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Question-216188

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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
$$\:{AC}:{BC}=\mathrm{5}:\mathrm{12} \\ $$$${if}\:{AC}=\mathrm{5}{cm}\:{BC}=\mathrm{12}{cm} \\ $$$${then}\:{AC}+{BC}=\mathrm{17}={AB} \\ $$$${not}\:{possible} \\ $$$${AC}=\mathrm{10}{cm} \\ $$$$\:\:{BC}=\mathrm{24}{cm} \\ $$$$\:{then}\:{AC}+{AB}=\mathrm{10}+\mathrm{17}=\mathrm{27}>{BC} \\ $$$${so}\:{possible}\:{perimeter} \\ $$$$\:\mathrm{10}+\mathrm{17}+\mathrm{24}=\mathrm{51}{cm} \\ $$
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
$${AB}={AD}+{BD}=\mathrm{5}+\mathrm{12}=\mathrm{17} \\ $$$${AC}:{BC}=\mathrm{5}:\mathrm{12} \\ $$$${let}\:{AC}=\mathrm{5}{m}\:\&\:{BC}=\mathrm{12}{m};\:{m}\in\mathbb{N} \\ $$$$\underline{{AC}+{BC}>{AB}_{\:} } \\ $$$$\mathrm{5}{m}+\mathrm{12}{m}>\mathrm{17}\Rightarrow\mathrm{17}{m}>\mathrm{17}\Rightarrow{m}>\mathrm{1} \\ $$$$\: \\ $$$$\underline{{AC}+{AB}>{BC}_{\:} } \\ $$$$\mathrm{5}{m}+\mathrm{17}>\mathrm{12}{m}\Rightarrow\mathrm{7}{m}<\mathrm{17}\Rightarrow{m}<\mathrm{3} \\ $$$$\therefore\:{m}=\mathrm{2} \\ $$$$\therefore\:{AC}=\mathrm{10}\:\&\:{BC}=\mathrm{24} \\ $$$${Perimeter}={AB}+{BC}+{AC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{17}+\mathrm{24}+\mathrm{10}=\mathrm{51}{cm} \\ $$
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
$$\frac{{AC}}{{BC}}\:=\:\frac{{AD}}{{BD}}\:=\:\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\frac{{AC}}{{AD}}\:=\:\frac{{BC}}{{BD}}\:=\:{x}\:\left({x}\geqslant\mathrm{1}\right)\: \\ $$$$\left.\begin{matrix}{\left({AC},{BC}\right)\:\in\:\mathbb{N}^{\mathrm{2}} }\\{\nexists\:{m},\:\mathrm{5}\:\mathrm{mod}\:{m}\:=\:\mathrm{12}\:\mathrm{mod}\:{m}\:=\:\mathrm{0}}\end{matrix}\right\}\:\Rightarrow\:{x}\:\in\:\mathbb{N} \\ $$$${p}\:=\:{AC}+{AD}+{BD}+{BC} \\ $$$${p}\:=\:\left({AD}+{BD}\right)\left(\mathrm{1}+{x}\right) \\ $$$${p}\:=\:\mathrm{17}\:\left(\mathrm{1}+{x}\right) \\ $$$${p}\:=\:\mathrm{17}\:{n}\:\left({n}\:=\:{x}+\mathrm{1}\:\geqslant\mathrm{2},\:{n}\:\in\:\mathbb{N}\right) \\ $$$${n}\:=\:\mathrm{2}\:\Rightarrow\:{p}\:=\:\mathrm{34} \\ $$$${n}\:=\:\mathrm{3}\:\Rightarrow\:\boldsymbol{{p}}\:=\:\mathrm{51} \\ $$$${n}\:=\:\mathrm{4}\:\Rightarrow\:\boldsymbol{{p}}\:=\:\mathrm{68} \\ $$
Commented by Rasheed.Sindhi last updated on 31/Jan/25
For p=34 , 68 what are the values of the sides of the triangle?
$${For}\:\boldsymbol{{p}}=\mathrm{34}\:,\:\mathrm{68}\:{what}\:{are}\:{the}\:{values} \\ $$$${of}\:{the}\:{sides}\:{of}\:{the}\:{triangle}? \\ $$
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
$${for}\:{p}\:=\:\mathrm{34} \\ $$$$\angle{CAB}\:=\:\angle{ABC}\:=\:\mathrm{0}\:;\:\angle{ACB}\:=\:\pi \\ $$$${AC}\:=\:\mathrm{5}\:;\:{BC}\:=\:\mathrm{12}\:;\:{AB}\:=\:\mathrm{17} \\ $$$$\angle{CAB}\:+\:\angle{ABC}\:+\:\angle{ACB}\:=\:\pi \\ $$$${AC}\:+\:{BC}\:+\:{AB}\:=\:\mathrm{34} \\ $$$${set}\:\Delta\:{the}\:{line}\:{bisector}\:{of}\:\angle{ACB} \\ $$$${D}={C}\:\Rightarrow\:{D}\:\in\:\Delta \\ $$$${and}\:\frac{{AC}}{{BC}}\:=\:\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$ \\ $$$${but}\:{p}\:=\:\mathrm{68}\:{is}\:{not}\:{a}\:{solution} \\ $$$$\frac{{AC}}{{BC}}\:=\:\frac{\mathrm{5}}{\mathrm{12}}\:{is}\:{not}\:{a}\:{sufficient}\:{pattern} \\ $$$${BC}\:\leqslant\:{AB}\:+\:{AC}\:{is}\:{mandatory}\:{also} \\ $$
Commented by som(math1967) last updated on 01/Feb/25
BC<AB+AC, not≤ only solution is 51
$${BC}<{AB}+{AC},\:{not}\leqslant \\ $$$${only}\:{solution}\:{is}\:\mathrm{51} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Feb/25
But ((AC)/(BC))=(5/(22))≠(5/(12))
$${But}\:\frac{{AC}}{{BC}}=\frac{\mathrm{5}}{\mathrm{22}}\neq\frac{\mathrm{5}}{\mathrm{12}} \\ $$
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.
$${BC}\:=\:{AB}+{AC}\:{remains}\:{correct} \\ $$$${This}\:{means}\:{that}\:{C},\:{A},\:{D},\:{B}\:{are}\:{aligned}\: \\ $$$${on}\:{the}\:{same}\:{line}\:\Delta \\ $$$${AC}\:=\:\mathrm{5},\:{BC}\:=\:\mathrm{22}\:{and}\:{AB}\:=\:\mathrm{17} \\ $$$${AC},\:{BC},\:{and}\:{AB}\:{are}\:{all}\:{integers}. \\ $$$${The}\:\mathrm{3}\:{angles}\:{of}\:{the}\:{triangle}\:{are}\: \\ $$$$\angle{ACB}\:=\:\angle{ABC}\:=\:\mathrm{0}\:{and}\:\angle{CAB}\:=\:\pi \\ $$$${Line}\:\Delta\:{is}\:{bisector}\:{of}\:\angle{ACB}\:{and}\:{D}\:\in\:\Delta \\ $$$${All}\:{the}\:{patterns}\:{of}\:{the}\:{initial}\:\:{problem}\: \\ $$$${are}\:{true},\:{but}\:{perimeter}\:{p}\:=\:\mathrm{44}\:{was}\: \\ $$$${not}\:{proposed}\:{in}\:{the}\:{choice}. \\ $$
Commented by dionigi last updated on 01/Feb/25
Right !
$${Right}\:! \\ $$

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