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Question-216188




Question Number 216188 by BaliramKumar last updated on 29/Jan/25
Answered by som(math1967) last updated on 29/Jan/25
 AC:BC=5:12  if AC=5cm BC=12cm  then AC+BC=17=AB  not possible  AC=10cm    BC=24cm   then AC+AB=10+17=27>BC  so possible perimeter   10+17+24=51cm
AC:BC=5:12ifAC=5cmBC=12cmthenAC+BC=17=ABnotpossibleAC=10cmBC=24cmthenAC+AB=10+17=27>BCsopossibleperimeter10+17+24=51cm
Answered by Rasheed.Sindhi last updated on 29/Jan/25
AB=AD+BD=5+12=17  AC:BC=5:12  let AC=5m & BC=12m; m∈N  AC+BC>AB_    5m+12m>17⇒17m>17⇒m>1     AC+AB>BC_    5m+17>12m⇒7m<17⇒m<3  ∴ m=2  ∴ AC=10 & BC=24  Perimeter=AB+BC+AC                     =17+24+10=51cm
AB=AD+BD=5+12=17AC:BC=5:12letAC=5m&BC=12m;mNAC+BC>AB5m+12m>1717m>17m>1AC+AB>BC5m+17>12m7m<17m<3m=2AC=10&BC=24Perimeter=AB+BC+AC=17+24+10=51cm
Answered by dionigi last updated on 31/Jan/25
((AC)/(BC)) = ((AD)/(BD)) = (5/(12))  ((AC)/(AD)) = ((BC)/(BD)) = x (x≥1)    {: (((AC,BC) ∈ N^2 )),((∄ m, 5 mod m = 12 mod m = 0)) } ⇒ x ∈ N  p = AC+AD+BD+BC  p = (AD+BD)(1+x)  p = 17 (1+x)  p = 17 n (n = x+1 ≥2, n ∈ N)  n = 2 ⇒ p = 34  n = 3 ⇒ p = 51  n = 4 ⇒ p = 68
ACBC=ADBD=512ACAD=BCBD=x(x1)(AC,BC)N2m,5modm=12modm=0}xNp=AC+AD+BD+BCp=(AD+BD)(1+x)p=17(1+x)p=17n(n=x+12,nN)n=2p=34n=3\boldsymbolp=51n=4\boldsymbolp=68
Commented by Rasheed.Sindhi last updated on 31/Jan/25
For p=34 , 68 what are the values  of the sides of the triangle?
For\boldsymbolp=34,68whatarethevaluesofthesidesofthetriangle?
Commented by dionigi last updated on 01/Feb/25
for p = 34  ∠CAB = ∠ABC = 0 ; ∠ACB = π  AC = 5 ; BC = 12 ; AB = 17  ∠CAB + ∠ABC + ∠ACB = π  AC + BC + AB = 34  set Δ the line bisector of ∠ACB  D=C ⇒ D ∈ Δ  and ((AC)/(BC)) = (5/(12))    but p = 68 is not a solution  ((AC)/(BC)) = (5/(12)) is not a sufficient pattern  BC ≤ AB + AC is mandatory also
forp=34CAB=ABC=0;ACB=πAC=5;BC=12;AB=17CAB+ABC+ACB=πAC+BC+AB=34setΔthelinebisectorofACBD=CDΔandACBC=512butp=68isnotasolutionACBC=512isnotasufficientpatternBCAB+ACismandatoryalso
Commented by som(math1967) last updated on 01/Feb/25
BC<AB+AC, not≤  only solution is 51
BC<AB+AC,notonlysolutionis51
Commented by Rasheed.Sindhi last updated on 01/Feb/25
But ((AC)/(BC))=(5/(22))≠(5/(12))
ButACBC=522512
Commented by dionigi last updated on 01/Feb/25
BC = AB+AC remains correct  This means that C, A, D, B are aligned   on the same line Δ  AC = 5, BC = 22 and AB = 17  AC, BC, and AB are all integers.  The 3 angles of the triangle are   ∠ACB = ∠ABC = 0 and ∠CAB = π  Line Δ is bisector of ∠ACB and D ∈ Δ  All the patterns of the initial  problem   are true, but perimeter p = 44 was   not proposed in the choice.
BC=AB+ACremainscorrectThismeansthatC,A,D,BarealignedonthesamelineΔAC=5,BC=22andAB=17AC,BC,andABareallintegers.The3anglesofthetriangleareACB=ABC=0andCAB=πLineΔisbisectorofACBandDΔAllthepatternsoftheinitialproblemaretrue,butperimeterp=44wasnotproposedinthechoice.
Commented by dionigi last updated on 01/Feb/25
Right !
Right!Right!

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