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.

@user-it6fh7hy6t

5 ай бұрын

Только так и следует решать.Ваше решение намного проще и короче авторского.

I can’t believe anybody missed that. 😉

Best geometry channel in youtube. I like you

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

@user-dz4ds6zf5e

5 ай бұрын

This is aright way

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

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.

The quick and easy way to solve this is to use the cosine (supplement) rule (which is what you use to derive Stewart’s Theorem). In any triangle, the cosine of any angle is (the sum of the squares of the adjacent sides, less the square of the opposite side) all divided by twice the product of the adjacent sides. The sum of the opposite angles of cyclic quadrilaterals add up to 180 degrees, and The cosine of an angle is minus the cosine of the supplement of that angle. Make AC = x, then: CosD = - cosB (5^2 +6^2 -x^2)/5*6*2 = (x^2 -3^2 -4^2)/2*3*4 x = 5.135 = AC Finding the radius of the circumscribed circle of a triangle is easy = 3.287, which agrees with the official answer.

@soli9mana-soli4953

6 ай бұрын

How did you find the radius? I've found x=sqrt(247/7) and cos(alpha)=3/7

@millenium9368

6 ай бұрын

​@@soli9mana-soli4953 x/sin(alpha)= 2R it's the law of sine

@juanalfaro7522

6 ай бұрын

The radius answer is correct (3.287).

@juanalfaro7522

6 ай бұрын

@@soli9mana-soli4953 Diameter = X / sin(D), X = sqrt (247/7), cos(D) = 3/7 => sin(D) = sqrt (40)/7. The diameter of a circle with an inscribed triangle is found by dividing a side of that triangle by its opposite angle.

Great demonstration !

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).

This channel certainly has a lot to teach.

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

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

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.

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))

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.

Parameshvara’s formula for finding circumradius

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.

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!

Whose formula or theorem( Which mathematician) plz show his name. No 3

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 √ )

@adamdosa

6 ай бұрын

en.wikipedia.org/wiki/Parameshvara_Nambudiri

Is it really impossible to find this radius without all these many theories?

@julinwu8064

6 ай бұрын

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

@Hayet-jb2sd

6 ай бұрын

Il faut qu'il fasse le tour du monde

why can’t u use Ptolemy's theorem?

Esiste la formula della circonferenza circoscritta...R=√1729/4√10

@zdrastvutye

6 ай бұрын

amadeus would say "alquel esta un desafio" if he spoke spanish

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?

@GWaters-xr1fv

5 ай бұрын

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 have a simple way for solve it

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

No wen no recaber gibme nbr plss ser

I HAVE A VERY SIMPLE METHOD

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.

@GWaters-xr1fv

5 ай бұрын

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.

@AllanAngusADA

5 ай бұрын

@@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)

@juanalfaro7522

6 ай бұрын

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))

You didn’t solve the problem. Answer is 6.4.😂

@miloradkadic

6 ай бұрын

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...

@dennisgannon

5 ай бұрын

No, that is the diameter, the radius is 3.2

If you say it's impossible then how did 1% of the students solve it?