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Question Number 158664 by mr W last updated on 07/Nov/21

Commented by mr W last updated on 07/Nov/21

$$AQ=CP=BR$$$$findthesmallestareaof\mathrm{\Delta }PQR$$$$intermsofa,b.$$

Commented by Tawa11 last updated on 07/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

Commented by Tawa11 last updated on 16/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

Answered by ajfour last updated on 07/Nov/21

$$Corigin.$$$$AQ=CP=BR=r$$$$a^2+b^2=s^2$$$$$$$$EquationofAB:\frac{x}{a}+\frac{y}{b}=1$$$$P(0,r);R(a-\frac{ar}{s},\frac{br}{s});Q(0,b-r)$$$$2A_{△PQR}=ab-r(b-r)$$$$-\frac{rb}{s}(a-r)-r(a-\frac{ar}{s})$$$$\frac{d\left(2A_{△PQR}\right)}{dr}=0\Rightarrow$$$$2r-b+\frac{2br}{s}-\frac{ab}{s}-a+\frac{2ar}{s}=0$$$$\Rightarrow 2r(1+\frac{a+b}{s})=a+b+\frac{ab}{s}$$$$2r=\frac{(a+b)\sqrt{a^2+b^2}+ab}{(\sqrt{a^2+b^2}+ab)}$$$$2A_{△PQR}=(\frac{a+b}{\sqrt{a^2+b^2}}+1)r^2$$$$-(\frac{ab}{\sqrt{a^2+b^2}}+a+b)r+ab$$$$=ab-(\frac{a+b}{s}+1)r^2$$$$hence$$$$A_{△{PQR}_{min}}=$$$$\frac{ab}{2}-\frac{1}{8}(\frac{a+b}{s}+1){\left(\frac{a+b+\frac{ab}{s}}{1+\frac{a+b}{s}}\right)}^2$$$$\mathrm{\forall }s=\sqrt{a^2+b^2}$$$$butletmecheck$$$$a=b=\sqrt{2}$$$$A_{min}=1-\frac{(\sqrt{2}+1)}{8}{\left(\frac{2\sqrt{2}+1}{1+\sqrt{2}}\right)}^2$$$$=1-\frac{(9+4\sqrt{2})}{8(1+\sqrt{2})}$$$$=1-\frac{(9+4\sqrt{2})(\sqrt{2}-1)}{8}$$$$=1-\frac{(5\sqrt{2}-1)}{8}$$$$=\frac{9-5\sqrt{2}}{8}$$

Commented by mr W last updated on 07/Nov/21

$$thankssir!$$$${it}^{\prime }scorrect!$$

Commented by ajfour last updated on 07/Nov/21

$$isntmax=\frac{ab}{2}itself?$$

Commented by mr W last updated on 07/Nov/21

$$yes!$$

Answered by mr W last updated on 07/Nov/21

$$c=\sqrt{a^2+b^2}$$$$AQ=BR=CP=t$$$$\frac{A_{\mathrm{\Delta }PQR}}{A_{\mathrm{\Delta }ABC}}=1-\frac{t(b-t)}{ab}-\frac{t(a-t)}{ac}-\frac{t(c-t)}{bc}$$$$\frac{A_{\mathrm{\Delta }PQR}}{A_{\mathrm{\Delta }ABC}}=1-\frac{t(ab+bc+ca)-t^2(a+b+c)}{abc}$$$$suchthat{A}_{\mathrm{\Delta }PQR}isminimum:$$$$\frac{d}{dt}\left(\frac{A_{\mathrm{\Delta }PQR}}{A_{\mathrm{\Delta }ABC}}\right)=0$$$$\Rightarrow (ab+bc+ca)-2(a+b+c)t=0$$$$\Rightarrow t=\frac{ab+bc+ca}{2(a+b+c)}$$$$\frac{max.A_{\mathrm{\Delta }PQR}}{A_{\mathrm{\Delta }ABC}}=1-\frac{{(ab+bc+ca)}^2}{4abc(a+b+c)}$$$$max.A_{\mathrm{\Delta }PQR}=\frac{ab}{2}-\frac{{[ab+(a+b)\sqrt{a^2+b^2}]}^2}{8\sqrt{a^2+b^2}(a+b+\sqrt{a^2+b^2})}$$

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