My solution was this: knowing that in a cyclic quadrilareral the sum of the opposite angles is 180°, I've found diagonal AC with cosine rule on ABC and ADC triangles, then compared the identities, and it is: AC = √ 247/7 than knowing that cos α = 3/7 I've found sin α = 2/7√ 10 In the sines law we know that every ratio between sides and sines of opposite angles are equal and the ratio is the diameter of the circumscribed circle, so r = (√ 247/7)/(2/7√ 10)*1/2
I solved the problem using the law of cosines and then the law of sines on an inscribed triangle. This is certainly an UPHILL walk (I'm strong in Algebra & Trigonometry but mediocre in Geometry), but it is very worth it. But I'll stick with the fundamental theorems for now and leave the advanced ones for later.
With O the centre of the circle OA= OB =OC =OD = radius r So drawing these we have four isosceles triangles. Each of them will be cut into two equal halves by lines from the midpoints to the centre. so there are eight smaller triangles with right angles . Letting the lengths of these heights be e, f, g, and h (shortest first etc.) We have r^2 = 3^2 + e^2 = 25/4 + f^2 = 4 +g^2 = 9/4 +g^2 ( Estimating.. The diameter of the circle will not differ a great amount from 13/2 so the radius will be just a bit more than 3) These eight triangles can be joined into four cyclic quadrilaterals (pretty!) with each radius equally half the radius of the first circle, because of the right angles that were made. Angle AOC is twice angle ADC , and three more statements like this about angles at the circumference and angles at the centre could be made. More useful ? The sine of Angle B = The sine of Angle D ( opposite angles in a cyclic quadrilateral) The sine of Angle A = The Sine of Angle C (oppostie angles in a cyclic quadrilateral) Area [ABCD] = sine A ( 3 . 6 + 4 . 5 )/2 = sine B ( 3 . 4 + 6 . 5 )/2 sine A / sine B = 21/ 19 = sine C / sine D Looks like I am one of the ninety nine percent , but my quadrilateral now has four cute little quads for company each approximately 1/4 of the original area.
Excellent! Now try to work out the net cost for a special woodruff key that is one quarter of an inch uniformly thick with its height to centre length of 1 inch. It is to be formed from a round bar of high tensile steel, whose cross sectional area = 25 pi squared inches. The costof the steel is 98 pence per cubic inch. Use Pythagous and simple trigonometry with area formulae to work out this.
If we draw a diagonal, it suffices to find an angle in one of the resulting triangles and the length of the opposite side. So draw AC. You have cos(ABC)=-cos(ADC) since the quadrilateral is cyclic. So call those x. Call AC y. So y^2=25-24x=61+60x from law of cosines. Now x=-36/84=-3/7 and y=sqrt(25+72/7)=sqrt(247/7). Finally the circumdiameter of either ABC or ADC is that divided by sqrt(40)/7, so sqrt(1729)/sqrt(40), and the circumradius is half that.
Let the radius of the circle be 'r' & AC = x Using the cosine formula - CosD = (6²+5²-x²)/2×6×5 = (61-x²)/60 CosB = (3²+4²-x²)/2×3×4 = (25-x²)/24 Since ABCD is inscribed in the circle B=180-D therefore CosB=Cos180-D= - CosD or (61-x²)/60=(x²-25)/24 7x²=247 or x²=247/7 CosD = (61- (247/7))/60= 3/7 Let the center of circle be O and OD divide angle D = k+l then CosD = Cos k+l =Cosk.Cosl - Sink.Sinl But since Cosk=5/2r & Cosl=6/2r, Sink=(1-(25/4r))^½ & Sinl=(1-(36/4r))^½ 3/7=30/4r² - 1/4r²[(4r²-25)(4r²-36)]^½ (105-6r²)²=49(4r⁴-61r²+15²) 36r⁴-1260r²=196r⁴-2989²r² r=¼(1729/10)^½
In General: Let BC = a, CD = b, AB = c, AD = d, BD = m, angle(DCB) = theta implies cos(theta) = (a^2 + b^2 - (c^2 + d^2))/(2 * (a * b + c * d)), m = sqrt(a^2 + b^2 - 2 * a * b * cos(theta)) and sin(theta) = sqrt(4 * (a * b + c * d)^2 - (a^2 + b^2 - (c^2 + d^2))^2)/(2 * (a * b + c * d)) implies R = m/(2 * sin(theta)).
