that is tricky if you do not know how to use clever geometry with chord length theorems!!! Also for yesterday's I have noticed that PreMath has hearted a ton of comments offering advanced explanations than the straightforward one given in yesterday's video. I think that PreMath could make a playlist of comments that showcase solutions that have been hearted. Maybe a compilation even!!!
Because OQ+ is a tangent tine to the small circle, PQO is a right triangle. PQ = 4 because it's r. OQ = 3 (given). OP = 5 because it is the hypotenuse of a 3,4,5. It it also R. Call AB 2x. Call its midpoint, M. It is a chord in both circles. In the red circle, intersecting chords give y(8 - y) = x^2 where y is the distance from M to the red circumference along a line MO. PM is 4 - y. OM is 5 - (4 - y) = y + 1. Imagine a full yellow circle. 2R = 10. Intersecting chords give (5 + y + 1)*(4 - y) = x^2 Simplify a bit for (6 + y)(4 - y) = x^2. In the red circle the chords are y(8 - y) = x^2. Therefore, (6 + y)(4 - y) = y(8 - y). Expand and tidy up: 24 - 2y - y^2 = 8y - y^2 24 - 2y = 8y. 24 = 10y. y = 2.4 or 12/5 if preferred. Red circle is now (12/5)*(8 - (12/5)) = x^2. Tidy up to (12/5)(28/5) = x^2. (336/25) = x^2 sqrt(336)/(sqrt(25)) = x, so sqrt(336)/5. As AB = 2x, AB = 2*sqrt(336)/5. In decimal, that approximates to 7.33. I now looked. Yes, our labelling differed, as you might expect, and I threw in a couple of extraneous calculations, but the essentials were all there.
1/ Focus on the triangle AOP of which the perimeter = 5+5+4= 14 so, by Heron theorem: the area of triangle APO = sqrt( 7x2x2x3)=2sqrt21 -> 1/2 AC.OP= AC.5/2= 2sqrt21-> AC= 4sqrt21/5 -- > AB = 8 sqrt21/5😅😅😅
OQ=3 ; QP=4---> OP=5 ---> Cuerda AB=2c; M es su punto medio y "s" la flecha ---> Potencia de M respecto a la circunferencia roja =(4+s)(4-s)=16-s² =c²= Potencia de M respecto a la circunferencia amarilla =s(2*5 -s)=10s-s² = 16-s²---> s=16/10=8/5---> c²=16-s²= 16-(8/5)² ---> c=4√21/5---> 2c=8√21/5 =AB. Gracias y saludos.
First I also found that pythagorean 3,4,5 triple. Then I divided the isosceles 5,5,4 triangle OPB into two equal right triangles with Hypothenuse 5 and one side 2. Then angle alpha = asn(2/5). Then using y/5 = sin (2 * alpha) leads to length of chord AB ≈ 7.33 Thanks for sharing this nice geometry question. Wish you a happy Sunday 😊
I'm pleased to see many different approaches towards this problem. Let me put mine: Well, AB is the common chord to both circles of radii 5 and 4. We know the formulla, squared length of a chord, AB**2=4(r**2-d**2) We know r1 and r2 =(5,4) . We also know d1+d2=5, hence we can determine d1/and/or/d2. By putting it back in AB**2 formulla we gan get the AB.
