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Nice video Your video on other concepts Is as better as your video on geometry question I really like your work

Ótima solução! 🎉🎉🎉

Nice video

Nice

Cos c = 34 - 4 r^2 / 30 C/ sinc = 1/ 4r 4r = sinc /c - 4rc = sin c Cos c = 34 - c^2 / 30 sin^2 c Cos c = 34 - c^2 / 30 sin^2 c Sin^2 c + cos^2 c = 1 16 r^2 c^2 + ( 34 - c^2 / 30 sin^2 c) ^2 = 1 16 r ^2 . 4r ^2 + ( 34 - 4r ^2 / 30 . 4 r^2 c^2 ) = 1 After solving quadratic eqn. Stated above we get the value of r I.e required in this question

Nice question sir An another good video with the different and nice concept

Nice job

21²+20²=29²: triangle rectangle avec 3 entiers

Good job😊

Nise video in geometry problem with trigonometry solution

Very nice question

h²=400-x² h²=225-(25-x)² h²=225-625+50x-x² h²=-400+50x-x² 400-x²=-400+50x-x² 800=50x x=16 h²=400-256=144 h=12 20=16-r1+12-r1 20=28-2r1 r1=4 15=9-r2+12-r2 r2=3 AB²=(4-3)²+(4+3)² AB²=1²+7² AB=√50 AB=5√2

it resulted in a repeated calculation: 10 print "quant circle-a geometry problem-solved using rotation" 20 dim x(2),y(2):l1=20:l2=21:l3=29:sw=sqr(l1^2+l2^2+l3^2)/39:goto 100 30 ys=sqr(l1^2-xs^2):dgu1=(xs-l4)^2/l2^2:dgu2=(ys/l2)^2:dg=dgu1+dgu2-1 40 return 50 xs=sw:gosub 30 60 dg1=dg:xs1=xs:xs=xs+sw/30:if xs>l1 then return 70 xs2=xs:gosub 30:if dg1*dg>0 then 60 80 xs=(xs1+xs2)/2:gosub 30:if dg1*dg>0 then xs1=xs else xs2=xs 90 if abs(dg)>1E-10 then 80 else return 100 l4=sw:goto 140 110 gosub 50:if xs>l1 then return 120 dfu1=(xs-l4/2)^2/l3^2:dfu2=(ys-sqr(3)/2*l4)^2/l3^2 130 df=dfu1+dfu2-1:return 140 gosub 110:if xs>l1 then else 160 150 l4=l4+sw:goto 140 160 l41=l4:df1=df:l4=l4+sw:l42=l4:gosub 110:if df1*df>0 then 160 170 l4=(l41+l42)/2:gosub 110:if df1*df>0 then l41=l4 else l42=l4 180 if abs(df)>1E-10 then 170 else 200 190 xbu=x*mass:ybu=y*mass:return 200 masy=1200*2/sqr(3)/l4:masx=850/l4:if masx<masy then mass=masx else mass=masy 210 x(0)=0:y(0)=0:x(1)=l4:y(1)=0:x(2)=l4/2:y(2)=sqr(3)*l4/2 220 x=x(0):y=y(0):gosub 190:xba=xbu:yba=ybu:for a=1 to 3:ia=a:if ia=3 then ia=0 230 x=x(ia):y=y(ia):gosub 190:xbn=xbu:ybn=ybu:goto 250 240 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 250 gosub 240:next a:gcol9:x=xs:y=ys:gosub 190:xbs=xbu:ybs=ybu:for a=0 to 2 260 xba=xbs:yba=ybs:x=x(a):y=y(a):gosub 190:xbn=xbu:ybn=ybu:gosub 240:next a 270 print "die gesuchte laenge des gleichseitigen dreiecks ist=";l4:ar=sqr(3)/4*l4^2 280 print "die dreiecksflaeche=";ar quant circle-a geometry problem-solved using rotation die gesuchte laenge des gleichseitigen dreiecks ist=39.6038046 die dreiecksflaeche=679.163682 > run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" at the beginning for full screen graphics

Triangle area is totally redundant, all u need is cosine theorem, cos x = 1/ sq 2, done

5:05-5:50 [AQR]=½AQ•ARsin150°= =½20•21•½=5•21=105 sq.un. 😁

Nice solution Your last two videos on geometry solution have an different ways to solve them Keep growing