As many have commented here, cosine law makes for a much less involved solution. It is preferable not to have to resort to so many arcane theorems about circles such as Brahmagupta's Law, Ptolemy's Theorem, etc. I solved using only cosine law and some algebra, AND the fact that the opposite angles inside a cyclic quadrilateral must add up to 180 degrees, AND the fact that the angle at the centre subtending a given arc is twice the angle subtended the same arc from any point on the circle circumference.
I have connected vertices with the center of the circle and recieved 4 iscoseles triangles in which central angles are proportional to the chords they stand on. Therefore, angles are 60,80,100 and 120 degrees. in a triangle with the 120 degrees angle two other angles are equal 30 degrees each and lateral sides are equal the Radius. We can now find the radius.
Name each vertex with a capital letter,for clarity reason and as everyone does, name the opposite side with the same letter but this time in lowercase . Please
Without rare formulas , with only the law of cosines: In ABD, the BAD angle is @ : BD^2=3^2 + 6^2 - 2*3*6*cos@= 45-36*cos@ In CBD, the BCD angle is 180-@, so : BD^2=4^2 + 5^2 - 2*4*5*cos (180-@)= 41+40*cos@ So, cos@ is 1/19, BD^2 = 819/19 = 9*7*13/ 19, BD^2 =~43, and , because BD^2 is smaller than AB^2+ AD^2=45, BAD angle is smaller than 90 degrees. If O is the center of the circle, BOD angle is twice of @, so cos(BOD) = cos(2@) = 2(cos@)^2 - 1 = - 359/19*19 In BOD triangle, BD^2 = r^2 + r^2 - 2*r*r*cos(BOD) 9*7*13/19 = 2*r^2 + 2*r^2*359/19*19, so 9*7*13/ 19 = 2*r^2 (1+359/19*19) , so 9*7*13/19*2 = r^2*(361+359)/(19*19), r^2 = 9*7*13*19 / 2*720= 7*13*19/ 16*10
The sum of opposite angles of cyclic quadrilateral is 180° Equating with cosine formula for ∆3,4,x and 5,6,x , we can get cos D= 3/7, Sin D = √40/7, x= √(247/7) ∆ACD=1/2 *6*5*sin D= 15√40/7 R=abc/4∆, R= 5*6*√(247/7) / 4*15√40/7 R= 1/4(√(1729/10)) ..Answer
Good day! Mr. presenter, this problem is solved finding the area of the quadrilateral given the lengths of its sides like this: (1). - From the formula for the area of a triangle given the lengths of two sides and the angle formed by them; the Law of Sines. and Ptolemy's theorem (relation between the sides and diagonals of a quadrilateral inscribed in a circle. (2).-- From the Heron/Brammaguta formula applied to find the area of the quadrilateral inscribed in the circle of radius R determined as P where P=[ABCD] of 1) i ) P = [ABC]+ [ACD] P= (1/2) ab sin(ABC+ + (1/2)sin (CDA) P=(ab/2)/(x/2R) + (cd/2)/(x/2R) So: P=(x/4R).(ab+cd) ... (I) ii) Furthermore S P=[ABD] +[BCD] analogously we obtain: P=(y/4R).(ad+cd) ...(II) from (I) and (II) we have to P²=(xy/16R²)(ab+cd)(ad+bc) of the quadrilateral inscribed in a circle theorem xy= ac+bd we have to: P²=(ab+cd)(ad+bc)(ac+bd). .(1/16R²) ... (3) of (2) in the quadrilateral: yes s=(a+b+c+d)/2 P²= (s-a)(s-b)(s-c)(s-d) ...(4) from (3) and (4) we obtain R. I wish you a good day!