In circle Point O Let R is the Radius of cirmicircle PQ=4 In ∆ OPQ OQ^2+PQ^2=OP^2 3^2+4^2=QP^2 So PQ=5 (4+x)((4-x)=(OA)OB) (1) x(10-x)=(OA)(OB) (2) (1)&(2) 16-x^2=10x-x^2 So x=16/10=8/5 M middle AB MP^2+OA^2=AP^2 (8/5)^2+OA^2=4^2 So OA=4√21/5 So AB=2(4√21/5)=8√21/5 units=7,33units.❤❤❤
As PQ is a radius of circle P, PQ = 4. As PQ = 4 and OQ = 3, ∆OQP is a 3:4:5 Pythagorean triple right triangle and OP = 5. As OP is a radius of semicircle O, OA = OB = OP = 5. Let ∠POB = θ. As AB is a chord for both semicircle O and circle P, and as a line that bisects a chord perpendicularly must be collinear with the center of the circle, OP bisects AB and thus ∠AOP = ∠POB = θ. By the law of cosines: cos(θ) = (OB²+OP²-PB²)/2(OB(OP) cos(θ) = (5²+5²-4²)/(2(5²)) cos(θ) = (25+25-16)/50 = 34/50 = 17/25 cos(2θ) = 2cos²(θ) - 1 cos(2θ) = 2(17/25)² - 1 cos(2θ) = 2(289/625) - 1 = 578/625 - 1 cos(2θ) = -47/625 AB² = OA² + OB² - 2OA(OB)cos(2θ) AB² = 5² + 5² - 2(5²)(-47/625) AB² = 25 + 25 + 50(47/625) AB² = 50 + 94/25 = (1250+94)/25 AB² = 1344/25 AB = √(1344/25) = (8√21)/5 ≈ 7.33 units
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle." Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
Построим треугольник OPQ, PQ=4, OQ=3, OP=5, соединим Р с В, получим два треугольника ОСВ и РСВ, у них общий катет , равный АВ/2, который мы ищем. обозначим СР как х и составим два уравнения (АВ/2)^2=16-x^2, (AB/2)^2=25-(5-x)^2. Приравняем правые части двух уравнений, возведём в квадрат слагаемое в скобках, и получим, что х=16/10. Теперь найдём АВ, которое найдём, подставив в любое из двух уравнений значение х. АВ=(8\/21)/5=7,33212111192...
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle." Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
My way of solution ▶ The tangent [OB] is perpendicular to radius of the red circle: 90° By considering the right trinagle Δ(POQ) [OQ]= 3 [QP] is equal to the radius of the red trinagle: [QP]= r [QP]= 4 ⇒ according to the Pythagorean theorem we can write: [OQ]²+[QP]²= [PO]² 3²+4²= [PO]² [PO]= 5 ii) there is a point between the intersection of [PO] and [AB], let's call it "R" [OP] will divide the length [AB] in two equal parts: [AB]/2 = y iii) for the red triangle, by writing the intersecting chords theorem: [AR]*[RB]= [RS]*[RT] [PS]= [PT]= r [PS]= [PT]= 4 [PR]= x [RS]= 4-x [AR]=[RB]= y ⇒ y²= (4+x)*(4-x) y²= 16-x².............Eq-1 iv) By considering the right triangle ΔAOR: [AO]= R [AO]= 5 [OR]= 5-x [RA]= y by applying the Pythagorean theorem: [AO]²= [OR]²+[RA]² 5²= (5-x)² + y² 25= 25-10x+x²+y² x²-10x+y²=0.........Eq-2 we know that: y= 16-x² ⇒ x²-10x+16-x²=0 10x= 16 x= 8/5 y²= 16-x² y²= 16- (8/5)² y= √336/25 y= 4√21/5 [AB]=2y [AB]= 8√21/5 [AB]≈ 7,332 length units
Good question, it looks like PreMath drew the perpendicular symbol without proving AB really is perpendicular to OP. As @DaRealNoobKing says, triangle AOP is congruent to POB. That leads to proving APC is congruent to BPC*. That proves that angles ACP and BCP (which add to 180) are equal, so they are right angles. Using this proof you don't even need to show that AB is perpendicular to OP as AC being equal to BC was proven along the way*.
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle." Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
Well I may not have it right but gave it a go. I think AB = 7.332. I figured big the big radius to be 5 and drew a parallelagram with sides 5, 5, 4, and 4 and presumed that line OP splits AB exactly in 2. Thats where the big question mark is but it seems right. And based on optics, I knew the answer had to be close to √(50).
@@alastairjames8723 Agree, for both APO and BPO have AO=BO and OC the same for both but this doesn't mean that AC is equal to CB. I asked my question as the author didn't clarify this
STEP-BY-STEP RESOLUTION PROPOSAL : 01) OP = R 02) OQ = 3 03) PQ = 4 04) R^2 = 3^2 +4^2 ; R^2 = 9 + 16 ; R^2 = 25 ; R = sqrt(25) ; R = 5 05) Now we know that ; PA = PB = 4 lin un and OA = OB = 5 lin un 06) OP = R = 5 lin un 07) Now I can built a Kite : Quadrilateral [OAPB] with External Sides equal to 4 (PA ; PB) and 5 (OB ; OA) lin un. 08) Notice that OP = 5 lin un 09) Using Heron's Formula I can get the Area of an Isosceles Triangle [OPB] (Sides = (5 ; 5 ; 4)): A = 9,165 sq un 10) Now, dividing : (2 * 9,165) / 5, I get the Distance from the Middle Point of AB (M) to B. 11) MB = 18,33 / 5 = 3,666 12) AB = 2 * 3,666 ; AB = 7,332 lin un Therefore, OUR BEST ANSWER IS : In an Euclidian Affine Space, Line AB must be equal to 7,332 Linear Units.
