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Question Number 189558 by normans last updated on 18/Mar/23
Answered by a.lgnaoui last updated on 19/Mar/23
•(BCEF)est un trapeze Aire(BCEF)=(((BC+EF)/2))×CE =(((2,5+y)(x+2))/2)=A_1 +A_2 +9(1) A_1 =((2,5x)/2) ⇒(((2,5+y)(x+2))/2)=((2,5)/2)x+9+A_2 ⇒ [y(x+2)=13+2A_2 ] 2•(ABGF) Aire=A_4 +A_5 +9 (Aire(HIF)=9) =(((AB+FG))/2)×AG=(((y+7,5)(x+2))/2) Aire(ABGF)=((xy+2y+7,5x+15)/2) (2) A_4 =Aire[(AHCD)−(A_1 +9)] Aire(AHCD)=(((3+x)(y+5))/2) ⇒A_4 =Aire(AHBI)=(((3+x)(y+5))/2)−(9+((2,5)/2)x) =((3y+15+xy+5x)/2)−((18+2,5x)/2)=(((2,5x+3y−3)+xy)/2) A_4 =((3y+xy+2,5x−3)/2) A_3 =((5(x−1))/2) Aire(ABGF)=((2y+7,5x+15+xy)/2)=9+((7,5x+3y+xy−8)/2) y−5=0 ⇒ y=5 Aire du rectangle=(2+x)(y+5) =Aire(ABGF)+ire(BCFE) : (1+2) =((2y+xy+7,5x+15)/2)+((2,5x+5x+5+xy)/2) =((15x+20+2xy)/2)=((25x+20)/2) Aire rectangle=12,5x+10 (3) 12,5x+10=10x+20 2,5x=10⇒x=4 donc (x,y)=(4,5) ⇒ Aire(rectsngle)= (2+x)(5+y)= (4+2)(5+5)=60 −−−−−− Aire(Rectangle)=60
$$\bullet\left({BCEF}\right){est}\:{un}\:{trapeze} \\ $$$$\:\:{Aire}\left({BCEF}\right)=\left(\frac{{BC}+{EF}}{\mathrm{2}}\right)×{CE} \\ $$$$=\frac{\left(\mathrm{2},\mathrm{5}+{y}\right)\left({x}+\mathrm{2}\right)}{\mathrm{2}}={A}_{\mathrm{1}} +{A}_{\mathrm{2}} +\mathrm{9}\left(\mathrm{1}\right) \\ $$$$\:\:{A}_{\mathrm{1}} =\frac{\mathrm{2},\mathrm{5}{x}}{\mathrm{2}}\:\: \\ $$$$ \\ $$$$\Rightarrow\frac{\left(\mathrm{2},\mathrm{5}+{y}\right)\left({x}+\mathrm{2}\right)}{\mathrm{2}}=\frac{\mathrm{2},\mathrm{5}}{\mathrm{2}}{x}+\mathrm{9}+{A}_{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\left[{y}\left({x}+\mathrm{2}\right)=\mathrm{13}+\mathrm{2}{A}_{\mathrm{2}} \:\right]\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{2}\bullet\left({ABGF}\right) \\ $$$$\:\:{Aire}={A}_{\mathrm{4}} +{A}_{\mathrm{5}} +\mathrm{9}\:\:\left({Aire}\left({HIF}\right)=\mathrm{9}\right) \\ $$$$=\frac{\left({AB}+{FG}\right)}{\mathrm{2}}×{AG}=\frac{\left({y}+\mathrm{7},\mathrm{5}\right)\left({x}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$$ \\ $$$${Aire}\left({ABGF}\right)=\frac{{xy}+\mathrm{2}{y}+\mathrm{7},\mathrm{5}{x}+\mathrm{15}}{\mathrm{2}}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$${A}_{\mathrm{4}} ={Aire}\left[\left({AHCD}\right)−\left({A}_{\mathrm{1}} +\mathrm{9}\right)\right] \\ $$$${Aire}\left({AHCD}\right)=\frac{\left(\mathrm{3}+{x}\right)\left({y}+\mathrm{5}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{A}_{\mathrm{4}} ={Aire}\left({AHBI}\right)=\frac{\left(\mathrm{3}+{x}\right)\left({y}+\mathrm{5}\right)}{\mathrm{2}}−\left(\mathrm{9}+\frac{\mathrm{2},\mathrm{5}}{\mathrm{2}}{x}\right) \\ $$$$=\frac{\mathrm{3}{y}+\mathrm{15}+{xy}+\mathrm{5}{x}}{\mathrm{2}}−\frac{\mathrm{18}+\mathrm{2},\mathrm{5}{x}}{\mathrm{2}}=\frac{\left(\mathrm{2},\mathrm{5}{x}+\mathrm{3}{y}−\mathrm{3}\right)+{xy}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{A}_{\mathrm{4}} =\frac{\mathrm{3}{y}+{xy}+\mathrm{2},\mathrm{5}{x}−\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{A}_{\mathrm{3}} =\frac{\mathrm{5}\left({x}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${Aire}\left({ABGF}\right)=\frac{\mathrm{2}{y}+\mathrm{7},\mathrm{5}{x}+\mathrm{15}+{xy}}{\mathrm{2}}=\mathrm{9}+\frac{\mathrm{7},\mathrm{5}{x}+\mathrm{3}{y}+{xy}−\mathrm{8}}{\mathrm{2}} \\ $$$${y}−\mathrm{5}=\mathrm{0}\:\:\:\:\Rightarrow\:\:{y}=\mathrm{5} \\ $$$${Aire}\:{du}\:{rectangle}=\left(\mathrm{2}+{x}\right)\left({y}+\mathrm{5}\right) \\ $$$$={Aire}\left({ABGF}\right)+{ire}\left({BCFE}\right)\::\:\left(\mathrm{1}+\mathrm{2}\right) \\ $$$$=\frac{\mathrm{2}{y}+{xy}+\mathrm{7},\mathrm{5}{x}+\mathrm{15}}{\mathrm{2}}+\frac{\mathrm{2},\mathrm{5}{x}+\mathrm{5}{x}+\mathrm{5}+{xy}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{15}{x}+\mathrm{20}+\mathrm{2}{xy}}{\mathrm{2}}=\frac{\mathrm{25}{x}+\mathrm{20}}{\mathrm{2}} \\ $$$$ \\ $$$${Aire}\:{rectangle}=\mathrm{12},\mathrm{5}{x}+\mathrm{10}\:\:\:\:\left(\mathrm{3}\right)\: \\ $$$$\mathrm{12},\mathrm{5}{x}+\mathrm{10}=\mathrm{10}{x}+\mathrm{20} \\ $$$$\mathrm{2},\mathrm{5}{x}=\mathrm{10}\Rightarrow{x}=\mathrm{4} \\ $$$${donc}\:\:\:\left({x},{y}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$$ \\ $$$$\:\Rightarrow\:\:\:{Aire}\left({rectsngle}\right)= \\ $$$$\:\:\:\:\:\:\left(\mathrm{2}+{x}\right)\left(\mathrm{5}+{y}\right)= \\ $$$$\:\:\:\:\:\left(\mathrm{4}+\mathrm{2}\right)\left(\mathrm{5}+\mathrm{5}\right)=\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−− \\ $$$$\:\boldsymbol{{Aire}}\left(\boldsymbol{{Rectangle}}\right)=\mathrm{60} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 19/Mar/23
Answered by mr W last updated on 19/Mar/23
Commented by mr W last updated on 19/Mar/23
