Question Number 47010 by ajfour last updated on 03/Nov/18
Commented by ajfour last updated on 03/Nov/18
$${Find}\:{maximum}\:{rectangular} \\ $$$${area}\:{and}\:\boldsymbol{{l}}\:\&\:\boldsymbol{{h}}\:\left({then}\right),\:{in}\:{terms} \\ $$$${of}\:{sides}\:{a},{b},{c}\:\:{of}\:\bigtriangleup{ABC}. \\ $$$$\left({with}\:{QR}\:\:{along}\:{BC}\right) \\ $$
Answered by ajfour last updated on 04/Nov/18
![area(△ABC)=△_0 ((Ar(△APS))/△_0 )=((l/a))^2 ((Ar(△PBQ)+Ar(△SRC))/△_9 )=(((a−l)/a))^2 ⇒ lh=△_0 [1−((l/a))^2 −(1−(l/a))^2 ] = 2△_0 ((l/a))(1−(l/a)) ⇒ max (lh) when (l/a) = (1/2) max(PQRS)=(△_0 /2) , l= (a/2) h= ((altitude of △ABC)/2) .](Q47025.png)
$$\:{area}\left(\bigtriangleup{ABC}\right)=\bigtriangleup_{\mathrm{0}} \\ $$$$\frac{{Ar}\left(\bigtriangleup{APS}\right)}{\bigtriangleup_{\mathrm{0}} }=\left(\frac{{l}}{{a}}\right)^{\mathrm{2}} \\ $$$$\frac{{Ar}\left(\bigtriangleup{PBQ}\right)+{Ar}\left(\bigtriangleup{SRC}\right)}{\bigtriangleup_{\mathrm{9}} }=\left(\frac{{a}−{l}}{{a}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{lh}=\bigtriangleup_{\mathrm{0}} \left[\mathrm{1}−\left(\frac{{l}}{{a}}\right)^{\mathrm{2}} −\left(\mathrm{1}−\frac{{l}}{{a}}\right)^{\mathrm{2}} \:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\bigtriangleup_{\mathrm{0}} \left(\frac{{l}}{{a}}\right)\left(\mathrm{1}−\frac{{l}}{{a}}\right) \\ $$$$\:\Rightarrow\:{max}\:\left({lh}\right)\:\:{when}\:\frac{{l}}{{a}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:{max}\left({PQRS}\right)=\frac{\bigtriangleup_{\mathrm{0}} }{\mathrm{2}}\:,\:{l}=\:\frac{{a}}{\mathrm{2}} \\ $$$$\:\:{h}=\:\frac{{altitude}\:{of}\:\bigtriangleup{ABC}}{\mathrm{2}}\:. \\ $$