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle." Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
Thanks. I love it. Challenging❤
Glad to hear that!
You are very welcome!
Thanks for the feedback ❤️
OP = 5 as calculated from 3,4,5 triangle.
Triangle APO with height y.
y^2 = 4^2 - x^2.
Also, y^2 = 5^2 - (5 - x)^2.
Therefore 16 - x^2 = 25 - 25 +10x - x^2.
16 = 10x.
x = 1.6
Then y^2 = 4^2 - 1.6^2 in triangle ACP.
y^2 = 13.44.
y = 3.666.
AB = 2y = 7.332.
🎉
Thank you!
We use an orthonormal center O and first axis (OQ). The radius of the yellow semi circle is 5 (see triangle OQP), so its equation is x^2 + y^2 = 25.
The equaton of the red circle is (x -3)^2 + (y -4)^2 = 16 or x^2 +y^2 - 6.x -8.y +9 = 0. We surch the intersection. By difference we have 6.x + 8.y -34 = 0
or y = (-3.x +17)/4, then we replace y by this value in x^2 + y^2 = 25 and obtain 25.x^2 -102.x -111 = 0. Deltaprime = 5376 = (2^8).21
So x = (51 -16.sqrt(21))/25 which is the abscissa of A, or x = (51 + 16.sqrt(21))/25 which is the abscissa of B
We obtain the ordinates of these two points with y = (-3.x +17)/25.
Finally: A((51 -16.sqrt(21))/25; (68 +12.sqrt(21))/25) and B((51 + 16.sqrt(21))/25; (68 - 12.sqrt(21))/25)
Then VectorAB((32.sqrt(21))/25; (-24.sqrt(21))/25) = (8.sqrt(21))/25.Vector U with VectorU(4; -3) and norm(VectorU) = 5
Then AB = [(8.sqrt(21))/25].5 = (8/5).sqrt(21).
Merci beaucoup pour votre effort
شكرا لكم على المجهودات
Pytagorean theorem:
R² = 3²+r² = 3²+4² --> R= 5cm
Cosine law for isosceles triangle AOP:
r² = 2R²(1-cosα)
cosα = 1- r²/ 2R² = 1-4²/(2*5²)
α = 47,15636°
Chord AB:
c = 2 R sinα = 7,332 cm (Solved √)
Pytagorean theorem:
R² = 3²+r² = 3²+4² --> R= 5cm
Cosine law for isosceles triangle AOP:
r² = 2R²(1-cosα)
cosα = 1- r²/ 2R² = 1-4²/(2*5²)
α = 47,15636°
β = ½(180°-α) = 66,42182°
Chord AB:
c = 2 R sinα = 7,332 cm
c = 2 r sinβ = 7,332 cm
I am glad that I have learned of a geometry problem that is tricky if you do not know how to use clever geometry with chord length theorems!!! Also for yesterday's I have noticed that PreMath has hearted a ton of comments offering advanced explanations than the straightforward one given in yesterday's video. I think that PreMath could make a playlist of comments that showcase solutions that have been hearted. Maybe a compilation even!!!
Love it!!!!!!!!!!
Señor profesor mil gracias por su dedicación y tiempo.
Buen ejercicio
Very tricky question.
Because OQ+ is a tangent tine to the small circle, PQO is a right triangle.
PQ = 4 because it's r.
OQ = 3 (given).
OP = 5 because it is the hypotenuse of a 3,4,5. It it also R.
Call AB 2x. Call its midpoint, M.
It is a chord in both circles.
In the red circle, intersecting chords give y(8 - y) = x^2 where y is the distance from M to the red circumference along a line MO.
PM is 4 - y.
OM is 5 - (4 - y) = y + 1.
Imagine a full yellow circle.
2R = 10.
