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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

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

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

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

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

$$Howdidyoufind4pq?$$

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

$${(q-p)}^2=q^2-2qp+p^2=q^2+2qp+p^2-4qp$$$$={(q+p)}^2-4qp$$

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

$$youmeanthowpq=h^2?$$

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