Nice magic

Linda solução! 🎉🎉🎉

zum beispiel x^4+1827^2=2048^2 x^2=925.405 x=30.42 es scheint aber kein rechtwinkliges dreieck zu sein

Bravo! Bravo! 🎉🎉🎉

hi at least it was not a nested repeated calculation: 10 print "quant circle-not as easy as it looks-a challenging geometry problem" 20 dim x(2,2),y(2,2):l2=7:l3=5:l4=3:ym=0:sw=l4/67:goto 50 30 xa=sqr(l3^2-h^2):r=(sqr(l3^2-h^2)+sqr(l4^2-h^2))/2:l1=2*r 40 l5=2*r-xa:dgu1=sqr(l4^2-l5^2)/l2:dgu2=sqr(l2^2-l5^2)/l2:dg=dgu1+dgu2-1:return 50 h=sw:gosub 30 60 h1=h:dg1=dg:h=h+sw:if h>l4 then stop 70 h2=h:gosub 30:if dg1*dg>0 then 60 80 h=(h1+h2)/2:gosub 30:if dg1*dg>0 then h1=h else h2=h 90 if abs(dg)>1E-10 then 80:rem den tangentenpunkt berechnen 100 print "der radius=";r:xm=r:xt=sw:goto 130 110 yt=ym+sqr(r*r-(xt-xm)^2):dgu1=(xm-xt)*(2*r-xt)/r/r:dgu2=(ym-yt)*(l1-yt)/r/r 120 dg=dgu1+dgu2:return 130 gosub 110 140 xt1=xt:dg1=dg:xt=xt+sw:if xt>20*r then stop 150 xt2=xt:gosub 110:if dg1*dg>0 then 140 160 xt=(xt1+xt2)/2:gosub 110:if dg1*dg>0 then xt1=xt else xt2=xt 170 if abs(dg)>1E-10 then 160 180 masx=1200/2/r:masy=850/l1:if masx<masy then mass=masx else mass=masy 190 goto 210 200 xbu=x*mass:ybu=y*mass:return 210 x(0,0)=0:y(0,0)=0:x(0,1)=2*r:y(0,1)=0:x(0,2)=xa:y(0,2)=h 220 x(1,0)=xa:y(1,0)=h:x(1,1)=2*r:y(1,1)=l1:x(1,2)=xt:y(1,2)=yt 230 for a=0 to 1:x=x(a,0):y=y(a,0):gosub 200:xba=xbu:yba=ybu:for b=1 to 3:ib=b 240 if ib=3 then ib=0 250 x=x(a,ib):y=y(a,ib):gosub 200:xbn=xbu:ybn=ybu:goto 270 260 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 270 gosub 260:next b:next a:x=xm:y=ym:gosub 200:gcol 3:circle xbu,ybu,r*mass:gcol8 280 x=2*r:y=l1:gosub 200:xba=xbu:yba=ybu:y=0:gosub 200:xbn=xbu:ybn=ybu:gosub 260 quant circle-not as easy as it looks-a challenging geometry problem der radius=3.94440716 > run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" at the beginning for the graphics display.

Very good and useful question for general knowledge 👏🏻😊

Thank you so much for your support

Amazing 😊

2.7959+3.7279=5.523867733

(2)^2=4 180ABC/4=4.20 4.2^10 4.2^2^5 2^2.1^1^1 1^2 .1 2(ABC ➖ 2ABC+1).

