1st in the classroom = 1st like + 1st comment 🤓

@PreMath

Ай бұрын

Excellent! Glad to hear that! You are very welcome! Thanks ❤️

An excellent exposition.

@PreMath

Ай бұрын

Glad to hear that! Thanks ❤️

Just amazing! Thanks so much for posting this beautiful problem.

So neat! Thanks!

@PreMath

Ай бұрын

You are so welcome! Thanks ❤️

1) Green Area = 3 2) Purple Area = 5 3) Yellow Area = 4 4) Red Area = Green Area + Purple Area + Yellow area 5) Red Area = 3 + 5 + 4 = 12 5) So, by The Carpet Theorem, Red Area is Equal to 12 Square Units.

I looked at that and thought no way. Yet you made it look so easy and simple. Very good!

@PreMath

Ай бұрын

Glad you liked it! Thanks ❤️

Thank you! Excellent application of a basic concept!

Lots of regions in the diagram but no real difficulty. First method may be the best.

@PreMath

Ай бұрын

Thanks for sharing❤️

Like for premath 👇🏻

@PreMath

Ай бұрын

Excellent! Thanks ❤️

Muito bom. Difícil.

@PreMath

Ай бұрын

Glad to hear that! Thanks ❤️

11:45 this took me lots of repeated calculations. my idea was, to choose lb and lh and keep the proportions of a1/a2 and a2/a3 with a nested calculation: 10 print "premath-do you know these skills? can you find area of the red region?" 20 lh=10:lb=10:xd=0:yd=0:xc=lb:yc=0:xa=0:ya=lh:xb=lb:yb=lh:dim x(8),y(8),a(7) 30 a1=3:a2=5:a3=4:sw=a1/(a1+a2+a3):ym=sw:yn=0:n=a1^2+a2^2:goto 270 40 xg11=0:yg11=ym:xg12=lb:yg12=0:xg21=0:yg21=lh:xg22=xn:yg22=0:rem mc mit an schneiden 50 gosub 130:if ngl=0 then return 60 xo=xl:yo=yl:xg21=xn:yg21=0:xg22=lb:yg22=lh:gosub 130:if ngl=0 then return 70 xp=xl:yp=yl:rem mc mit nb 80 xg11=0:yg11=ym:xg12=lb:yg12=lh:xg21=0:yg21=lh:xg22=xn:yg22=0:gosub 130 90 if ngl=0 then return :rem mb mit an 100 xq=xl:yq=yl:a1r=(lh-ym)*xq/2:au21=xo*ym/2:au22=yo*xn/2 110 a2r=au21+au22:rem if a2r=0 then return 120 a3r=yp*(lb-xn)/2:dgu1=a2r*a3:dgu2=a2*a3r:dg=(dgu1-dgu2)/n:return 130 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 140 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 150 a13=a131+a132:a23=a231+a232:ngl1=a12*a21:ngl2=a22*a11 160 ngl=ngl1-ngl2:if ngl=0 then return:then print "keine loesung":end 170 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 180 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 190 xl=zx/ngl:yl=zy/ngl:return 200 xn=sw: gosub 40 210 dg1=dg:xn1=xn:xn=xn+sw:gosub 40:xn2=xn:if dg1*dg>0 then 210 220 xn=(xn1+xn2)/2:gosub 40:if dg1*dg>0 then xn1=xn else xn2=xn 230 if abs(dg)>1E-10 then 220 240 rem print a1/a3;"%";a1r/a3r: 250 return 260 gosub 200:dfu1=a1r*a2:dfu2=a1*a2r:df=(dfu1-dfu2)/n:return 270 gosub 260:if ngl=0 then else 290 280 ym=ym+sw:goto 270 290 df1=df:ym1=ym:ym=ym+sw:ym2=ym:gosub 260:if df1*df>0 then 290 300 ym=(ym1+ym2)/2:gosub 260:if df1*df>0 then ym1=ym else ym2=ym 310 if abs(df)>1E-10 then 300 320 a(0)=a1r:a(1)=a2r:a(2)=a3r:x1=xo:y1=yo:x2=xp:y2=yp:xpu=xn:ypu=yn:goto 360 330 dx=x2-x1:dy=y2-y1:zx=dx*(xpu-x1):zy=dy*(ypu-y1):k=(zx+zy)/(dx^2+dy^2) 340 dxk=k*dx:dyk=k*dy:xlo=x1+dxk:ylo=y1+dyk:lo=sqr((xlo-xpu)^2+(ylo-ypu)^2) 350 au=sqr(dx^2+dy^2)*lo/2:return 360 gosub 330:a(3)=au:xpu=xq:ypu=yq:x1=xo:y1=yo:x2=xb:y2=yb: gosub 330:au1=au 370 xpu=xp:ypu=yp:gosub 330:au2=au:a(4)=au1+au2 380 a(5)=(lb-xp)*lh/2:a(6)=(lh-yq)*lb/2:x1=xq:y1=yq:x2=xo:y2=yo:xpu=0:ypu=ym 390 gosub 330:a(7)=au:a7r=xn*lh/2:a7r=a7r-a1r-a2r:print a(7),"%",a7r 400 fe=a(7)/a7r:fe=(1-fe)*100:print "der fehler=";fe;"%" 410 sua=0:for a=0 to 7:sua=sua+a(a):next a:print sua:fe=lb*lh/sua:fe=(1-fe)*100 420 print "der fehler=";fe;"%" 430 a8p=xn/2*lh-a2r-a1r:print a(7),"%",a8p:a4g=a(4)*a1/a1r:print "die flaeche=";a4g 440 masx=1200/lb:masy=900/lh:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

Los triángulos ABN y BCM tienen la misma superficie y juntos suman el área de ABCD→ En la figura se solapan en el cuadrilátero rojo→ Todos los polígonos que, estando dentro del rectángulo, quedan fuera de los límites de esos dos triángulos superpuestos deben sumar un área igual al solapo de ambos para completar el área de ABCD → Área del solapo = Área roja =3+5+4=12. Gracias a "Rabota Akk" por advertir el error de mi razonamiento y respuesta anterior. Un saludo cordial.

@rabotaakk-nw9nm

Ай бұрын

👍🥰

Is there enough information to find the other empty areas labeled x,y,z, and w, individually?

Two of the method are exactly the same (1 and 2) I would put the third method this way: MDC=MD*DC/2=MDB ADN=DN*AD/2=DNB MDC+ADN=MDB+NDB=NDMB (5+z+4)+(3+y+5)=5+y+z+a a=12

Let's call the white areas A, C, M and N. X is the red area we are looking for. Triangles ABN and BCM are both half the area of the rectangle and, therefore, the same area. So, we can say: A + X + N = C + X + M The parts that don't belong to the triangles are the same as well. So, we can say: 3 + M + 5 + C + 4 = A + 3 + 5 + N + 4 M + C + 12 = A + N + 12 Since all the areas in the equations are half the rectangle, we can switch them: A + N + X = A + N + 12 C + M + X = C + M + 12 In both cases, we can come up with: X (red area) = 12 (3 + 5 + 4) A (red) = A (green) + A (purple) + A (yellow) = 12 square units

2hr

Very difficult. I should better quit.😅😅😅😅😅 0:04

@PreMath

Ай бұрын

Keep watching... Thanks ❤️

Nice sir Plss pin

The Diagram may not be 100% True to the Scale, Can you provide the Original Diagram!! I think this diagram is very... very, inaccurate!! So tell us, please. If we draw a Diagonal Line from B to D, does it divides the Red Area in two Equal Parts? This is to use the CARPET THEOREM, am I right? Thank you.