*Sugestão professor para o seu próximo vídeo:* Seja ABCD um quadrilátero qualquer. Sejam AB=2, BC=3, CD=8 e AD=5 e que os ângulos B + D = 120°. Qual a área do quadrilátero ABCD?
An alternative method to solve the problem without Brahmagoutpta formula is to use the complementary of the opposite angle in cyclic quadrilaterals together with the cosine law applied twice. We arrive to the same answer (3/4)*Sqrt(5). The details can be posted if you are interested.
in germany this could be called "a system with one degree of liberty". but, i got a different result for ri. however, see the graphics of the following calculation. this will not always give an accurate result if l1,l2,l3 or l4 have been changed: 10 print "premath-can you calculate the radius of the inscribed circle":dim x(3),y(3) 20 xu0=0:yu0=0:l1=2:l2=3:l3=6:l4=5:sw2=sqr(l1^2+l2^2+l3^2+l4^2)/180:sw1=sw2/9 30 x(3)=l4:y(3)=0:goto 200 40 ym1=sqr(l1^2-xm1^2):y2=sqr(abs(l3^2-(x2-l4)^2)) 50 dgu1=(x2-xm1)^2/l4^2:dgu2=(y2-ym1)^2/l4^2:dgu3=l2^2/l4^2 60 dg=dgu1+dgu2-dgu3 :return 70 x2=l4-l3+sw1:gosub 40 80 dg1=dg:x21=x2:x2=x2+sw2:x22=x2:gosub 40:if xm1>l1 then stop 90 if dg1*dg>0 then 80 100 x2=(x21+x22)/2:gosub 40:if dg1*dg>0 then x21=x2 else x22=x2 110 if abs(dg)>1E-8 then 100:rem den mittelpunkt des aussenkreises berechnen *** 120 xu1=xm1:yu1=ym1:xu2=x2:yu2=y2:xu3=l4:yu3=0:a11=2*(xu3-xu1):a12=2*(yu3-yu1) a13=xu3^2+yu3^2-xu1^2-yu1^2:a21=2*(xu2-xu1):a22=2*(xu2-yu1):a23=xu2^2+yu2^2-xu1^2-yu1^2 140 ngl1=a12*a21:ngl2=a22*a11 150 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 160 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 170 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 180 xl=zx/ngl:yl=zy/ngl:xma=xl:yma=yl:rua=sqr((xu1-xma)^2+(yu1-yma)^2):rem print "x=";xl;"y=";yl 190 return 200 xm1=-l1+sw2:gosub 210:goto 230 210 gosub 70:dfu1=(xma-xu2)^2/l4^2:dfu2=(yma-yu2)^2/l4^2:dfu3=rua^2/l4^2 220 df=dfu1+dfu2-dfu3:print df:return 230 df1=df:xm11=xm1:xm1=xm1+sw1:xm12=xm1:if xm1>l1 then stop 240 gosub 210:if df1*df>0 then 230 250 xm1=(xm11+xm12)/2:gosub 210:if df1*df>0 then xm11=xm1 else xm12=xm1 260 if abs(df)>1E-8 then 250:rem den radius des innenkreises berechnen *** 270 x(0)=0:y(0)=0:x(1)=xm1:y(1)=ym1:x(2)=x2:y(2)=y2:x(3)=l4:y(3)=0 280 xu4=l4:yu4=0:goto 370 290 xg1=xu1:yg1=yu1:xg2=xu0:yg2=yu0:xpu=xmi:ypu=ymi:gosub 300:goto 320 300 dx=xg2-xg1:dy=yg2-yg1:zx=dx*(xpu-xg1):zy=dy*(ypu-yg1):ku=(zx+zy)/(dx^2+dy^2) 310 dxk=ku*dx:dyk=ku*dy:xlo=xg1+dxk:ylo=yg1+dyk:ru=sqr((xlo-xpu)^2+(ylo-ypu)^2):return 320 ru1=ru:xg1=xu2:yg1=yu2:xg2=xu3:yg2=yu3:gosub 300:dg=(ru-ru1)/l3:return 330 xmi=sw1:gosub 290 340 dg1=dg:xmi1=xmi:xmi=xmi+sw1:xmi2=xmi:gosub 290:if dg1*dg>0 then 340 350 xmi=(xmi1+xmi2)/2:gosub 290:if dg1*dg>0 then xmi1=xmi