i just applied the fact that in an isosceles triangle the angle bisector is intersecting the midpoint of the opposite length of the angle: 10 print "quant circle-a nice geometry question from india cat exam":l=1 20 dim x(1,5),y(1,5):x(0,2)=0:y(0,2)=0:x(0,3)=l/2:y(0,3)=0 30 x(0,4)=l:y(0,4)=0:x(0,0)=l*.5:y(0,0)=sqr(1-.5^2)*l:x(0,5)=(x(0,4)+x(0,0))/2 40 y(0,5)=(y(0,0)+y(0,4))/2:x(0,1)=x(0,0)/2:y(0,1)=y(0,5):dx=x(0,5):dy=y(0,5) 50 k=l/(2*dx+4*dy):xm=k*dx:ym=k*dy:ru=ym:print "der kreisradius ist=";ru/l;"*l":dx=l/2:dy=y(0,0) 60 x(1,0)=xm:y(1,0)=ym:for a=1 to 2:x(1,a)=x(1,a-1)+2*ru:y(1,a)=y(1,a-1):next a 70 x(1,3)=x(1,0)+ru:y(1,3)=y(1,0)+sqr(3)*ru:dl=sqr((x(1,3)-x(1,0))^2+(y(1,3)-y(1,0))^2) 80 fe=(2*ru-dl)/2/ru*100:print "der fehler=";fe;"%" 90 x(1,4)=x(1,3)+2*ru:y(1,4)=y(1,3):x(1,5)=x(1,4)-ru:y(1,5)=y(1,4)+sqr(3)*ru 100 masx=1200/l:masy=850/y(0,0):if masx<masy then mass=masx else mass=masy 110 goto 130 120 xbu=x*mass:ybu=y*mass:return 130 x=x(0,0):y=y(0,0):gosub 120:xba=xbu:yba=ybu:for a=1 to 6:ia=a:if ia=6 then ia=0 140 x=x(0,ia):y=y(0,ia):gosub 120:xbn=xbu:ybn=ybu:goto 160 150 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 160 gosub 150:next a:for a=0 to 5:x=x(1,a):y=y(1,a):gosub 120:circle xbu,ybu,ru*mass:next a quant circle-a nice geometry question from india cat exam der kreisradius ist=0.133974596*l der fehler=0% > run in bbcbasic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" at the beginning, just for the graphics

^=read as to the power *=read as square root QR=*(20^2 + 15^2) =*(400+225)=*625=25 Let SR=x So QS=25- x In triangle PQS, angle S =90 So, PS^2 =PQ^2 - QS^2 =(20^2) - (25 - x)^2 =400-{625 -50x+x^2) =400-625+50x- x^2 =50x -x^2 -225.....eqn1 Again in triangle PSR, angle S=90 So, PS^2=PR^2 - SR^2 =15^2 - X^2=225 -X^2....eqn2 Eqn1 & eqn2 are equal due to PS^2 SO, 50X -X^ -225 =225-X^2 50X=225+225=450 X= 450/50 = 9 SO, SR=9 QS= 25-9 =16 Now in triangle PQS PS=*(PQ^2 -QS^2) =*(20^2-16^2) =*(400-256)=*144=12 QS=16 SO area of PQS=(12×16)/2=96 Let R= radious of circle which stands in the interior of PQS triangle R=(PQ×QR×QS)/(4×area) =(20×16×12)/(4×96) )=10 Let r= radious of the circle which stands in the interior region of the triangle PSR Area of PSR =(PS×SR)/2 =(12×9)/2=54 r=(PR×SR×PS)/(4×Area ) =(15×9×12)/(4×54)=7.5 AB=R+r =10+7.5=17.5 Hence AB=17.5 Wrongly i have aplied the formula for exterior circle Hope this question will be as below In triangle PRS the circle intersects PS, SR, PR at the point m, n, o respectively Let PO=y So, RO=15-y Ro=Rn=15-y So Sn=9-(15-y)=y-6 Again po=Pm=y Sm=12-y Sm=Sn 12-y=y-6 2y=18 Y=18/2=9 Sm=r=12-y=12-9=3 In the triangle PQS the circle intersects at a, b, c to PQ, QS, PS respectively Let Pa=t Pa=Pc=t So, Sc=12-t Qa=Qb=20-t So, Sb=16-(20-t)=t-4 Sb=Sc t-4=12-t 2t=16 t=18/2=8 Sc=R=12-t=12-8=4 AB=R+r=4+3=7 Hence AB=7 Note :Draw the parallel line segment to SC from A to PS, simillarly from B to Ps with parallel to Sn( where they touch the circle are also the radious of the circle

