# Question-68524

Question Number 68524 by Maclaurin Stickker last updated on 13/Sep/19
Answered by mr W last updated on 13/Sep/19
Commented by Maclaurin Stickker last updated on 13/Sep/19
$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{right}! \\$$
Commented by mr W last updated on 13/Sep/19
$$\frac{{p}}{{h}}=\frac{{h}}{{q}} \\$$$$\Rightarrow{pq}={h}^{\mathrm{2}} \\$$$${q}−{p}={h}\:\:\:\:…\left({i}\right) \\$$$$\left({q}+{p}\right)^{\mathrm{2}} =\left({q}−{p}\right)^{\mathrm{2}} +\mathrm{4}{pq}={h}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} =\mathrm{5}{h}^{\mathrm{2}} \\$$$$\Rightarrow{q}+{p}=\sqrt{\mathrm{5}}{h}\:\:\:\:…\left({ii}\right) \\$$$$\Rightarrow{q}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}{h} \\$$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}{h} \\$$$$\Rightarrow{a}=\sqrt{{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }={h}\sqrt{\mathrm{1}+\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}{h} \\$$$$\Rightarrow{b}=\sqrt{{h}^{\mathrm{2}} +{q}^{\mathrm{2}} }={h}\sqrt{\mathrm{1}+\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}{h} \\$$$$\Rightarrow{c}={p}+{q}=\sqrt{\mathrm{5}}{h} \\$$
Commented by Maclaurin Stickker last updated on 01/Oct/19
$${How}\:{did}\:{you}\:{find}\:\mathrm{4}{pq}? \\$$
Commented by mr W last updated on 01/Oct/19
$$\left({q}−{p}\right)^{\mathrm{2}} ={q}^{\mathrm{2}} −\mathrm{2}{qp}+{p}^{\mathrm{2}} ={q}^{\mathrm{2}} +\mathrm{2}{qp}+{p}^{\mathrm{2}} −\mathrm{4}{qp} \\$$$$=\left({q}+{p}\right)^{\mathrm{2}} −\mathrm{4}{qp} \\$$
Commented by mr W last updated on 01/Oct/19
$${you}\:{meant}\:{how}\:{pq}={h}^{\mathrm{2}} ? \\$$