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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

Your answer is right!

$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{right}! \\ $$

Commented by mr W last updated on 13/Sep/19

(p/h)=(h/q) ⇒pq=h^2 q−p=h ...(i) (q+p)^2 =(q−p)^2 +4pq=h^2 +4h^2 =5h^2 ⇒q+p=(√5)h ...(ii) ⇒q=(((√5)+1)/2)h ⇒p=(((√5)−1)/2)h ⇒a=(√(h^2 +p^2 ))=h(√(1+((((√5)−1)/2))^2 ))=((√(10−2(√5)))/2)h ⇒b=(√(h^2 +q^2 ))=h(√(1+((((√5)+1)/2))^2 ))=((√(10+2(√5)))/2)h ⇒c=p+q=(√5)h

$$\frac{{p}}{{h}}=\frac{{h}}{{q}} \\ $$$$\Rightarrow{pq}={h}^{\mathrm{2}} \\ $$$${q}−{p}={h}\:\:\:\:...\left({i}\right) \\ $$$$\left({q}+{p}\right)^{\mathrm{2}} =\left({q}−{p}\right)^{\mathrm{2}} +\mathrm{4}{pq}={h}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} =\mathrm{5}{h}^{\mathrm{2}} \\ $$$$\Rightarrow{q}+{p}=\sqrt{\mathrm{5}}{h}\:\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{q}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}{h} \\ $$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}{h} \\ $$$$\Rightarrow{a}=\sqrt{{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }={h}\sqrt{\mathrm{1}+\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}{h} \\ $$$$\Rightarrow{b}=\sqrt{{h}^{\mathrm{2}} +{q}^{\mathrm{2}} }={h}\sqrt{\mathrm{1}+\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}{h} \\ $$$$\Rightarrow{c}={p}+{q}=\sqrt{\mathrm{5}}{h} \\ $$

Commented by Maclaurin Stickker last updated on 01/Oct/19

How did you find 4pq?

$${How}\:{did}\:{you}\:{find}\:\mathrm{4}{pq}? \\ $$

Commented by mr W last updated on 01/Oct/19

(q−p)^2 =q^2 −2qp+p^2 =q^2 +2qp+p^2 −4qp =(q+p)^2 −4qp

$$\left({q}−{p}\right)^{\mathrm{2}} ={q}^{\mathrm{2}} −\mathrm{2}{qp}+{p}^{\mathrm{2}} ={q}^{\mathrm{2}} +\mathrm{2}{qp}+{p}^{\mathrm{2}} −\mathrm{4}{qp} \\ $$$$=\left({q}+{p}\right)^{\mathrm{2}} −\mathrm{4}{qp} \\ $$

Commented by mr W last updated on 01/Oct/19

you meant how pq=h^2 ?

$${you}\:{meant}\:{how}\:{pq}={h}^{\mathrm{2}} ? \\ $$

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