yes it is tricky but can be solved. the repeated calculation changes the centerpoint of the circle with its known radius until the quotient "fl=" as written in line 20, is reached: 10 print "quant circle-how to solve this tricky geometry problem with 3 circles" 20 dim x(1,2),y(1,2),r(2):r(0)=2:sw=r(0)/10:fl=10/3:y(1,0)=r(0):ym=r(0):y(1,0)=ym:nr=30 30 nr=30 :goto 150 40 ys=y(1,0)+sqr(abs(r(0)^2-(xs-xm)^2)):dgu1=(xs^2+ys^2)/r(0)^2:dgu2=xs*xm/r(0)^2 50 dgu3=ys*ym/r(0)^2:dg=dgu1-dgu2-dgu3:return 60 xm=r(0)+sw:gosub 40 70 dg1=dg:xm1=xm:xm=xm+sw:if xm>20*r(0) then stop 80 xm2=xm:gosub 40:if dg1*dg>0 then 70 90 xm=(xm1+xm2)/2:gosub 40:if dg1*dg>0 then xm1=xm else xm2=xm 100 if abs(dg)>1E-10 then 90 110 return 120 gosub 60:ye=ys/xs*(xm+r(0)):lbh=sqr(xs^2+ys^2):lah=sqr((xm+r(0)-xs)^2+(ye-ys)^2) 130 lac=ye:df=lbh/lah: df=df-fl:return 140 return 150 xs=sw:gosub 120 160 xs1=xs:df1=df:xs=xs+sw:if xs>nr*r(0) then stop 170 xs2=xs:gosub 120:if df1*df>0 then 160 180 xs=(xs1+xs2)/2:gosub 120:if df1*df>0 then xs1=xs else xs2=xs 190 if abs(df)>1E-10 then 180 200 lbc=sqr((lbh+lah)^2-lac^2):lab=lbh+lah:print "lbh=";lbh:print"lah=";lah:print "lab=";lab 210 print "lbc=";lbc;"lac";lac: print xs,"%",ys 220 x(0,0)=0:y(0,0)=0:x(0,1)=lbc:x(0,2)=x(0,1):y(0,2)=ye:x(1,0)=xm 230 xmu=sw:goto 260:rem den berechneten mittelpunkt verwenden *** 240 ymu=xmu/xm*ym:ru=ymu:dgu1=(xm-xmu)^2/r(0)^2:dgu2=(ym-ymu)^2/r(0)^2:dgu3=(ru+r(0))^2/r(0)^2 250 dg=dgu1+dgu2-dgu3 :return 260 gosub 240 270 dg1=dg:xmu1=xmu:xmu=xmu+sw:xmu2=xmu:gosub 240:if dg1*dg>0 then 270 280 xmu=(xmu1+xmu2)/2:gosub 240:if dg1*dg>0 then xmu1=xmu else xmu2=xmu 290 if abs(dg)>1E-10 then 280 300 x(1,1)=xmu:y(1,1)=ymu:r(1)=ru:dx=xm-lbc:dy=ym-ye:k=sw:goto 330 310 dxk=k*dx:dyk=k*dy:xmu=lbc+dxk:ymu=ye+dyk:ru=lbc-xmu:dgu1=(xmu-xm)^2/r(0)^2 320 dgu2=(ymu-ym)^2/r(0)^2:dgu3=(ru+r(0))^2/r(0)^2:dg=dgu1+dgu2-dgu3:return 330 gosub 310 340 dg1=dg:k1=k:k=k+sw:k2=k:gosub 310:if dg1*dg>0 then 340 350 k=(k1+k2)/2:gosub 310:if dg1*dg>0 then k1=k else k2=k 360 if abs(dg)>1E-10 then 350 370 x(1,2)=xmu:y(1,2)=ymu:r(2)=ru:masx=1200/x(0,1):masy=850/y(0,2) 380 if masx<masy then mass=masx else mass=masy 390 goto 410 400 xbu=x*mass:ybu=y*mass:return 410 x=x(0,0):y=y(0,0):gosub 400:xba=xbu:yba=ybu:for a=1 to 3:ia=a:if ia=3 then ia=0 420 x=x(0,ia):y=y(0,ia):gosub 400:xbn=xbu:ybn=ybu:goto 440 430 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 440 gosub 430:next a:for a=0 to 2:x=x(1,a):y=y(1,a):gosub 400:circle xbu,ybu,r(a)*mass:next a quant circle-how to solve this tricky geometry problem with 3 circles lbh=10 lah=3 lab=13 lbc=12lac5 9.23076923% 3.84615385 > run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" at the start.

Nice question Your videos on geometry question Always have an another concept Your videos on geometry question always give the better experience Keep going sir 👍

Linda resolução! 🎉🎉🎉

Thank you for your support solve this problem

👍

Keep supporting the community

great approach....

