Question Number 188267 by normans last updated on 27/Feb/23
Answered by mr W last updated on 02/Mar/23
Commented by mr W last updated on 02/Mar/23
$${xy}=\mathrm{128} \\ $$$$\frac{\mathrm{16}{a}−{x}}{\mathrm{2}}=\mathrm{3}{a}\:\Rightarrow{x}=\mathrm{10}{a} \\ $$$$\frac{{x}}{{y}}=\frac{{c}}{{a}} \\ $$$${yc}={ax}=\mathrm{10}{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{12}{c}−\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{8}{a}\right)^{\mathrm{2}} =\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{144}{c}^{\mathrm{2}} −\mathrm{12}×\mathrm{10}{a}^{\mathrm{2}} +\mathrm{39}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{c}=\frac{\mathrm{3}{a}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{xy}=\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{128}\:\Rightarrow{y}=\frac{\mathrm{16}\sqrt{\mathrm{6}}}{\mathrm{3}}\:\Rightarrow{x}=\mathrm{4}\sqrt{\mathrm{6}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{10}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}}\:\Rightarrow{c}=\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{10}} \\ $$$${A}_{{triangle}} =\frac{\left(\mathrm{25}{a}\right)\left(\mathrm{25}{c}\right)}{\mathrm{2}}=\frac{\mathrm{25}^{\mathrm{2}} {ac}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}}×\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{10}} \\ $$$$\:\:\:\:\:=\mathrm{25}×\mathrm{9}=\mathrm{225} \\ $$
Answered by a.lgnaoui last updated on 01/Mar/23
![posons posons DM=y HI=x Area(DEMN)=xy=128 x=((128)/y) Area of teiangle(ADM)=((MD×AL)/2) =Area(ALD)+(ALM) △LCD cos α=((25a)/(25b))=(a/b)⇒ sin α=((√(b^2 −a^2 ))/b) DL=12(√(b^2 −a^2 )) CL=12bcos α=12a ⇒AL=25a−12a AL=13a △ADM ∡AOM=2∡ADM =2ϕ AD^2 =AL^2 +DL^2 =(13a)^2 +12^2 (b^2 −a^2 ) AD=(√(12^2 b^2 +25a^2 )) cos ϕ=((DL)/(AD))=((12(√(b^2 −a^2 )))/( (√(DL^2 +AL^2 ))))= cos ϕ=((12(√(b^2 −a^2 )))/( (√(12^2 b^2 +25a^2 )))) AM^2 =DM^2 +AD^2 −2AD×DMcos ϕ (1) △AMD (KM=((AM)/2);∡MDK=(ϕ/2)) sin (ϕ/2)= ((MK)/(MD))=((AM)/(2MD)) AM=2MDsin (ϕ/2)(2) (1)⇒ 4MD^2 sin^2 (ϕ/2)= DM^2 +AD^2 −2AD×DMcos ϕ DM^2 (1−4sin^2 (ϕ/2))−2AD×DMcos ϕ+AD^2 =0 y^2 −((24(√(b^2 −a^2 )))/( (24(√(b^2 −a^2 )) −3(√(12^2 b^2 +25a^2 )))))y+ +(((12^2 b^2 +25a^2 )(√(12^2 b^2 +25a^2 )))/(24(√(b^2 −a^2 )) −3(√(12^2 b^2 +25a^2 )))) (y−((4(√(b^2 −a^2 )))/(8(√(b^2 −a^2 )) −(√(12^2 b^2 +25a^2 )))))^2 + ((√((12^2 b^2 +25a^2 )^3 ))/(24(√(b^2 −a^2 )) −3(√(12^2 b^2 +25a^2 )))) − ((16(b^2 −a^2 ))/((24(√(b^2 −a^2 )) −(√(12^2 b^2 +25a^2 )) )^2 )) y=((4(√(b^2 −a^2 )))/(8(√(b^2 −a^2 )) −(√(12^2 b^2 +25a^2 ))))± ((√(16(b^2 −a^2 ) −(√((12b^2 +25a^2 )^3 ))[(24(√(b^2 −a^2 )) −3(√(12b^2 +25a^2 )))))/((24(√(b^2 −a^2 ))− (√(12b^2 +25a^2 ))))) [Area=((13a)/2)y] (3) Methode geometrique: △DEP tan α= ((EP)/(DE))=((BI)/(DI))=((AB)/(AC)) ((EP)/x)=((√(b^2 −a^2 ))/a) (1) ((HP)/(CH))=((AB)/(AC)) ⇒(((12(√(b^2 −a^2 )) )+FP)/(x+12a))(2) FP=((x(√(b^2 −a^2 )))/a) ((FP+12(√(b^2 −a^2 )))/(x+12a)) =((√(b^2 −a^2 ))/a) ((((x(√(b^2 −a^2 )))/a)+12(√(b^2 −a^2 )))/(x+12a))=((√(b^2 −a^2 ))/a) ((x(√(b^2 −a^2 )) +12a(√(b^2 −a^2 )))/(x+12a)) =((√(b^2 −a^2 ))/a) ax(√(b^2 −a^2 )) +12a^2 (√(b^2 −a^2 )) = x(√(b^2 −a^2 )) +12a(√(b^2 −a^2 )) x(a(√(b^2 −a^2 )) −(√(b^2 −a^2 )) )= 12a((√(b^2 −a^2 )) −a(√(b^2 −a^2 )) x((√(b^2 −a^2 )) )(a−1)= 12a(√(b^2 −a^2 )) (1−a) x=12a xy=128 y=((32)/(3a))= (3) Area(triangle)=((13a)/2)y Area=69,33](https://www.tinkutara.com/question/Q188420.png)
