Question Number 180256 by mr W last updated on 09/Nov/22
Commented by HeferH last updated on 09/Nov/22
Answered by HeferH last updated on 10/Nov/22
$$\:{d}\:+\:{c}\:=\:? \\ $$$$\:\bullet\:{a}\:+\:{b}\:=\:\mathrm{8} \\ $$$$\:\bullet\:{b}+\:{c}\:=\:\mathrm{11}\:\:\:\left(+\right) \\ $$$$\:\bullet\:{a}+\:{d}\:=\:\mathrm{9} \\ $$$$\:{a}\:+\:{b}\:+\:{d}\:+\:{c}\:=\:\mathrm{20} \\ $$$$\:\mathrm{8}\:+\:{d}\:+\:{c}\:=\:\mathrm{20} \\ $$$$\:{d}\:+\:{c}\:=\:\mathrm{12}^{\ast\:} \\ $$$$\: \\ $$
Commented by mr W last updated on 10/Nov/22
$${right}\:{sir}!\:{thanks}! \\ $$
Answered by a.lgnaoui last updated on 10/Nov/22
![AC∣∣EH EHA=90 Aire [ACEH]=(((AC+EH)/2))×AH=a(((2c+a(√3))/2))=ac+(a^2 /2)(√3) Aire [EHG]=((a^2 (√3))/2) Aire totale=ac+a^2 (√3) =28+x (x=Aire (FIHG)) calul de x x=HI×a+FH×IF(∡FGH=60) (1) Aire[ABHI] =11=((AB+HI)/2)×a tan 30=((HI)/a) HI=(a/3)(√(3 )) x=((2a^2 )/3)(√3) (2×Aire [HIH]) (2) (2) ac+a^2 (√3) =28+2(a^2 /3)(√3) 11=(((((a(√3))/3)+c))/2)a⇒22=((a^2 (√3))/3) +ac ac=22−((a^2 (√3))/3) 28+x=(22−((a^2 (√3))/3) )+a^2 (√3) x=2((a^2 (√3))/3)−6=((a^2 (√3))/3) ((a^2 (√3))/3)=6 a^2 =((18(√3))/3)=6(√3) Donc Aire =6](https://www.tinkutara.com/question/Q180320.png)
$${AC}\mid\mid{EH}\:\:\:{EHA}=\mathrm{90} \\ $$$${Aire}\:\left[{ACEH}\right]=\left(\frac{{AC}+{EH}}{\mathrm{2}}\right)×{AH}={a}\left(\frac{\mathrm{2}{c}+{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)={ac}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$${Aire}\:\left[{EHG}\right]=\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${Aire}\:{totale}={ac}+{a}^{\mathrm{2}} \sqrt{\mathrm{3}}\:=\mathrm{28}+{x}\:\left({x}={Aire}\:\left({FIHG}\right)\right) \\ $$$${calul}\:{de}\:{x} \\ $$$${x}={HI}×{a}+{FH}×{IF}\left(\measuredangle{FGH}=\mathrm{60}\right)\:\:\left(\mathrm{1}\right) \\ $$$${Aire}\left[{ABHI}\right]\:\:\:=\mathrm{11}=\frac{{AB}+{HI}}{\mathrm{2}}×{a} \\ $$$$\mathrm{tan}\:\mathrm{30}=\frac{{HI}}{{a}}\:\:\:{HI}=\frac{{a}}{\mathrm{3}}\sqrt{\mathrm{3}\:}\:\:\:\:\:\:{x}=\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}\sqrt{\mathrm{3}}\:\left(\mathrm{2}×{Aire}\:\left[{HIH}\right]\right)\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:\:{ac}+{a}^{\mathrm{2}} \sqrt{\mathrm{3}}\:\:=\mathrm{28}+\mathrm{2}\frac{{a}^{\mathrm{2}} }{\mathrm{3}}\sqrt{\mathrm{3}} \\ $$$$\mathrm{11}=\frac{\left(\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{3}}+{c}\right)}{\mathrm{2}}{a}\Rightarrow\mathrm{22}=\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}}\:+{ac} \\ $$$${ac}=\mathrm{22}−\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\mathrm{28}+{x}=\left(\mathrm{22}−\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}}\:\right)+{a}^{\mathrm{2}} \sqrt{\mathrm{3}} \\ $$$$\:\:{x}=\mathrm{2}\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}}−\mathrm{6}=\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:\:\:\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{3}}=\mathrm{6}\:\:\:\:\:{a}^{\mathrm{2}} =\frac{\mathrm{18}\sqrt{\mathrm{3}}}{\mathrm{3}}=\mathrm{6}\sqrt{\mathrm{3}}\:\:\: \\ $$$$\:\:\:\:{Donc}\:\:\:\:\:\:\:\:{Aire}\:=\mathrm{6} \\ $$$$ \\ $$
Answered by mr W last updated on 10/Nov/22
Commented by mr W last updated on 10/Nov/22
$${A}_{\mathrm{1}} ={B}_{\mathrm{1}} \\ $$$${A}_{\mathrm{2}} ={D}_{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} ={B}_{\mathrm{2}} \\ $$$${C}_{\mathrm{1}} ={D}_{\mathrm{1}} \\ $$$$\Rightarrow{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{C}_{\mathrm{1}} +{C}_{\mathrm{2}} ={B}_{\mathrm{1}} +{B}_{\mathrm{2}} +{D}_{\mathrm{1}} +{D}_{\mathrm{2}} \\ $$$$\Rightarrow{A}+{C}={B}+{D} \\ $$$$\mathrm{8}+{x}=\mathrm{9}+\mathrm{11} \\ $$$$\Rightarrow{x}=\mathrm{12}\:\checkmark \\ $$
Commented by a.lgnaoui last updated on 10/Nov/22
![thanks aplication: the low of mediane in triangle Aire (A1)=Aire (B1)... with Aire[ABI]=[Aire (A1)]+[Aire(B1)] finaly x=C1+C2](https://www.tinkutara.com/question/Q180332.png)
$${thanks}\:\: \\ $$$${aplication}:\:{the}\:{low}\:{of}\:{mediane}\:{in} \\ $$$${triangle}\:{Aire}\:\left({A}\mathrm{1}\right)={Aire}\:\left({B}\mathrm{1}\right)… \\ $$$${with}\:{Aire}\left[{ABI}\right]=\left[{Aire}\:\left({A}\mathrm{1}\right)\right]+\left[{Aire}\left({B}\mathrm{1}\right)\right] \\ $$$${finaly}\:\:{x}={C}\mathrm{1}+{C}\mathrm{2} \\ $$$$ \\ $$