Intersecting chords give (5 + y + 1)*(4 - y) = x^2
Simplify a bit for (6 + y)(4 - y) = x^2.
In the red circle the chords are y(8 - y) = x^2.
Therefore, (6 + y)(4 - y) = y(8 - y).
Expand and tidy up:
24 - 2y - y^2 = 8y - y^2
24 - 2y = 8y.
24 = 10y.
y = 2.4 or 12/5 if preferred.
Red circle is now (12/5)*(8 - (12/5)) = x^2.
Tidy up to (12/5)(28/5) = x^2.
(336/25) = x^2
sqrt(336)/(sqrt(25)) = x, so sqrt(336)/5.
As AB = 2x, AB = 2*sqrt(336)/5.
In decimal, that approximates to 7.33.
I now looked. Yes, our labelling differed, as you might expect, and I threw in a couple of extraneous calculations, but the essentials were all there.
1/ Focus on the triangle AOP of which the perimeter = 5+5+4= 14
so, by Heron theorem: the area of triangle APO = sqrt( 7x2x2x3)=2sqrt21
-> 1/2 AC.OP= AC.5/2= 2sqrt21-> AC= 4sqrt21/5
-- > AB = 8 sqrt21/5😅😅😅
It's great that you saw this solution! 👏🏻👏🏻👏🏻
@@imetroangola17 Thank you so much!
I did it the same way !
OQ=3 ; QP=4---> OP=5 ---> Cuerda AB=2c; M es su punto medio y "s" la flecha ---> Potencia de M respecto a la circunferencia roja =(4+s)(4-s)=16-s² =c²= Potencia de M respecto a la circunferencia amarilla =s(2*5 -s)=10s-s² = 16-s²---> s=16/10=8/5---> c²=16-s²= 16-(8/5)² ---> c=4√21/5---> 2c=8√21/5 =AB.
Gracias y saludos.
First I also found that pythagorean 3,4,5 triple.
Then I divided the isosceles 5,5,4 triangle OPB into two equal right triangles with Hypothenuse 5 and one side 2.
Then angle alpha = asn(2/5).
Then using y/5 = sin (2 * alpha) leads to length of chord AB ≈ 7.33
Thanks for sharing this nice geometry question.
Wish you a happy Sunday 😊
I'm pleased to see many different approaches towards this problem. Let me put mine:
Well, AB is the common chord to both circles of radii 5 and 4.
We know the formulla, squared length of a chord, AB**2=4(r**2-d**2)
We know r1 and r2 =(5,4) . We also know d1+d2=5, hence we can determine d1/and/or/d2.
By putting it back in AB**2 formulla we gan get the AB.
In circle Point O
Let R is the Radius of cirmicircle
PQ=4
In ∆ OPQ
OQ^2+PQ^2=OP^2
3^2+4^2=QP^2
So PQ=5
(4+x)((4-x)=(OA)OB) (1)
x(10-x)=(OA)(OB) (2)
(1)&(2)
16-x^2=10x-x^2
So x=16/10=8/5
M middle AB
MP^2+OA^2=AP^2
(8/5)^2+OA^2=4^2
So OA=4√21/5
So AB=2(4√21/5)=8√21/5 units=7,33units.❤❤❤
As PQ is a radius of circle P, PQ = 4. As PQ = 4 and OQ = 3, ∆OQP is a 3:4:5 Pythagorean triple right triangle and OP = 5. As OP is a radius of semicircle O, OA = OB = OP = 5. Let ∠POB = θ. As AB is a chord for both semicircle O and circle P, and as a line that bisects a chord perpendicularly must be collinear with the center of the circle, OP bisects AB and thus ∠AOP = ∠POB = θ.
By the law of cosines:
cos(θ) = (OB²+OP²-PB²)/2(OB(OP)
cos(θ) = (5²+5²-4²)/(2(5²))
cos(θ) = (25+25-16)/50 = 34/50 = 17/25
cos(2θ) = 2cos²(θ) - 1
cos(2θ) = 2(17/25)² - 1
cos(2θ) = 2(289/625) - 1 = 578/625 - 1
cos(2θ) = -47/625
AB² = OA² + OB² - 2OA(OB)cos(2θ)
AB² = 5² + 5² - 2(5²)(-47/625)
AB² = 25 + 25 + 50(47/625)
AB² = 50 + 94/25 = (1250+94)/25
AB² = 1344/25
AB = √(1344/25) = (8√21)/5 ≈ 7.33 units
How do we know that the chord is perpendicular to OP?