$${posons}\:\: \\ $$$${posons}\:{DM}={y}\:\:\:{HI}={x} \\ $$$${Area}\left({DEMN}\right)={xy}=\mathrm{128}\:\:{x}=\frac{\mathrm{128}}{{y}} \\ $$$${Area}\:{of}\:{teiangle}\left({ADM}\right)=\frac{{MD}×{AL}}{\mathrm{2}} \\ $$$$={Area}\left({ALD}\right)+\left({ALM}\right) \\ $$$$\bigtriangleup{LCD} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{25}{a}}{\mathrm{25}{b}}=\frac{{a}}{{b}}\Rightarrow\:\:\mathrm{sin}\:\alpha=\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{b}} \\ $$$${DL}=\mathrm{12}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\:{CL}=\mathrm{12}{b}\mathrm{cos}\:\alpha=\mathrm{12}{a} \\ $$$$\Rightarrow{AL}=\mathrm{25}{a}−\mathrm{12}{a}\:\:\:\:\:\:\:\:\:\:{AL}=\mathrm{13}{a} \\ $$$$\bigtriangleup{ADM}\:\: \\ $$$$\measuredangle{AOM}=\mathrm{2}\measuredangle{ADM}\:=\mathrm{2}\varphi \\ $$$$ \\ $$$${AD}^{\mathrm{2}} ={AL}^{\mathrm{2}} +{DL}^{\mathrm{2}} =\left(\mathrm{13}{a}\right)^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$${AD}=\sqrt{\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\:\:\varphi=\frac{{DL}}{{AD}}=\frac{\mathrm{12}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\:\sqrt{{DL}^{\mathrm{2}} +{AL}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\mathrm{cos}\:\varphi=\frac{\mathrm{12}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\:\sqrt{\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} }} \\ $$$${AM}^{\mathrm{2}} ={DM}^{\mathrm{2}} +{AD}^{\mathrm{2}} −\mathrm{2}{AD}×{DM}\mathrm{cos}\:\varphi\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{AMD}\:\left({KM}=\frac{{AM}}{\mathrm{2}};\measuredangle{MDK}=\frac{\varphi}{\mathrm{2}}\right)\:\: \\ $$$$\mathrm{sin}\:\frac{\varphi}{\mathrm{2}}=\:\frac{{MK}}{{MD}}=\frac{{AM}}{\mathrm{2}{MD}}\:\:\:{AM}=\mathrm{2}{MD}\mathrm{sin}\:\frac{\varphi}{\mathrm{2}}\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\:\mathrm{4}{MD}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}= \\ $$$${DM}^{\mathrm{2}} +{AD}^{\mathrm{2}} −\mathrm{2}{AD}×{DM}\mathrm{cos}\:\varphi \\ $$$${DM}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{4sin}\:^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}\right)−\mathrm{2}{AD}×{DM}\mathrm{cos}\:\varphi+{AD}^{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{{y}}^{\mathrm{2}} −\frac{\mathrm{24}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\:\left(\mathrm{24}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:−\mathrm{3}\sqrt{\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} }\right)}\boldsymbol{{y}}+ \\ $$$$+\frac{\left(\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} \right)\sqrt{\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} }}{\mathrm{24}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}\:−\mathrm{3}\sqrt{\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} }} \\ $$$$\left(\boldsymbol{{y}}−\frac{\mathrm{4}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }}{\mathrm{8}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:−\sqrt{\mathrm{12}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{25}{a}^{\mathrm{2}} }}\right)^{\mathrm{2}} + \\ $$$$\frac{\sqrt{\left(\mathrm{12}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} \right)^{\mathrm{3}} \:}}{\mathrm{24}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \:}\:−\mathrm{3}\sqrt{\mathrm{12}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} }}\:\:\:\:\:− \\ $$$$\frac{\mathrm{16}\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)}{\left(\mathrm{24}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \:}\:−\sqrt{\mathrm{12}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} }\:\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\boldsymbol{{y}}=\frac{\mathrm{4}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }}{\mathrm{8}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:−\sqrt{\mathrm{12}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} }}\pm \\ $$$$\frac{\sqrt{\mathrm{16}\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)\:−\sqrt{\left(\mathrm{12}\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} \right)^{\mathrm{3}} }\left[\left(\mathrm{24}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:−\mathrm{3}\sqrt{\left.