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Question Number 177128 by cortano1 last updated on 01/Oct/22

Commented by cortano1 last updated on 01/Oct/22

all angles =120° . Find the perimeter

$$\mathrm{all}\:\mathrm{angles}\:=\mathrm{120}°\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{perimeter} \\ $$

Answered by mr W last updated on 01/Oct/22

Commented by cortano1 last updated on 01/Oct/22

what is the formula sir?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{sir}? \\ $$

Commented by mr W last updated on 02/Oct/22

6+4=b+9 ⇒b=1 a+b=4+8 ⇒a+b=12 ⇒a=11 perimeter=6+4+8+9+a+b=39 ✓

$$\mathrm{6}+\mathrm{4}={b}+\mathrm{9}\:\Rightarrow{b}=\mathrm{1} \\ $$$${a}+{b}=\mathrm{4}+\mathrm{8}\:\Rightarrow{a}+{b}=\mathrm{12}\:\Rightarrow{a}=\mathrm{11} \\ $$$${perimeter}=\mathrm{6}+\mathrm{4}+\mathrm{8}+\mathrm{9}+{a}+{b}=\mathrm{39}\:\checkmark \\ $$

Commented by Rasheed.Sindhi last updated on 02/Oct/22

Excellent sir!

$$\mathcal{E}{xcellent}\:\boldsymbol{{sir}}! \\ $$

Commented by cortano1 last updated on 02/Oct/22

nice

$$\mathrm{nice} \\ $$

Commented by Tawa11 last updated on 02/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 01/Oct/22

Commented by mr W last updated on 01/Oct/22

BC=4+8+9=21 AB=4+6+a=21 ⇒a=11 AC=a+b+9=21 ⇒b=1 perimeter=6+4+8+9+a+b=39 ✓

$${BC}=\mathrm{4}+\mathrm{8}+\mathrm{9}=\mathrm{21} \\ $$$${AB}=\mathrm{4}+\mathrm{6}+{a}=\mathrm{21}\:\Rightarrow{a}=\mathrm{11} \\ $$$${AC}={a}+{b}+\mathrm{9}=\mathrm{21}\:\Rightarrow{b}=\mathrm{1} \\ $$$${perimeter}=\mathrm{6}+\mathrm{4}+\mathrm{8}+\mathrm{9}+{a}+{b}=\mathrm{39}\:\checkmark \\ $$

Commented by cortano1 last updated on 01/Oct/22

(a/(21)) = (a/(a+b+9)) ⇒a+b+9=21 (9/(10+a)) = (9/(9+a+b)) ⇒10+a=9+a+b ⇒b=1 & a=11 perimeter = 6+8+4+11+1+9=39 it′ correct ?

$$\:\frac{\mathrm{a}}{\mathrm{21}}\:=\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{9}}\:\Rightarrow\mathrm{a}+\mathrm{b}+\mathrm{9}=\mathrm{21} \\ $$$$\frac{\mathrm{9}}{\mathrm{10}+\mathrm{a}}\:=\:\frac{\mathrm{9}}{\mathrm{9}+\mathrm{a}+\mathrm{b}}\:\Rightarrow\mathrm{10}+\mathrm{a}=\mathrm{9}+\mathrm{a}+\mathrm{b} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{1}\:\&\:\mathrm{a}=\mathrm{11} \\ $$$$\mathrm{perimeter}\:=\:\mathrm{6}+\mathrm{8}+\mathrm{4}+\mathrm{11}+\mathrm{1}+\mathrm{9}=\mathrm{39} \\ $$$$\mathrm{it}'\:\mathrm{correct}\:? \\ $$

Commented by mr W last updated on 01/Oct/22

ok. but you have directly 21=a+b+9, no need from (a/(21)) = (a/(a+b+9)). since ΔABC is equilateral.

$${ok}.\: \\ $$$${but}\:{you}\:{have}\:{directly}\:\mathrm{21}={a}+{b}+\mathrm{9},\: \\ $$$${no}\:{need}\:{from}\:\frac{\mathrm{a}}{\mathrm{21}}\:=\:\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{9}}. \\ $$$${since}\:\Delta{ABC}\:{is}\:{equilateral}. \\ $$