I have the same doubt
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle."
Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
Построим треугольник OPQ, PQ=4, OQ=3, OP=5, соединим Р с В, получим два треугольника ОСВ и РСВ, у них общий катет , равный АВ/2, который мы ищем. обозначим СР как х и составим два уравнения (АВ/2)^2=16-x^2, (AB/2)^2=25-(5-x)^2. Приравняем правые части двух уравнений, возведём в квадрат слагаемое в скобках, и получим, что х=16/10. Теперь найдём АВ, которое найдём, подставив в любое из двух уравнений значение х. АВ=(8\/21)/5=7,33212111192...
R=5..AB=2*4*sinarccos(2/5)=8√(1-4/25)=8√21/5
How AB and OP are perpendicular?
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle."
Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
My way of solution ▶
The tangent [OB] is perpendicular to radius of the red circle: 90°
By considering the right trinagle Δ(POQ)
[OQ]= 3
[QP] is equal to the radius of the red trinagle:
[QP]= r
[QP]= 4
⇒
according to the Pythagorean theorem we can write:
[OQ]²+[QP]²= [PO]²
3²+4²= [PO]²
[PO]= 5
ii) there is a point between the intersection of [PO] and [AB], let's call it "R"
[OP] will divide the length [AB] in two equal parts:
[AB]/2 = y
iii) for the red triangle, by writing the intersecting chords theorem:
[AR]*[RB]= [RS]*[RT]
[PS]= [PT]= r
[PS]= [PT]= 4
[PR]= x
[RS]= 4-x
[AR]=[RB]= y
⇒
y²= (4+x)*(4-x)
y²= 16-x².............Eq-1
iv) By considering the right triangle ΔAOR:
[AO]= R
[AO]= 5
[OR]= 5-x
[RA]= y
by applying the Pythagorean theorem:
[AO]²= [OR]²+[RA]²
5²= (5-x)² + y²
25= 25-10x+x²+y²
x²-10x+y²=0.........Eq-2
we know that:
y= 16-x²
⇒
x²-10x+16-x²=0
10x= 16
x= 8/5
y²= 16-x²
y²= 16- (8/5)²
y= √336/25
y= 4√21/5
[AB]=2y
[AB]= 8√21/5
[AB]≈ 7,332 length units
@ 3:10, how do we know AB is perpendicular to OP?
because AOP and POB are congruent because their side lengths are R, R and r and they share PO
Good question, it looks like PreMath drew the perpendicular symbol without proving AB really is perpendicular to OP. As @DaRealNoobKing says, triangle AOP is congruent to POB. That leads to proving APC is congruent to BPC*. That proves that angles ACP and BCP (which add to 180) are equal, so they are right angles. Using this proof you don't even need to show that AB is perpendicular to OP as AC being equal to BC was proven along the way*.
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle."
Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
Well I may not have it right but gave it a go. I think AB = 7.332. I figured big the big radius to be 5 and drew a parallelagram with sides 5, 5, 4, and 4 and presumed that line OP splits AB exactly in 2. Thats where the big question mark is but it seems right. And based on optics, I knew the answer had to be close to √(50).
Let's face this challenge:
.
..
...
....
.....
