Question Number 112780 by ajfour last updated on 09/Sep/20
Commented by ajfour last updated on 09/Sep/20
$${If}\:\:{regions}\:{A},\:{B},\:{C},\:{D}\:{have}\:{equal} \\ $$$${areas},\:{find}\:{P}\:\left({h},{k}\right)\:{in}\:{terms}\:{of}\:{the} \\ $$$${square}\:{side}\:\boldsymbol{{a}}. \\ $$
Answered by mr W last updated on 09/Sep/20
Commented by mr W last updated on 09/Sep/20
$$\frac{{pa}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{p}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\frac{{q}+{r}}{\mathrm{2}}×{a}=\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{q}+{r}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\frac{{h}}{{k}}=\frac{{p}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${h}=\frac{\mathrm{2}{k}}{\mathrm{3}} \\ $$$$\frac{{q}−{r}}{{a}}=\frac{{h}−{a}+{q}}{{k}}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{{a}−{q}}{{k}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{q}\right){k}=\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${k}=\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}\left({a}−{q}\right)} \\ $$$$\frac{{q}−\frac{\mathrm{2}{a}}{\mathrm{3}}+{q}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}\left({a}−{q}\right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$$\frac{{q}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{1}−\frac{{q}}{{a}}\right)^{\mathrm{2}} \\ $$$$−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{1}−\mathrm{2}\lambda+\lambda^{\mathrm{2}} \right)+\lambda=\mathrm{0} \\ $$$$\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{6}\lambda−\mathrm{7}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{3}}=\frac{{q}}{{a}} \\ $$$$\Rightarrow\frac{{k}}{{a}}=\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{3}}\right)}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{{h}}{{a}}=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{3}} \\ $$
Commented by ajfour last updated on 10/Sep/20
$${Thank}\:{you}\:{Sir},\:{nice}\:{solution},\:{errorless}. \\ $$
Commented by ajfour last updated on 10/Sep/20
$${p}={a}−\left({p}+{r}\right)+\left({a}−{q}\right)={q}+{r}\:=\:\frac{\left({a}^{\mathrm{2}} /\mathrm{3}\right)}{\left({a}/\mathrm{2}\right)} \\ $$$$\Rightarrow\:{p}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:,\:\:{q}+{r}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:\:\:\:\:\:…\left({i}\right),\:\left({ii}\right) \\ $$$${Also}\:\:\:\left({a}−{q}\right)\frac{{k}}{\mathrm{2}}=\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}\:\:\:\: \\ $$$$\Rightarrow\:\:\:{q}={a}−\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}{k}}\:\:\:\:\:….\left({iii}\right) \\ $$$$\:\:\:\frac{{k}}{{h}}=\frac{{a}}{\left({p}=\frac{\mathrm{2}{a}}{\mathrm{3}}\right)}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:…\left({iv}\right) \\ $$$$\:\:\frac{{q}−{r}}{{a}}=\frac{{h}−\left({a}−{q}\right)}{{k}}\:\:\:\:\:\:\:\:…\left({v}\right) \\ $$$$\left({ii}\right)+{a}×\left({v}\right)\:\:{using}\:\left({iii}\right),\left({iv}\right)\:{alongwith} \\ $$$$\mathrm{2}\left({a}−\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}{k}}\right)=\frac{\mathrm{2}{a}}{\mathrm{3}}+\frac{\mathrm{2}{a}}{\mathrm{3}}−{a}\left(\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}{k}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\frac{\mathrm{2}{a}}{\mathrm{3}}\left(\frac{{k}}{{a}}\right)^{\mathrm{2}} −\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{3}{k}}\left(\frac{{k}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)+\frac{\mathrm{4}{a}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\frac{{k}}{{a}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{k}}{{a}}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{{k}}{{a}}=\mathrm{2}+\sqrt{\mathrm{2}}\:\:\:,\:\:\:{h}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right). \\ $$$$\:\:\:……………………………………… \\ $$