I am from Brasil. Let me try first. Afterwards, I watch the video. Let the low point at left be A, the low oint of right be B, the upper point of right be C and the other be D. let AB=a=3, BC=b=4, CD=c=5 and DA=d=6. And Let d1=AC and d2=BD 2S(ABCD)= sin(D)*(cd+ab)=sin(A)*(ad+bc) (i) As the circumscribed circumference is the same for the triangles ABD and ACD we have that d1/sin(D)=d2/sin(A) So sin(D)/sin(A)=di/d2 (ii) (i) ==> sin(D)/sin(A)=(ad+bc)/(ab+cd) (iii) (ii) and (iii) ==> d1/d2= (ad+bc)/(ab+cd) multiplying by d1*d2 both sides we have: d1^2= (ad+bc)/(ab+cd)*d1d2 (iv) But as we kow d1*d2=(ac+bd) (v) (iv) and (v) d1^2=(ad+bc)*(ac+bd)/(ab+cd) (vi) By (vi) d1^2=(18+20)*(15+24)/(30+12) ...d1^2=19*13/7 (vii) (vii) and triangle ACD d1^2=19*13/7=5^2+36^2-2*5*6cos(D) ... cos(D)=180/420=3/7 ==> sin(D)=2*sqr(10)/7 (viii) (viii) and triangle ACD ==> 2R=d1/sen(D) 2R=sqr(19*13/7)*7/(2*sqr(10)) R=1/4*sqr(19*13*7/10)=1/4sqr(1729/10) I guess. The ansewr is the same as the video.
Not many people know these advanced theorems. Certainly not me. I would stick with the cosine rule. This is like solving a geometry problem on a triangle with trigonometry (a faster solution but misses the logic used to solve the problem), or like solving a difficult limits problem with L'Hospitale rule (may not be allowed at that point).
I solved it using a non-linear system of 6 equations (with 6 variables). Of course I had no idea how to solve this system of equations myself (I tried but gave up after many pages) so I entered it in Wolfram Alpha and amazingly it delivered the correct solutions in like one second. AI is starting to really scare me. Thanks for the simple and concise explanation. Ive studied math and never heard of any of these formulas before.
Yes, we can. First to find the cosine of angle A by considering triangle ABD and triangle CBD. Then use the law of Sine for triangles to find the diameter of the circumscribing circle of triangle ABD
I ask the professor to take the time from his busy schedule to kindly explain why the following method which produces a slightly different answer is incorrect. I can find nothing wrong. The entire class may learn something from the exercise. Make point “O” the center of the circle. Extend radii to points A, B, C, and D. Lines AD, DC, CB, and AB will generate central angles proportionate to their length: 6, 5, 4 and 3. Their sum is 18. Divide 360 by 18 to get 20. Then angle AOD = 6 x 20 = 120 degrees. Now draw a perpendicular bisector from O to “E,” the center of chord AD. That produces two 30, 60, 90 right triangles because the bisector splits the central angle 120 degrees, intersects the chord at a right angle, and the remaining angle must be 180-(60 + 90) = 30. Focus on right triangle OED. OD is the radius (hypotenuse), ED, the smallest side, = 3 because it is 1/2 of 6 split by the perpendicular bisector. Now applying the familiar 1, 2, sq rt of 3 ratios of the sides of 30, 60, 90 right triangles, 2/sq rt 3 = radius (OD) /3. A little algebra reveals the radius is 2 x square rt 3 = 3.464… Your answer is 3.287… What’s “off,” if anything?
Hi Monroe : Your error is almost certainly the assumption that "Lines AD, DC, CB, and AB will generate central angles proportionate to their length". That is, you are assuming here that in a given circle, the length of any chord is directly proportional to the actual angle subtended by the chord. This is false. Recall the "Sine Law" for triangles : The meaning of sinA/a = sinB/b is that the sides in any given triangle are proportional NOT to the vertex angles themselves, but to the SINES of the vertex angles. The same thing is true here with chords in a circle. The lengths of chords in any given circle are proportional not to the angles themselves, but to the SINES of the half-angles subtended by the chord.