else xmi2=xmi 360 if abs(dg)>1E-10 then 350 else return 370 ymi=sw1:goto 390 380 gosub 330:xg1=xu3:yg1=yu3:xg2=xu0:yg2=yu0:gosub 300:df=(ru-ru1)/l4:return 390 gosub 380 400 ymi1=ymi:df1=df:ymi=ymi+sw1:ymi2=ymi:gosub 380:if df1*df>0 then 400 410 ymi=(ymi1+ymi2)/2:gosub 380:if df1*df>0 then ymi1=ymi else ymi2=ymi 420 if abs(df)>1E-10 then 410 else rui=ru:print "rui=";rui 430 xmin=(xu1+xu2+xu3)/3:ymin=(yu1+yu2+yu3)/3:xmax=xmin:ymax=ymin 440 for a=0 to 3:if xmin>x(a) then xmin=x(a) 450 if xmaxy(a) then ymin=y(a) 470 if ymax
That's an interesting point! To be complete, we would need to prove that the quadrilateral is cyclic and that a circle can also be inscribed in it, in other words, it is a tangential quadrilateral as well. For the quadrilateral to be cyclic, we need to prove that a quadrilateral can be drawn with the given side lengths and
Ok!! Using B. Formula Online Calculator , it gives me : 01) Area of Quadrilateral [ABCD] (A) = 13,42 sq un 02) 2R + 3R + 5R + 6R = 2A 03) 16R = 2A 04) R = 2A / 16 05) R = A / 8 06) R = 13,42 / 8 07) R = 1,6775 lin un Therefore, MY FINAL ANSWER : Radius r equal 1,6775 Linear Units.
Let's find the radius r: . .. ... .... ..... By drawing lines from the corners of the quadrilateral ABCD to the center O of the inscribed circle we divide the quadrilateral into four triangles. The total sum of their areas corresponds to the area of the quadrilateral: A(ABCD) = A(ABO) + A(BCO) + A(CDO) + A(DAO) = (1/2)*AB*h(AB) + (1/2)*BC*h(BC) + (1/2)*CD*h(CD) + (1/2)*DA*h(DA) = (1/2)*AB*r + (1/2)*BC*r + (1/2)*CD*r + (1/2)*DA*r = (1/2)*r*(AB + BC + CD + DA) = (1/2)*r*P ⇒ r = 2*A(ABCD)/P P=AB+BC+CD+DA=2+3+6+5=16 is the perimeter of the quadrilateral. Since this quadrilateral does not only have an inscribed circle, but also a circumscribed one, its area can be calculated by applying the formula of Brahmagupta: A(ABCD) = √[(s − AB)(s − BC)(s − CD)(s − DA)] = √[(P/2 − AB)(P/2 − BC)(P/2 − CD)(P/2 − DA)] = √[(16/2 − 2)(16/2 − 3)(16/2 − 6)(16/2 − 5)] = √[(8 − 2)(8 − 3)(8 − 6)(8 − 5)] = √(6*5*2*3) = 6√5 Now we are able to calculate the value of the radius r: r = 2*A(ABCD)/P = 2*6√5/16 = (3/4)√5 Best regards from Germany
RUSSIA! Радиус вписанной в многоугольник равен площади днлённой на полупериметр. r=A/p; А=\/(р-6)(р-5)(р-3)(р-2)=6\/5=8r, r=6\/5/8= 3\/5/4=1,67705098311...линейных единиц.
Very interesting formula and solution to the problem. Thank you.
Glad you liked it
You are very welcome!
Thanks for the feedback ❤️
Very interesting formula! Keep bringing us good quality math questions Sir! 👌👍
Good Morning Master
Forte Abraço do Brasil 🇧🇷
Very good.
Excellent!
Thanks for the feedback ❤️
Thank you for introducing me Brahmagupta's formula! 📐
You are very welcome!
