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Question Number 222479 by mr W last updated on 28/Jun/25
Commented by mr W last updated on 28/Jun/25
find the area of the rectangle.
$${find}\:{the}\:{area}\:{of}\:{the}\:{rectangle}. \\ $$
Answered by ajfour last updated on 28/Jun/25
bp=10 bq=14 (s+t)h=30 (b/p)=(h/s) & (b/q)=(h/t) & ⇒ ((s+t)/h)=((p+q)/b)=((24)/b^2 ) ((30)/h^2 )=((24)/b^2 ) ⇒ A=b(p+q+s+t)=ba (recrangle area) =24+(b/h)(30) A=24+30×(√((24)/(30))) A=24+12(√5)
$${bp}=\mathrm{10} \\ $$$${bq}=\mathrm{14} \\ $$$$\left({s}+{t}\right){h}=\mathrm{30} \\ $$$$\frac{\boldsymbol{{b}}}{{p}}=\frac{{h}}{{s}}\:\:\&\:\:\frac{\boldsymbol{{b}}}{{q}}=\frac{{h}}{{t}}\:\&\: \\ $$$$\Rightarrow\:\frac{{s}+{t}}{{h}}=\frac{{p}+{q}}{{b}}=\frac{\mathrm{24}}{{b}^{\mathrm{2}} } \\ $$$$\:\:\frac{\mathrm{30}}{{h}^{\mathrm{2}} }=\frac{\mathrm{24}}{{b}^{\mathrm{2}} }\:\:\Rightarrow\:\: \\ $$$${A}=\boldsymbol{{b}}\left({p}+{q}+{s}+{t}\right)=\boldsymbol{{ba}}\:\:\left({recrangle}\:{area}\right) \\ $$$$\:\:=\mathrm{24}+\frac{\boldsymbol{{b}}}{{h}}\left(\mathrm{30}\right) \\ $$$$\:\:\:\boldsymbol{{A}}=\mathrm{24}+\mathrm{30}×\sqrt{\frac{\mathrm{24}}{\mathrm{30}}} \\ $$$$\:\:\:\boldsymbol{{A}}=\mathrm{24}+\mathrm{12}\sqrt{\mathrm{5}} \\ $$
Commented by mr W last updated on 28/Jun/25
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Commented by ajfour last updated on 28/Jun/25
https://youtu.be/V4nZrQmPzeE?si=Nawvpp-UlSkUXAoJ
Answered by A5T last updated on 28/Jun/25
Commented by A5T last updated on 28/Jun/25
[s−(((10)/t)+((14)/t))]=s−((24)/t)=((st−24)/t) s−((24)/t)>0⇒A=st>24 (((((st−24)/t))^2 )/s^2 )=((15)/(15+A−12))⇒(st−24)^2 =((15s^2 t^2 )/(st+3)) ⇒(A^2 +24^2 −48A)(A+3)=15A^2 ⇒(A−12)(A^2 −48A−144)=0 ⇒A=st=24+12(√5) (since A> 24)
$$\left[\mathrm{s}−\left(\frac{\mathrm{10}}{\mathrm{t}}+\frac{\mathrm{14}}{\mathrm{t}}\right)\right]=\mathrm{s}−\frac{\mathrm{24}}{\mathrm{t}}=\frac{\mathrm{st}−\mathrm{24}}{\mathrm{t}} \\ $$$$\mathrm{s}−\frac{\mathrm{24}}{\mathrm{t}}>\mathrm{0}\Rightarrow\mathrm{A}=\mathrm{st}>\mathrm{24} \\ $$$$\frac{\left(\frac{\mathrm{st}−\mathrm{24}}{\mathrm{t}}\right)^{\mathrm{2}} }{\mathrm{s}^{\mathrm{2}} }=\frac{\mathrm{15}}{\mathrm{15}+\mathrm{A}−\mathrm{12}}\Rightarrow\left(\mathrm{st}−\mathrm{24}\right)^{\mathrm{2}} =\frac{\mathrm{15s}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} }{\mathrm{st}+\mathrm{3}} \\ $$$$\Rightarrow\left(\mathrm{A}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} −\mathrm{48A}\right)\left(\mathrm{A}+\mathrm{3}\right)=\mathrm{15A}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{A}−\mathrm{12}\right)\left(\mathrm{A}^{\mathrm{2}} −\mathrm{48A}−\mathrm{144}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{A}=\mathrm{st}=\mathrm{24}+\mathrm{12}\sqrt{\mathrm{5}}\:\left(\mathrm{since}\:\mathrm{A}>\:\mathrm{24}\right) \\ $$
Commented by mr W last updated on 28/Jun/25
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Answered by A5T last updated on 28/Jun/25
Commented by A5T last updated on 28/Jun/25
(b_s ,h_s )=( base,height) of smaller triangle(△_s =15) b_s t=st−24⇒b_s =((st−24)/t)...(i) Let bigger triangle(△_b ) be partitioned into areas A_1 and A_2 with altitude to s. (5/A_1 )=(t^2 /h_b ^2 ); (7/A_2 )=(t^2 /h_b ^2 )⇒A_1 +A_2 =((12h_b ^2 )/t^2 )=((s×h_b )/2) ⇒st^2 =24h_b ⇒h_b =((st^2 )/(24)) ((15)/((sh_b )/2))=(h_s ^2 /h_b ^2 )=((30)/(sh_b ))⇒h_s ^2 =((30h_b )/s)=((30)/s)×((st^2 )/(24))=((5t^2 )/4)⇒h_s =((t(√5))/2) b_s ×h_s =30⇒b_s =((12(√5))/t)...(ii) (i)&(ii)⇒((st−24)/t)=((12(√5))/t)⇒st=24+12(√5)