Answered by a.lgnaoui last updated on 01/Oct/22

Posons y=CD x=DE y=CD=HH′=HF+FH^′ d apres la figure ∡ ABO=∡BAO=60 donc OB=OA=AB=4 ⇒OC=10 sin 60=((CH)/(10))=((√3)/2)⇒CH=5(√3) cos 60=(1/2)=((4+AH)/(10))⇒AH=1A ⇒HF=7 calcul de FH^′ FEO^′ =EO^′ F=60 ⇒EF=EO^′ =9 OD=x+9 ⇒sin 60=((√3)/2)=((5(√3))/(9+x))⇒x=1 donc DE=1 FD^2 =9^2 +x^2 −18cos 120=82+9=91 ⇒FD=(√(91 )) FD=(√(FD^2 −FF′^2 )) =(√(91−75)) =4 ⇒F′D=4 alors Donc CD=7+4=11 finalement Peeimetre=4+6+8+9+1+11=39

$$\mathrm{P}{osons}\:\:{y}=\mathrm{CD}\:\:\:\:{x}=\mathrm{DE} \\ $$$${y}=\mathrm{CD}=\mathrm{HH}'=\mathrm{HF}+\mathrm{FH}^{'} \\ $$$${d}\:{apres}\:{la}\:{figure}\:\:\measuredangle\:\mathrm{ABO}=\measuredangle\mathrm{BAO}=\mathrm{60}\:{donc}\:\:\:\mathrm{OB}=\mathrm{OA}=\mathrm{AB}=\mathrm{4}\:\:\Rightarrow\mathrm{OC}=\mathrm{10} \\ $$$$\mathrm{sin}\:\mathrm{60}=\frac{\mathrm{CH}}{\mathrm{10}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\mathrm{CH}=\mathrm{5}\sqrt{\mathrm{3}} \\ $$$$\mathrm{cos}\:\mathrm{60}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{4}+\mathrm{AH}}{\mathrm{10}}\Rightarrow\mathrm{AH}=\mathrm{1A}\:\:\:\:\Rightarrow\mathrm{HF}=\mathrm{7} \\ $$$${calcul}\:{de}\:\mathrm{FH}^{'} \\ $$$$\mathrm{FEO}^{'} =\mathrm{EO}^{'} \mathrm{F}=\mathrm{60}\:\:\Rightarrow\mathrm{EF}=\mathrm{EO}^{'} =\mathrm{9} \\ $$$$\mathrm{OD}={x}+\mathrm{9}\:\:\:\Rightarrow\mathrm{sin}\:\mathrm{60}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{9}+{x}}\Rightarrow{x}=\mathrm{1} \\ $$$${donc}\:\:\:\mathrm{DE}=\mathrm{1} \\ $$$$\mathrm{FD}^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{18cos}\:\mathrm{120}=\mathrm{82}+\mathrm{9}=\mathrm{91}\:\Rightarrow\mathrm{FD}=\sqrt{\mathrm{91}\:}\: \\ $$$$\mathrm{FD}=\sqrt{\mathrm{FD}^{\mathrm{2}} −\mathrm{FF}'^{\mathrm{2}} \:}\:=\sqrt{\mathrm{91}−\mathrm{75}}\:=\mathrm{4}\:\:\Rightarrow\mathrm{F}'\mathrm{D}=\mathrm{4} \\ $$$${alors} \\ $$$$\mathrm{D}{onc}\:\:\mathrm{CD}=\mathrm{7}+\mathrm{4}=\mathrm{11} \\ $$$${finalement}\:\: \\ $$$${Peeimetre}=\mathrm{4}+\mathrm{6}+\mathrm{8}+\mathrm{9}+\mathrm{1}+\mathrm{11}=\mathrm{39} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 01/Oct/22

Commented by a.lgnaoui last updated on 01/Oct/22

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