Thanks for the feedback ❤️
*Sugestão professor para o seu próximo vídeo:*
Seja ABCD um quadrilátero qualquer. Sejam AB=2, BC=3, CD=8 e AD=5 e que os ângulos B + D = 120°. Qual a área do quadrilátero ABCD?
Radius =
(2*area of the quadrilatera)/ perimeter of the Quadrilateral
= 2*6√5/16=3√5/4 unit
Excellent!
Thanks for sharing ❤️
An alternative method to solve the problem without Brahmagoutpta formula is to use the complementary of the opposite angle in cyclic quadrilaterals together with the cosine law applied twice. We arrive to the same answer (3/4)*Sqrt(5). The details can be posted if you are interested.
in germany this could be called "a system with one degree of liberty". but, i got a different result for ri. however, see the graphics of the following calculation. this will not always give an accurate result if l1,l2,l3 or l4 have been changed:
10 print "premath-can you calculate the radius of the inscribed circle":dim x(3),y(3)
20 xu0=0:yu0=0:l1=2:l2=3:l3=6:l4=5:sw2=sqr(l1^2+l2^2+l3^2+l4^2)/180:sw1=sw2/9
30 x(3)=l4:y(3)=0:goto 200
40 ym1=sqr(l1^2-xm1^2):y2=sqr(abs(l3^2-(x2-l4)^2))
50 dgu1=(x2-xm1)^2/l4^2:dgu2=(y2-ym1)^2/l4^2:dgu3=l2^2/l4^2
60 dg=dgu1+dgu2-dgu3 :return
70 x2=l4-l3+sw1:gosub 40
80 dg1=dg:x21=x2:x2=x2+sw2:x22=x2:gosub 40:if xm1>l1 then stop
90 if dg1*dg>0 then 80
100 x2=(x21+x22)/2:gosub 40:if dg1*dg>0 then x21=x2 else x22=x2
110 if abs(dg)>1E-8 then 100:rem den mittelpunkt des aussenkreises berechnen ***
120 xu1=xm1:yu1=ym1:xu2=x2:yu2=y2:xu3=l4:yu3=0:a11=2*(xu3-xu1):a12=2*(yu3-yu1)
a13=xu3^2+yu3^2-xu1^2-yu1^2:a21=2*(xu2-xu1):a22=2*(xu2-yu1):a23=xu2^2+yu2^2-xu1^2-yu1^2
140 ngl1=a12*a21:ngl2=a22*a11
150 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
160 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2
170 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2
180 xl=zx/ngl:yl=zy/ngl:xma=xl:yma=yl:rua=sqr((xu1-xma)^2+(yu1-yma)^2):rem print "x=";xl;"y=";yl
190 return
200 xm1=-l1+sw2:gosub 210:goto 230
210 gosub 70:dfu1=(xma-xu2)^2/l4^2:dfu2=(yma-yu2)^2/l4^2:dfu3=rua^2/l4^2
220 df=dfu1+dfu2-dfu3:print df:return
230 df1=df:xm11=xm1:xm1=xm1+sw1:xm12=xm1:if xm1>l1 then stop
240 gosub 210:if df1*df>0 then 230
250 xm1=(xm11+xm12)/2:gosub 210:if df1*df>0 then xm11=xm1 else xm12=xm1
260 if abs(df)>1E-8 then 250:rem den radius des innenkreises berechnen ***
270 x(0)=0:y(0)=0:x(1)=xm1:y(1)=ym1:x(2)=x2:y(2)=y2:x(3)=l4:y(3)=0
280 xu4=l4:yu4=0:goto 370
290 xg1=xu1:yg1=yu1:xg2=xu0:yg2=yu0:xpu=xmi:ypu=ymi:gosub 300:goto 320
300 dx=xg2-xg1:dy=yg2-yg1:zx=dx*(xpu-xg1):zy=dy*(ypu-yg1):ku=(zx+zy)/(dx^2+dy^2)
310 dxk=ku*dx:dyk=ku*dy:xlo=xg1+dxk:ylo=yg1+dyk:ru=sqr((xlo-xpu)^2+(ylo-ypu)^2):return