i applied the cos formula for isosceles triangles (2 lengths are equal to r): 10 l1=3:l2=4:l3=5:l4=6:dim x(3),y(3),w(3):sw=.01:r=sw:goto 90 20 rl=2*r^2:sp=0:cw1=(rl-l1^2)/rl:if abs(cw1)>1 then sp=1:goto 80 30 cw2=(rl-l2^2)/rl:if abs(cw2)>1 then sp=1:goto 80 40 cw3=(rl-l3^2)/rl:if abs(cw3)>1 then sp=1:goto 80 50 cw4=(rl-l4^2)/rl:if abs(cw4)>1 then sp=1:goto 80 60 w(0)=acs(cw1):w(1)=acs(cw2):w(2)=acs(cw3):w(3)=acs(cw4) 70 dg=acs(cw1)+acs(cw2)+acs(cw3)+acs(cw4)-2*pi 80 return 90 gosub 20:if sp=1 then else 110 100 r=r+sw:goto 90 110 r1=r:dg1=dg:r=r+sw:if r>100*l1 then stop 120 r2=r:gosub 20:if dg1*dg>0 then 110 130 r=(r1+r2)/2:gosub 20:if dg1*dg>0 then r1=r else r2=r 140 if abs(dg)>1E-10 then 130 150 print r:mass=500/r:wa=rad(10)+w(0):for a=0 to 3:x(a)=r*cos(wa):y(a)=r*sin(wa) 160 wa=wa+w(a):next a:goto 180 170 xb=(x+r)*mass:yb=(y+r)*mass:return 180 x=x(0):y=y(0):gosub 170:xba=xb:yba=yb:for a=1 to 4:ia=a:if ia=4 then ia=0 190 x=x(ia):y=y(ia):gosub 170:xbn=xb:ybn=yb:line xba,yba,xbn,ybn:xba=xbn:yba=ybn 200 next a:x=0:y=0:gosub 170:circle xb,yb,r*mass 3.28728611 > run in bbc basic sdl and hit ctrl tab to copy
Is it not the case that the two halves of the quadrilateral form two triangles joined along a "hypotenuse" that contacts the circle's circumference at two points. It appears that the only solution for that length is 9, where both triangles collapse to a line segment, and we get it as both the sum of 3+6 and 4+5? Then there is no quadrilateral at all and no real way to decide if 9 is a diameter.
Sorry Alan, but your solution is gibberish. In this problem, the sides of the quadrilateral are fixed in length. "Collapsing" them by changing the angle between the sides does not correspond to the problem at hand at all. And, " dia = 9 " is definitely an incorrect solution.
@@GWaters-xr1fv Don't you think that opening your reply with "gibberish" is rather the work of a troll? To repeat, consider either triangle with the 3-6 sides or the 4-5 sides. Connect each with a line segment. Then postulate bringing these two triangles together along these lines such that they must be equal. Ignore the question of any circle within which they may or may not be contained, simply consider the constraint of equality of the connecting line. Call the angle between the 3-6 lines alpha and the angle between the 4-5 lines beta. Solve using the cosine law for the two opposite sides being of the same length. I fine both angles are pi radians and the two triangles have collapsed to line segments of 9 each. I never claimed that this would be a diameter. In fact, one could draw any old circle around this line segment.
According to the Problem, AB + AD = 3 + 6 = 9 and CB + CD = 4 + 5 = 9. BD divides the circle into two equal parts and BD will be the diameter of the circle. Triangle BAD has a right angle at A and triangle BCD has a right angle at C. According to the Pythagorean theorem : 1. BD^2 = AB^2 + AD^2 = 3^2 + 6^2 = 45 2. BD^2 = CB^2 + CD^2 = 4^2 + 5^2 = 41 R = 1/2BD Is the above problem correct? (ĐN)
I only try to check for me new formula for cyclic quadrilatefral.... not solve problem... You are right, cos(A) is not 4/5776 but 4/76... My fault... but I don't know how... I assume a copy paste problem with an unreliable mouse on my laptop... or something in this moment unknown... I apologize...
My solution was this:
knowing that in a cyclic quadrilareral the sum of the opposite angles is 180°, I've found diagonal AC with cosine rule on ABC and ADC triangles, then compared the identities, and it is:
AC = √ 247/7
than knowing that cos α = 3/7 I've found sin α = 2/7√ 10
In the sines law we know that every ratio between sides and sines of opposite angles are equal and the ratio is the diameter of the circumscribed circle, so
r = (√ 247/7)/(2/7√ 10)*1/2
I solved the problem using the law of cosines and then the law of sines on an inscribed triangle. This is certainly an UPHILL walk (I'm strong in Algebra & Trigonometry but mediocre in Geometry), but it is very worth it. But I'll stick with the fundamental theorems for now and leave the advanced ones for later.
Только так и следует решать.Ваше решение намного проще и короче авторского.
With O the centre of the circle OA= OB =OC =OD = radius r
So drawing these we have four isosceles triangles. Each of them will be cut into two equal halves by lines from the midpoints to the centre.
so there are eight smaller triangles with right angles . Letting the lengths of these heights be e, f, g, and h (shortest first etc.)
We have r^2 = 3^2 + e^2 = 25/4 + f^2 = 4 +g^2 = 9/4 +g^2 ( Estimating.. The diameter of the circle will not differ a great amount from 13/2
so the radius will be just a bit more than 3)
These eight triangles can be joined into four cyclic quadrilaterals (pretty!) with each radius equally half the radius of the first circle, because of the right angles that were made.