Since OQ is a tangent to the circle, we know that the triangle OPQ is a right triangle. So we can apply the Pythagorean theorem:
OP² = OQ² + PQ² = 3² + 4² = 9 + 16 = 25 ⇒ OP = √25 = 5
Now let's assume that O is the center of the coordinate system and that OQ is located on the x-axis. Then we obtain the following coordinates:
O: ( 0 ; 0 )
A: ( xA ; yA )
B: ( xB ; yB )
P: ( 3 ; 4 )
Q: ( 3 ; 0 )
Since A and B are located on the semicircle and the circle, we can conclude:
(x − xO)² + (y − yO)² = OP²
(x − xP)² + (y − yP)² = PQ²
(x − 0)² + (y − 0)² = 5²
(x − 3)² + (y − 4)² = 4²
x² + y² = 25
x² − 6*x + 9 + y² − 8*y + 16 = 16
x² + y² = 25
x² + y² − 6*x − 8*y = −9
6*x + 8*y = 34
6*x = 34 − 8*y
⇒ x = 34/6 − 8*y/6 = 17/3 − (4/3)*y
x² + y² = 25
[17/3 − (4/3)*y]² + y² = 25
289/9 − (136/9)*y + (16/9)*y² + y² = 25
289 − 136*y + 16*y² + 9*y² = 225
25*y² − 136*y + 64 = 0
y = [136 ± √(136² − 4*25*64)]/(2*25)
y = [136 ± √(18496 − 6400)]/50
y = (136 ± √12096)/50
Since yA > yB, we can conclude:
yA = (136 + √12096)/50
yB = (136 − √12096)/50
Now we are able to calculate the length of AB:
AB²
= (xB − xA)² + (yB − yA)²
= {[17/3 − (4/3)*yB] − [17/3 − (4/3)*yA]}² + (yB − yA)²
= [17/3 − (4/3)*yB − 17/3 + (4/3)*yA]² + (yB − yA)²
= [(4/3)*yA − (4/3)*yB]² + (yB − yA)²
= (16/9)*(yA − yB)² + (yB − yA)²
= (16/9)*(yA − yB)² + (yA − yB)²
= (16/9)*(yA − yB)² + (9/9)*(yA − yB)²
= (25/9)*(yA − yB)²
= (25/9)*[(136 + √12096)/50 − (136 − √12096)/50]²
= (25/9)*(136/50 + √12096/50 − 136/50 + √12096/50)²
= (25/9)*(√12096/25)²
= (25/9)*(12096/625)
= 1344/25
⇒ AB = √(1344/25) = 8√21/5 ≈ 7.332
Best regards from Germany
Wow!!
Interesting, why is a radius OP perpendicular to a chord AB?
I was wondering the same thing. It seemed like circular reasoning especially since they always disclsim that this may not be drawn to scale
Why not? You can always rotate radius OP anyway you want to make it perpendicular to AB
@@rey-dq3nx Sure, but P is fixed on a circumference and we cant know an angle between AB and OP
I think APO and BPO are congruent as they both have the same three side lengths and they are reflected about PO
@@alastairjames8723 Agree, for both APO and BPO have AO=BO and OC the same for both but this doesn't mean that AC is equal to CB. I asked my question as the author didn't clarify this
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OP = R
02) OQ = 3
03) PQ = 4
04) R^2 = 3^2 +4^2 ; R^2 = 9 + 16 ; R^2 = 25 ; R = sqrt(25) ; R = 5
05) Now we know that ; PA = PB = 4 lin un and OA = OB = 5 lin un
06) OP = R = 5 lin un
07) Now I can built a Kite : Quadrilateral [OAPB] with External Sides equal to 4 (PA ; PB) and 5 (OB ; OA) lin un.
08) Notice that OP = 5 lin un
09) Using Heron's Formula I can get the Area of an Isosceles Triangle [OPB] (Sides = (5 ; 5 ; 4)): A = 9,165 sq un
10) Now, dividing : (2 * 9,165) / 5, I get the Distance from the Middle Point of AB (M) to B.
11) MB = 18,33 / 5 = 3,666
12) AB = 2 * 3,666 ; AB = 7,332 lin un
Therefore,
OUR BEST ANSWER IS :
In an Euclidian Affine Space, Line AB must be equal to 7,332 Linear Units.
Is it right??
OQ=3, QP=4,then OP is obviously 5
Then OB =5, OA= 5 also
Then AB will be. √50 or 5√2
Sir plz answer..... 🙋🏻♀️🙏🏻
Why? What is the math behind it?
@@AllmondISP is it actually right way or wrong ?? Cause 5√2 or 7.07🙃
@@AllmondISP like pythagorean theory
@@JinnirasFlairit's seems wrong.
how did you got to 5 sqrt(2) exactly? What are the numbers you used on your pythagoryean theory?
AB = 5√2 only if ∠AOB = 90°. In this case, ∠AOB is not 90°.
Who could explain why the Radius "OP" is perpendicular to the chord "AB" please....?
Perpendicular bisector theorem of circle says that "The perpendicular bisector of any chord of a circle will pass through the center of the circle."
Now the OP is the common Radius and AB is the common chord. So they OP bisects AB and perpendicular to each other.
😢