\mathrm{12}\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} \:\right)}\right.\right.}}{\left(\mathrm{24}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \:}−\:\sqrt{\left.\mathrm{12}\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{25}\boldsymbol{{a}}^{\mathrm{2}} \right)}\right.} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{Area}=\frac{\mathrm{13}{a}}{\mathrm{2}}\boldsymbol{{y}}\right]\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\:\boldsymbol{{Methode}}\:\:\boldsymbol{{geometrique}}: \\ $$$$\bigtriangleup{DEP}\:\:\:\:\mathrm{tan}\:\alpha=\:\frac{{EP}}{{DE}}=\frac{{BI}}{{DI}}=\frac{{AB}}{{AC}} \\ $$$$\frac{{EP}}{{x}}=\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\frac{{HP}}{{CH}}=\frac{{AB}}{{AC}}\:\:\Rightarrow\frac{\left(\mathrm{12}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\right)+{FP}}{{x}+\mathrm{12}{a}}\left(\mathrm{2}\right) \\ $$$$ \\ $$$${FP}=\frac{{x}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}} \\ $$$$\frac{{FP}+\mathrm{12}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{x}+\mathrm{12}{a}}\:\:\:=\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}} \\ $$$$\:\:\frac{\frac{{x}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}+\mathrm{12}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{x}+\mathrm{12}{a}}=\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}} \\ $$$$\frac{{x}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:+\mathrm{12}{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{x}+\mathrm{12}{a}}\:=\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}} \\ $$$${ax}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}\:+\mathrm{12}{a}^{\mathrm{2}} \sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}\:= \\ $$$${x}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}\:+\mathrm{12}{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}\: \\ $$$$\:\:\:\boldsymbol{{x}}\left(\boldsymbol{{a}}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:−\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:\right)= \\ $$$$\mathrm{12}\boldsymbol{{a}}\left(\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \:}\:−\boldsymbol{{a}}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \:}\right. \\ $$$$\:\:\:\:\:\boldsymbol{{x}}\left(\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:\right)\left(\boldsymbol{{a}}−\mathrm{1}\right)= \\ $$$$\:\:\:\:\:\mathrm{12}\boldsymbol{{a}}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }\:\left(\mathrm{1}−\boldsymbol{{a}}\right) \\ $$$$\:\:\:\:\:\:\boldsymbol{{x}}=\mathrm{12}\boldsymbol{{a}}\:\:\:\:\:\:\:{xy}=\mathrm{128} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\frac{\mathrm{32}}{\mathrm{3}\boldsymbol{{a}}}= \\ $$$$\:\:\left(\mathrm{3}\right)\:{Area}\left({triangle}\right)=\frac{\mathrm{13}{a}}{\mathrm{2}}{y} \\ $$$$\:\:\:\:\:\:\boldsymbol{{Area}}=\mathrm{69},\mathrm{33} \\ $$$$\: \\ $$
Commented by a.lgnaoui last updated on 01/Mar/23
Commented by mr W last updated on 02/Mar/23
$${clearly}\:{wrong}!\:{the}\:{triangle}\:{looks} \\ $$$${almost}\:{double}\:{so}\:{large}\:{as}\:{the} \\ $$$${rectangle},\:{but}\:{your}\:{answer}\:{says}\:{the} \\ $$$${triangle}\:{is}\:{only}\:{half}\:{so}\:{large}\:{as}\:{the}\: \\ $$$${rectangle}! \\ $$$${you}\:{seem}\:{to}\:{have}\:{no}\:{sense}\:{of}\:{geometric} \\ $$$${size}. \\ $$