Angle AOC is twice angle ADC , and three more statements like this about angles at the circumference and angles at the centre could be made.
More useful ? The sine of Angle B = The sine of Angle D ( opposite angles in a cyclic quadrilateral)
The sine of Angle A = The Sine of Angle C (oppostie angles in a cyclic quadrilateral)
Area [ABCD] = sine A ( 3 . 6 + 4 . 5 )/2 = sine B ( 3 . 4 + 6 . 5 )/2
sine A / sine B = 21/ 19 = sine C / sine D
Looks like I am one of the ninety nine percent , but my quadrilateral now has four cute little quads for company each approximately 1/4 of the original area.
Excellent! Now try to work out the net cost for a special woodruff key that is one quarter of an inch uniformly thick with its height to centre length of 1 inch. It is to be formed from a round bar of high tensile steel, whose cross sectional area = 25 pi squared inches. The costof the steel is 98 pence per cubic inch. Use Pythagous and simple trigonometry with area formulae to work out this.
If we draw a diagonal, it suffices to find an angle in one of the resulting triangles and the length of the opposite side. So draw AC. You have cos(ABC)=-cos(ADC) since the quadrilateral is cyclic. So call those x. Call AC y. So y^2=25-24x=61+60x from law of cosines. Now x=-36/84=-3/7 and y=sqrt(25+72/7)=sqrt(247/7). Finally the circumdiameter of either ABC or ADC is that divided by sqrt(40)/7, so sqrt(1729)/sqrt(40), and the circumradius is half that.
Nice! φ = 30°; ∎ABCD → AB = 3; BC = 4; CD = 5; AD = 6; r = AO = BO = CO = DO = ?
BD = y; DAB = α → BCD = 6φ - α; y^2 = 45 - 36cos(α) ↔ cos(6φ - α) = -cos(α) ↔ y^2 = 41 + 40cos(α) →
cos(α) = 1/19 → y^2 = 45 - 36cos(α) = 819/19 → sin(α) = √((1 - cos^2(α))) = 6√10/19 →
cos(2α) = cos^2(α) - sin^2(α) = -359/361 → y^2 = 2r^2(1 - cos(2α)) = 819/19 → r = (1/40)√17290 ≈ 3,2873
Let the radius of the circle be 'r' & AC = x
Using the cosine formula -
CosD = (6²+5²-x²)/2×6×5 = (61-x²)/60
CosB = (3²+4²-x²)/2×3×4 = (25-x²)/24
Since ABCD is inscribed in the circle B=180-D therefore
CosB=Cos180-D= - CosD or
(61-x²)/60=(x²-25)/24
7x²=247 or x²=247/7
CosD = (61- (247/7))/60= 3/7
Let the center of circle be O and OD divide angle D = k+l then CosD = Cos k+l =Cosk.Cosl - Sink.Sinl
But since Cosk=5/2r & Cosl=6/2r, Sink=(1-(25/4r))^½ & Sinl=(1-(36/4r))^½
3/7=30/4r² - 1/4r²[(4r²-25)(4r²-36)]^½
(105-6r²)²=49(4r⁴-61r²+15²)
36r⁴-1260r²=196r⁴-2989²r²
r=¼(1729/10)^½
This is aright way
In General: Let BC = a, CD = b, AB = c, AD = d, BD = m, angle(DCB) = theta implies cos(theta) = (a^2 + b^2 - (c^2 + d^2))/(2 * (a * b + c * d)),
m = sqrt(a^2 + b^2 - 2 * a * b * cos(theta)) and sin(theta) = sqrt(4 * (a * b + c * d)^2 - (a^2 + b^2 - (c^2 + d^2))^2)/(2 * (a * b + c * d)) implies
R = m/(2 * sin(theta)).
As many have commented here, cosine law makes for a much less involved solution. It is preferable not to have to resort to so many arcane theorems about circles such as Brahmagupta's Law, Ptolemy's Theorem, etc. I solved using only cosine law and some algebra, AND the fact that the opposite angles inside a cyclic quadrilateral must add up to 180 degrees, AND the fact that the angle at the centre subtending a given arc is twice the angle subtended the same arc from any point on the circle circumference.
