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1-Evaluer-Aire-A-B-C-D-2-En-deduire-Aire-A-B-C-D-Aire-ABCD-




Question Number 189233 by a.lgnaoui last updated on 13/Mar/23
1•Evaluer :Aire(A′B′C′D′)  2•En deduire:((Aire(A′B′C′D′))/(Aire(ABCD)))
$$\mathrm{1}\bullet{Evaluer}\::\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right) \\ $$$$\mathrm{2}\bullet{En}\:{deduire}:\frac{\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right)}{\boldsymbol{{Aire}}\left(\boldsymbol{{ABCD}}\right)} \\ $$
Commented by a.lgnaoui last updated on 13/Mar/23
[A ′B′ D]alignes
$$\left[\boldsymbol{{A}}\:'\boldsymbol{{B}}'\:\boldsymbol{{D}}\right]{alignes} \\ $$
Commented by a.lgnaoui last updated on 13/Mar/23
Answered by HeferH last updated on 14/Mar/23
Commented by HeferH last updated on 14/Mar/23
 { ((3a = (b/( (√3))))),((a(√3) + b = 5(√3))) :}   •  3a(√3) + a(√3) =5(√3)   a = (5/4)⇒ b = ((15(√3))/4)   • a×b = ((75(√3))/(16))   • ((Area(A′B′C′D′))/(Area(ABCD))) = ((75(√3))/(16)) × (1/(25(√3))) = (3/(16))
$$\begin{cases}{\mathrm{3a}\:=\:\frac{\mathrm{b}}{\:\sqrt{\mathrm{3}}}}\\{\mathrm{a}\sqrt{\mathrm{3}}\:+\:\mathrm{b}\:=\:\mathrm{5}\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\:\bullet\:\:\mathrm{3a}\sqrt{\mathrm{3}}\:+\:\mathrm{a}\sqrt{\mathrm{3}}\:=\mathrm{5}\sqrt{\mathrm{3}} \\ $$$$\:\mathrm{a}\:=\:\frac{\mathrm{5}}{\mathrm{4}}\Rightarrow\:\mathrm{b}\:=\:\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:\bullet\:\mathrm{a}×\mathrm{b}\:=\:\frac{\mathrm{75}\sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$$$\:\bullet\:\frac{\mathrm{Area}\left(\mathrm{A}'\mathrm{B}'\mathrm{C}'\mathrm{D}'\right)}{\mathrm{Area}\left(\mathrm{ABCD}\right)}\:=\:\frac{\mathrm{75}\sqrt{\mathrm{3}}}{\mathrm{16}}\:×\:\frac{\mathrm{1}}{\mathrm{25}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{3}}{\mathrm{16}} \\ $$
Commented by a.lgnaoui last updated on 14/Mar/23
exactly.thank you
$${exactly}.{thank}\:{you} \\ $$

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