$$\left(\mathrm{b}_{\mathrm{s}} ,\mathrm{h}_{\mathrm{s}} \right)=\left(\:\mathrm{base},\mathrm{height}\right)\:\mathrm{of}\:\mathrm{smaller}\:\mathrm{triangle}\left(\bigtriangleup_{\mathrm{s}} =\mathrm{15}\right) \\ $$$$\mathrm{b}_{\mathrm{s}} \mathrm{t}=\mathrm{st}−\mathrm{24}\Rightarrow\mathrm{b}_{\mathrm{s}} =\frac{\mathrm{st}−\mathrm{24}}{\mathrm{t}}…\left(\mathrm{i}\right) \\ $$$$\mathrm{Let}\:\mathrm{bigger}\:\mathrm{triangle}\left(\bigtriangleup_{\mathrm{b}} \right)\:\mathrm{be}\:\mathrm{partitioned}\:\mathrm{into}\:\mathrm{areas}\: \\ $$$$\mathrm{A}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{A}_{\mathrm{2}} \:\mathrm{with}\:\mathrm{altitude}\:\mathrm{to}\:\mathrm{s}. \\ $$$$\frac{\mathrm{5}}{\mathrm{A}_{\mathrm{1}} }=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{h}_{\mathrm{b}} ^{\mathrm{2}} };\:\frac{\mathrm{7}}{\mathrm{A}_{\mathrm{2}} }=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{h}_{\mathrm{b}} ^{\mathrm{2}} }\Rightarrow\mathrm{A}_{\mathrm{1}} +\mathrm{A}_{\mathrm{2}} =\frac{\mathrm{12h}_{\mathrm{b}} ^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }=\frac{\mathrm{s}×\mathrm{h}_{\mathrm{b}} }{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{st}^{\mathrm{2}} =\mathrm{24h}_{\mathrm{b}} \Rightarrow\mathrm{h}_{\mathrm{b}} =\frac{\mathrm{st}^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\frac{\mathrm{15}}{\frac{\mathrm{sh}_{\mathrm{b}} }{\mathrm{2}}}=\frac{\mathrm{h}_{\mathrm{s}} ^{\mathrm{2}} }{\mathrm{h}_{\mathrm{b}} ^{\mathrm{2}} }=\frac{\mathrm{30}}{\mathrm{sh}_{\mathrm{b}} }\Rightarrow\mathrm{h}_{\mathrm{s}} ^{\mathrm{2}} =\frac{\mathrm{30h}_{\mathrm{b}} }{\mathrm{s}}=\frac{\mathrm{30}}{\mathrm{s}}×\frac{\mathrm{st}^{\mathrm{2}} }{\mathrm{24}}=\frac{\mathrm{5t}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow\mathrm{h}_{\mathrm{s}} =\frac{\mathrm{t}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{b}_{\mathrm{s}} ×\mathrm{h}_{\mathrm{s}} =\mathrm{30}\Rightarrow\mathrm{b}_{\mathrm{s}} =\frac{\mathrm{12}\sqrt{\mathrm{5}}}{\mathrm{t}}…\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right)\Rightarrow\frac{\mathrm{st}−\mathrm{24}}{\mathrm{t}}=\frac{\mathrm{12}\sqrt{\mathrm{5}}}{\mathrm{t}}\Rightarrow\mathrm{st}=\mathrm{24}+\mathrm{12}\sqrt{\mathrm{5}} \\ $$
Commented by mr W last updated on 28/Jun/25
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Answered by mr W last updated on 28/Jun/25
Commented by mr W last updated on 28/Jun/25
((15)/X)=(X/(7+5)) ⇒X=(√(15×12))=6(√5) ((area of rectangle)/2)=5+X+7=12+6(√5) ⇒area of rectangle =12(2+(√5))
$$\frac{\mathrm{15}}{{X}}=\frac{{X}}{\mathrm{7}+\mathrm{5}}\:\Rightarrow{X}=\sqrt{\mathrm{15}×\mathrm{12}}=\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\frac{{area}\:{of}\:{rectangle}}{\mathrm{2}}=\mathrm{5}+{X}+\mathrm{7}=\mathrm{12}+\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{area}\:{of}\:{rectangle}\:=\mathrm{12}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$
Commented by ajfour last updated on 28/Jun/25
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