I have connected vertices with the center of the circle and recieved 4 iscoseles triangles in which central angles are proportional to the chords they stand on. Therefore, angles are 60,80,100 and 120 degrees. in a triangle with the 120 degrees angle two other angles are equal 30 degrees each and lateral sides are equal the Radius. We can now find the radius.
Name each vertex with a capital letter,for clarity reason and as everyone does, name the opposite side with the same letter but this time in lowercase . Please
Best geometry channel in youtube. I like you
Without rare formulas , with only the law of cosines:
In ABD, the BAD angle is @ : BD^2=3^2 + 6^2 - 2*3*6*cos@= 45-36*cos@
In CBD, the BCD angle is 180-@, so : BD^2=4^2 + 5^2 - 2*4*5*cos (180-@)= 41+40*cos@
So, cos@ is 1/19,
BD^2 = 819/19 = 9*7*13/ 19, BD^2 =~43, and , because BD^2 is smaller than AB^2+ AD^2=45, BAD angle is smaller than 90 degrees.
If O is the center of the circle, BOD angle is twice of @, so
cos(BOD) = cos(2@) = 2(cos@)^2 - 1 = - 359/19*19
In BOD triangle, BD^2 = r^2 + r^2 - 2*r*r*cos(BOD)
9*7*13/19 = 2*r^2 + 2*r^2*359/19*19, so
9*7*13/ 19 = 2*r^2 (1+359/19*19) , so 9*7*13/19*2 = r^2*(361+359)/(19*19),
r^2 = 9*7*13*19 / 2*720= 7*13*19/ 16*10
The sum of opposite angles of cyclic quadrilateral is 180°
Equating with cosine formula for ∆3,4,x and 5,6,x , we can get cos D= 3/7, Sin D = √40/7, x= √(247/7)
∆ACD=1/2 *6*5*sin D= 15√40/7
R=abc/4∆,
R= 5*6*√(247/7) / 4*15√40/7
R= 1/4(√(1729/10)) ..Answer
Good day!
Mr. presenter, this problem is solved
finding the area of the quadrilateral given the lengths of its sides like this:
(1). - From the formula for the area of a triangle given the lengths of two sides and the angle formed by them; the Law of Sines.
and Ptolemy's theorem (relation between the sides and diagonals of a quadrilateral inscribed in a circle.
(2).-- From the Heron/Brammaguta formula applied to find the area of the quadrilateral inscribed in the circle of radius R determined as P
where P=[ABCD]
of 1)
i ) P = [ABC]+ [ACD]
P= (1/2) ab sin(ABC+
+ (1/2)sin (CDA)
P=(ab/2)/(x/2R) +
(cd/2)/(x/2R)
So:
P=(x/4R).(ab+cd) ... (I)
ii) Furthermore S
P=[ABD] +[BCD]
analogously we obtain:
P=(y/4R).(ad+cd) ...(II)
from (I) and (II) we have to
P²=(xy/16R²)(ab+cd)(ad+bc)
of the quadrilateral inscribed in a circle theorem
xy= ac+bd
we have to:
P²=(ab+cd)(ad+bc)(ac+bd).
.(1/16R²) ... (3)
of (2) in the quadrilateral: yes
s=(a+b+c+d)/2
P²= (s-a)(s-b)(s-c)(s-d) ...(4)
from (3) and (4) we obtain R.
I wish you a good day!
I am from Brasil. Let me try first. Afterwards, I watch the video.
Let the low point at left be A, the low oint of right be B, the upper point of right be C and the other be D.
let AB=a=3, BC=b=4, CD=c=5 and DA=d=6. And Let d1=AC and d2=BD
2S(ABCD)= sin(D)*(cd+ab)=sin(A)*(ad+bc) (i)
As the circumscribed circumference is the same for the triangles ABD and ACD we have that d1/sin(D)=d2/sin(A)
So sin(D)/sin(A)=di/d2 (ii)
(i) ==> sin(D)/sin(A)=(ad+bc)/(ab+cd) (iii)
(ii) and (iii) ==> d1/d2= (ad+bc)/(ab+cd) multiplying by d1*d2 both sides we have:
d1^2= (ad+bc)/(ab+cd)*d1d2 (iv) But as we kow d1*d2=(ac+bd) (v)
(iv) and (v) d1^2=(ad+bc)*(ac+bd)/(ab+cd) (vi)
By (vi) d1^2=(18+20)*(15+24)/(30+12) ...d1^2=19*13/7 (vii)
(vii) and triangle ACD d1^2=19*13/7=5^2+36^2-2*5*6cos(D) ... cos(D)=180/420=3/7 ==> sin(D)=2*sqr(10)/7 (viii)
(viii) and triangle ACD ==> 2R=d1/sen(D)
2R=sqr(19*13/7)*7/(2*sqr(10)) R=1/4*sqr(19*13*7/10)=1/4sqr(1729/10)
I guess.