320 ru1=ru:xg1=xu2:yg1=yu2:xg2=xu3:yg2=yu3:gosub 300:dg=(ru-ru1)/l3:return
330 xmi=sw1:gosub 290
340 dg1=dg:xmi1=xmi:xmi=xmi+sw1:xmi2=xmi:gosub 290:if dg1*dg>0 then 340
350 xmi=(xmi1+xmi2)/2:gosub 290:if dg1*dg>0 then xmi1=xmi else xmi2=xmi
360 if abs(dg)>1E-10 then 350 else return
370 ymi=sw1:goto 390
380 gosub 330:xg1=xu3:yg1=yu3:xg2=xu0:yg2=yu0:gosub 300:df=(ru-ru1)/l4:return
390 gosub 380
400 ymi1=ymi:df1=df:ymi=ymi+sw1:ymi2=ymi:gosub 380:if df1*df>0 then 400
410 ymi=(ymi1+ymi2)/2:gosub 380:if df1*df>0 then ymi1=ymi else ymi2=ymi
420 if abs(df)>1E-10 then 410 else rui=ru:print "rui=";rui
430 xmin=(xu1+xu2+xu3)/3:ymin=(yu1+yu2+yu3)/3:xmax=xmin:ymax=ymin
440 for a=0 to 3:if xmin>x(a) then xmin=x(a)
450 if xmaxy(a) then ymin=y(a)
470 if ymax
The radius is 3/4(sqrt(5)). By golly I really appreciated that refresher in cyclic quadrilaterals!!!
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
A=(somma di 4 triangoli)=5r/2+6r/2+3r/2+2r/2=8r...A=√(p-3)(p-2)(p-5)(p-6)=√180(formula di Braham..con p=2p/2=8)..r=√180/8=6√5/8=3√5/4
Excellent!
Thanks for sharing ❤️
A questão é muito bonita. É possível resolvê-la utilizando outros métodos. Parabéns pela escolha.
Se tem outros métodos, fico surpreso, pois a fórmula de Brahmagupta é a que temos!
Muitas abordagens são possíveis para encontrar a solução para este problema!
No entanto, este método é mais eficiente.
Obrigado pelo feedback ❤️
asnwer=3/4 isit gmm fmm
2:15-3:08 s=(a+b+c+d)/2 ???
4:45-9:28 A=A1+A2+A3+A4 ???
Tangential quadrilateral :
s=a+c=b+d; A=sr !!! 😁
The quadrilateral in the figure is both cyclic and tangential.
That's an interesting point! To be complete, we would need to prove that the quadrilateral is cyclic and that a circle can also be inscribed in it, in other words, it is a tangential quadrilateral as well. For the quadrilateral to be cyclic, we need to prove that a quadrilateral can be drawn with the given side lengths and
r=3√5/4 units.❤
Excellent!
Thanks for sharing ❤️
My way of solution ▶
Here in this circle, we have four Deltoids.
Let's consider the triangle ΔBAD and ΔDCB, also the angles:
∠BAD= θ
∠DCB= 180°- θ
By applying the cosine theorem for the triangle ΔBAD we get:
[BD]²= [BA]²+[AD]²-2*[BA]*[AD]*cos(θ)
[BA]= 2
[AD]= 5
⇒
[BD]²= 2²+5²-2*2*5*cos(θ)
[BD]²= 4+25-20cos(θ)
[BD]²= 29 - 20cos(θ)............Eq-1
By applying the cosine theorem for the triangle ΔDCB we get:
[BD]²= [BC]²+[CD]²-2*[BC]*[CD]*cos(180°-θ)
[BC]= 3
[CD]= 6
⇒
[BD]²= 3²+6²-2*3*6*cos(180°-θ)
[BD]²= 9+36-36cos(180°-θ)
cos(180°-θ)= cos(180°)*cos(θ) - sin(180°)*sin(θ)
cos(180°-θ)= -cos(θ)
⇒
[BD]²= 45 + 36cos(θ)............Eq-2
⇒
Equation-1 and equation-2 must be equal to each other, so:
29 - 20cos(θ) = 45 + 36cos(θ).