Sir BE is equal to 4√3

Good. Math animation sir❤

円Aと円Ｂは接してないんですね

Muito interessante! Parabéns!

Parabéns pelo trabalho! Linda questão!

Nice question sir Your work on geometry question always give an better experience Keep going sir

Nice question and thanks for support

Sir answer is wrong. Correct answer is 7cm. Mistake is : BE is not equal to 7.. Rather AB is equal to 7cm..

As we are given the ratio of BH to HA as 10:3, we know that the actual length of BH and HA will be 10k and 3k, with k being an as-yet unknown constant. As BH and FB are tangents to circle O that intersect at B, then by the two tangents theorem, FB = BH = 10k. Similarly, as HA and AD are tangents to circle O that intersect at A so AD = HA = 3k. As AC and CB are tangent to circle O at D and F respectively, then ∠ODC = ∠CFO = 90°, as OD and OF are radii of circle O. Since ∠DCF = 90° as well, then ∠FOD must equal 360°-3(90°) = 90°, and as OF and OD are equal in length and adjacent, quadrilateral ODCF is a square with side length of 2. The side lengths of triangle ∆ACB are thus as follows: BA = 10k+3k = 13k, AC = 3k+2, CB = 2+10k. Triangle ∆ACB: AC² + CB² = BA² (3k+2)² + (10k+2)² = (13k)² 9k² + 12k + 4 + 100k² + 40k + 4 = 169k² 109k² + 52k + 8 = 169k² 60k² - 52k - 8 = 0 15k² - 13k - 2 = 0 15k² - 15k + 2k - 2 = 0 15k(k-1) + 2(k-1) = 0 (k-1)(15k+2) = 0 k - 1 = 0 | 15k + 2 = 0 k = 1 | k = -2/15 ❌ k ≥ 0 BA = 13(1) = 13 AC = 3(1) + 2 = 5 CB = 2 + 10(1) = 12 As CB is tangent to circle Q at G, then ∠QGB = 90°, as QG is a radius of circle Q. As AC is tangent to circle P at E, then ∠AEP = 90°, as PE is a radius of circle P. As PE and OD are both perpendicular to AC and thus parallel to each other, then ∠EPA and ∠DOA are corresponding angles and thus congruent. As ∠PAE = ∠OAD, then ∆AEP and ∆ADO are similar triangles by angle-angle similarity. As the point of tangency between two circles is collinear with the two centers, then OP = 2+r₁. Triangle ∆ADO: AD² + OD² = OA² 3² + 2² = OA² OA² = 9 + 4 = 13 OA = √13 OD/OA = PE/PA 2/√13 = r₁/(√13-(2+r₁)) √13r₁ = 2(√13-2-r₁) √13r₁ = 2√13 - 4 - 2r₁ √13r₁ + 2r₁ = 2√13 - 4 r₁(√13+2) = 2√13 - 4 r₁ = (2√13-4)/(√13+2) r₁ = (2√13-4)(√13-2)/(√13+2)(√13-2) r₁ = (26-4√13-4√13+8)/(13-4) [ r₁ = (34-8√13)/9 ≈ 0.573cm ] As OF and QG are both perpendicular to CB and thus parallel to each other, then ∠BQG and ∠BOF are corresponding angles and thus congruent. As ∠GBQ = ∠FBO, then ∆QGB and ∆OFB are similar triangles by angle-angle similarity. As the point of tangency between two circles is collinear with the two centers, then OQ = 2+r₂. Triangle ∆OFB: OF² + FB² = OB² 2² + 10² = OB² OB² = 4 + 100 = 104 OB = √104 = 2√26 OF/OB = QG/QB 2/2√26 = r₂/(2√26-(2+r₂)) 1/√26 = r₂/(2√26-2-r₂) √26r₂ = 2√26 - 2 - r₂ √26r₂ + r₂ = 2√26 - 2 r₂(√26+1) = 2√26 - 2 r₂ = (2√26-2)/(√26+1) r₂ = (2√26-2)(√26-1)/(√26+1)(√26-1) r₂ = (52-2√26-2√26+2)/(26-1) [ r₂ = (54-4√26)/25 ≈ 1.344cm ]

Very nice

very smart thank you

asnwer=(b) isit

Nice question

👍

Nice video sir