The ansewr is the same as the video.
Not many people know these advanced theorems. Certainly not me. I would stick with the cosine rule. This is like solving a geometry problem on a triangle with trigonometry (a faster solution but misses the logic used to solve the problem), or like solving a difficult limits problem with L'Hospitale rule (may not be allowed at that point).
Great demonstration !
I solved it using a non-linear system of 6 equations (with 6 variables). Of course I had no idea how to solve this system of equations myself (I tried but gave up after many pages) so I entered it in Wolfram Alpha and amazingly it delivered the correct solutions in like one second. AI is starting to really scare me.
Thanks for the simple and concise explanation. Ive studied math and never heard of any of these formulas before.
Just write the role of cosine in 2 triangular on opsite sides to find the cos(A) and after that sin(A) then us the role of R=a/(2.sin(A))
Is it really impossible to find this radius without all these many theories?
Yes, we can. First to find the cosine of angle A by considering triangle ABD and triangle CBD. Then use the law of Sine for triangles to find the diameter of the circumscribing circle of triangle ABD
Il faut qu'il fasse le tour du monde
I can’t believe anybody missed that. 😉
Applying Paramesuara's formula:
(4R)²= (ab+cd)(ac+bd)(ad+bc)/[(s-a)(s-b)(s-c)(s-d)]
R = 3,2872 cm ( Solved √ )
en.wikipedia.org/wiki/Parameshvara_Nambudiri
This channel certainly has a lot to teach.
Whose formula or theorem( Which mathematician) plz show his name. No 3
I ask the professor to take the time from his busy schedule to kindly explain why the following method which produces a slightly different answer is incorrect. I can find nothing wrong. The entire class may learn something from the exercise. Make point “O” the center of the circle. Extend radii to points A, B, C, and D. Lines AD, DC, CB, and AB will generate central angles proportionate to their length: 6, 5, 4 and 3. Their sum is 18. Divide 360 by 18 to get 20. Then angle AOD = 6 x 20 = 120 degrees. Now draw a perpendicular bisector from O to “E,” the center of chord AD. That produces two 30, 60, 90 right triangles because the bisector splits the central angle 120 degrees, intersects the chord at a right angle, and the remaining angle must be
180-(60 + 90) = 30. Focus on right triangle OED. OD is the radius (hypotenuse), ED, the smallest side, = 3 because it is 1/2 of 6 split by the perpendicular bisector. Now applying the familiar 1, 2, sq rt of 3 ratios of the sides of 30, 60, 90 right triangles, 2/sq rt 3 = radius (OD) /3. A little algebra reveals the radius is 2 x square rt 3 = 3.464… Your answer is 3.287… What’s “off,” if anything?
Hi Monroe : Your error is almost certainly the assumption that "Lines AD, DC, CB, and AB will generate central angles proportionate to their length". That is, you are assuming here that in a given circle, the length of any chord is directly proportional to the actual angle subtended by the chord. This is false. Recall the "Sine Law" for triangles : The meaning of sinA/a = sinB/b is that the sides in any given triangle are proportional NOT to the vertex angles themselves, but to the SINES of the vertex angles. The same thing is true here with chords in a circle. The lengths of chords in any given circle are proportional not to the angles themselves, but to the SINES of the half-angles subtended by the chord.