29-45= 20cos(θ)+ 36cos(θ)
56cos(θ)= -16
cos(θ)= -16/56
cos(θ)= -2/7
⇒
θ= 106,6015°
180°-θ = 73,3985°
b) Let's consider the Deltoids: D(EAFO), D(FDGO), D(GCFO), and D(FBEO)
E ∈ [AB]
F ∈ [AD]
G ∈ [DC]
H ∈ [BC]
if [EB] = x
[EB]=[FB]
[FB]= x
[EA]= 2-x
[EA]=[AF]
[AF]= 2-x
⇒
[FD]= 5 - (2-x)
[FD]= 3+x
[FD]= [DG]
[DG]= 3+x
[GC]= 6-(3+x)
[GC]= 3-x
[GC]= [CF]
[CF]= 3-x
c) Let's write the equations for the Deltoid D(EAFO) and D(GCFO)
tan(θ/2)= r/(2-x)
tan[(180-θ)/2]= r/(3-x)
⇒
tan(106,6015°/2)= r/(2-x)
tan(73,3985°/2)= r/(3-x)
⇒
1,341639= r/(2-x)
0,7453= r/(3-x)
⇒
r= 2,6832 - 1,3416 x
⇒
0,7453= (2,6832 - 1,3416 x)/(3-x)
2,2359 - 0,7453 x= 2,6832 - 1,3416x
1,3416x - 0,7453x= 2,6832 - 2,2359
x= 0,7501 length units ✅
r= 2,6832 - 1,3416*0,7501
r= 1,6768
r≈ 1,677 length units ✅
You are implicitly using Brahmagupta's formula!However, more work!
Thanks for sharing ❤️
Once you have found cosine theta you could find sines theta and then the area of the two triangles
Ok!!
Using B. Formula Online Calculator , it gives me :
01) Area of Quadrilateral [ABCD] (A) = 13,42 sq un
02) 2R + 3R + 5R + 6R = 2A
03) 16R = 2A
04) R = 2A / 16
05) R = A / 8
06) R = 13,42 / 8
07) R = 1,6775 lin un
Therefore,
MY FINAL ANSWER :
Radius r equal 1,6775 Linear Units.
Thank Brahmagupta for nothin! ...🙂
😀
Thanks for the feedback ❤️
Let's find the radius r:
.
..
...
....
.....
By drawing lines from the corners of the quadrilateral ABCD to the center O of the inscribed circle we divide the quadrilateral into four triangles. The total sum of their areas corresponds to the area of the quadrilateral:
A(ABCD)
= A(ABO) + A(BCO) + A(CDO) + A(DAO)
= (1/2)*AB*h(AB) + (1/2)*BC*h(BC) + (1/2)*CD*h(CD) + (1/2)*DA*h(DA)
= (1/2)*AB*r + (1/2)*BC*r + (1/2)*CD*r + (1/2)*DA*r
= (1/2)*r*(AB + BC + CD + DA)
= (1/2)*r*P
⇒ r = 2*A(ABCD)/P
P=AB+BC+CD+DA=2+3+6+5=16 is the perimeter of the quadrilateral. Since this quadrilateral does not only have an inscribed circle, but also a circumscribed one, its area can be calculated by applying the formula of Brahmagupta:
A(ABCD)
= √[(s − AB)(s − BC)(s − CD)(s − DA)]
= √[(P/2 − AB)(P/2 − BC)(P/2 − CD)(P/2 − DA)]
= √[(16/2 − 2)(16/2 − 3)(16/2 − 6)(16/2 − 5)]
= √[(8 − 2)(8 − 3)(8 − 6)(8 − 5)]
= √(6*5*2*3)
= 6√5
Now we are able to calculate the value of the radius r:
r = 2*A(ABCD)/P = 2*6√5/16 = (3/4)√5
Best regards from Germany
Excellent!
Thanks for sharing ❤️
Why do you speak like that?
RUSSIA! Радиус вписанной в многоугольник равен площади днлённой на полупериметр. r=A/p; А=\/(р-6)(р-5)(р-3)(р-2)=6\/5=8r, r=6\/5/8= 3\/5/4=1,67705098311...линейных единиц.
Excellent!
Thanks for sharing ❤️