0.5×3×(2.5+x)+0.5×5y+A+9=0.5×(5+2.5+x)(3+y) ⇒2.5y+xy=3+2A ...(i) 0.5×2.5×(y+1)+9+B+0.5×2x=0.5×(2.5+x)(3+y) ⇒x+xy=13+2B ...(ii) 0.5×3×(2.5+x)+A+9+0.5×2.5×(y+1)=0.5×(3+y+1)(5+x) ⇒x+2.5y+xy=8+2A ...(iii) 0.5×5y+9+B+0.5×2x=0.5×(2+y)×(5+x) ⇒xy=8+2B ...(iv) (iii)−(i) or (ii)−(iv): ⇒x=5 from (i): 7.5y=3+2A ⇒A=((7.5y−3)/2) from (iv): 5y=8+2B ⇒B=((5y−8)/2) (A/9)=(9/B) ⇒A×B=9×9=81 ((7.5y−3)/2)×((5y−8)/2)=81 y^2 −2y−8=0 ⇒y=4 reactangle (5+x)(3+y)=10×7=70 ✓
$$\mathrm{0}.\mathrm{5}×\mathrm{3}×\left(\mathrm{2}.\mathrm{5}+{x}\right)+\mathrm{0}.\mathrm{5}×\mathrm{5}{y}+{A}+\mathrm{9}=\mathrm{0}.\mathrm{5}×\left(\mathrm{5}+\mathrm{2}.\mathrm{5}+{x}\right)\left(\mathrm{3}+{y}\right) \\ $$$$\Rightarrow\mathrm{2}.\mathrm{5}{y}+{xy}=\mathrm{3}+\mathrm{2}{A}\:\:\:…\left({i}\right) \\ $$$$\mathrm{0}.\mathrm{5}×\mathrm{2}.\mathrm{5}×\left({y}+\mathrm{1}\right)+\mathrm{9}+{B}+\mathrm{0}.\mathrm{5}×\mathrm{2}{x}=\mathrm{0}.\mathrm{5}×\left(\mathrm{2}.\mathrm{5}+{x}\right)\left(\mathrm{3}+{y}\right) \\ $$$$\Rightarrow{x}+{xy}=\mathrm{13}+\mathrm{2}{B}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{0}.\mathrm{5}×\mathrm{3}×\left(\mathrm{2}.\mathrm{5}+{x}\right)+{A}+\mathrm{9}+\mathrm{0}.\mathrm{5}×\mathrm{2}.\mathrm{5}×\left({y}+\mathrm{1}\right)=\mathrm{0}.\mathrm{5}×\left(\mathrm{3}+{y}+\mathrm{1}\right)\left(\mathrm{5}+{x}\right) \\ $$$$\Rightarrow{x}+\mathrm{2}.\mathrm{5}{y}+{xy}=\mathrm{8}+\mathrm{2}{A}\:\:\:…\left({iii}\right) \\ $$$$\mathrm{0}.\mathrm{5}×\mathrm{5}{y}+\mathrm{9}+{B}+\mathrm{0}.\mathrm{5}×\mathrm{2}{x}=\mathrm{0}.\mathrm{5}×\left(\mathrm{2}+{y}\right)×\left(\mathrm{5}+{x}\right) \\ $$$$\Rightarrow{xy}=\mathrm{8}+\mathrm{2}{B}\:\:\:…\left({iv}\right) \\ $$$$\left({iii}\right)−\left({i}\right)\:{or}\:\left({ii}\right)−\left({iv}\right): \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{7}.\mathrm{5}{y}=\mathrm{3}+\mathrm{2}{A}\:\:\Rightarrow{A}=\frac{\mathrm{7}.\mathrm{5}{y}−\mathrm{3}}{\mathrm{2}} \\ $$$${from}\:\left({iv}\right): \\ $$$$\mathrm{5}{y}=\mathrm{8}+\mathrm{2}{B}\:\Rightarrow{B}=\frac{\mathrm{5}{y}−\mathrm{8}}{\mathrm{2}} \\ $$$$\frac{{A}}{\mathrm{9}}=\frac{\mathrm{9}}{{B}}\:\Rightarrow{A}×{B}=\mathrm{9}×\mathrm{9}=\mathrm{81} \\ $$$$\frac{\mathrm{7}.\mathrm{5}{y}−\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{5}{y}−\mathrm{8}}{\mathrm{2}}=\mathrm{81} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{4} \\ $$$${reactangle}\:\left(\mathrm{5}+{x}\right)\left(\mathrm{3}+{y}\right)=\mathrm{10}×\mathrm{7}=\mathrm{70}\:\checkmark \\ $$

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