Parameshvara’s formula for finding circumradius
i applied the cos formula for isosceles triangles (2 lengths are equal to r):
10 l1=3:l2=4:l3=5:l4=6:dim x(3),y(3),w(3):sw=.01:r=sw:goto 90
20 rl=2*r^2:sp=0:cw1=(rl-l1^2)/rl:if abs(cw1)>1 then sp=1:goto 80
30 cw2=(rl-l2^2)/rl:if abs(cw2)>1 then sp=1:goto 80
40 cw3=(rl-l3^2)/rl:if abs(cw3)>1 then sp=1:goto 80
50 cw4=(rl-l4^2)/rl:if abs(cw4)>1 then sp=1:goto 80
60 w(0)=acs(cw1):w(1)=acs(cw2):w(2)=acs(cw3):w(3)=acs(cw4)
70 dg=acs(cw1)+acs(cw2)+acs(cw3)+acs(cw4)-2*pi
80 return
90 gosub 20:if sp=1 then else 110
100 r=r+sw:goto 90
110 r1=r:dg1=dg:r=r+sw:if r>100*l1 then stop
120 r2=r:gosub 20:if dg1*dg>0 then 110
130 r=(r1+r2)/2:gosub 20:if dg1*dg>0 then r1=r else r2=r
140 if abs(dg)>1E-10 then 130
150 print r:mass=500/r:wa=rad(10)+w(0):for a=0 to 3:x(a)=r*cos(wa):y(a)=r*sin(wa)
160 wa=wa+w(a):next a:goto 180
170 xb=(x+r)*mass:yb=(y+r)*mass:return
180 x=x(0):y=y(0):gosub 170:xba=xb:yba=yb:for a=1 to 4:ia=a:if ia=4 then ia=0
190 x=x(ia):y=y(ia):gosub 170:xbn=xb:ybn=yb:line xba,yba,xbn,ybn:xba=xbn:yba=ybn
200 next a:x=0:y=0:gosub 170:circle xb,yb,r*mass
3.28728611
>
run in bbc basic sdl and hit ctrl tab to copy
why can’t u use Ptolemy's theorem?
AC =5 (Pythagoras / 3 - 4 - 5). The angle in point B has 90°. Then you can use the formula for the circumference radius.
I think not 3,4,5 triangle needs to have a right angle. You can draw it to check.
If you say it's impossible then how did 1% of the students solve it?
Is it not the case that the two halves of the quadrilateral form two triangles joined along a "hypotenuse" that contacts the circle's circumference at two points. It appears that the only solution for that length is 9, where both triangles collapse to a line segment, and we get it as both the sum of 3+6 and 4+5? Then there is no quadrilateral at all and no real way to decide if 9 is a diameter.
Sorry Alan, but your solution is gibberish. In this problem, the sides of the quadrilateral are fixed in length. "Collapsing" them by changing the angle between the sides does not correspond to the problem at hand at all. And, " dia = 9 " is definitely an incorrect solution.
@@GWaters-xr1fv Don't you think that opening your reply with "gibberish" is rather the work of a troll? To repeat, consider either triangle with the 3-6 sides or the 4-5 sides. Connect each with a line segment. Then postulate bringing these two triangles together along these lines such that they must be equal. Ignore the question of any circle within which they may or may not be contained, simply consider the constraint of equality of the connecting line. Call the angle between the 3-6 lines alpha and the angle between the 4-5 lines beta. Solve using the cosine law for the two opposite sides being of the same length. I fine both angles are pi radians and the two triangles have collapsed to line segments of 9 each. I never claimed that this would be a diameter. In fact, one could draw any old circle around this line segment.
In the case of triángles
Esiste la formula della circonferenza circoscritta...R=√1729/4√10
amadeus would say "alquel esta un desafio" if he spoke spanish
According to the Problem, AB + AD = 3 + 6 = 9 and CB + CD = 4 + 5 = 9. BD divides the circle into two equal parts and BD will be the diameter of the circle. Triangle BAD has a right angle at A and triangle BCD has a right angle at C. According to the Pythagorean theorem :
1. BD^2 = AB^2 + AD^2 = 3^2 + 6^2 = 45
2. BD^2 = CB^2 + CD^2 = 4^2 + 5^2 = 41
R = 1/2BD
Is the above problem correct?
(ĐN)
Your argument is incorrect. (1) You need to consider the cosine of A or C in the BD formula, (2) R = BD / (2* sin (A))
I have a simple way for solve it
I HAVE A VERY SIMPLE METHOD
Sorry, this is a ridiculous solution. See the comments below.
No wen no recaber gibme nbr plss ser
You didn’t solve the problem. Answer is 6.4.😂
I only try to check for me new formula for cyclic quadrilatefral.... not solve problem...
You are right, cos(A) is not 4/5776 but 4/76... My fault... but I don't know how... I assume a copy paste problem with an unreliable mouse on my laptop... or something in this moment unknown... I apologize...
No, that is the diameter, the radius is 3.2