Riješite zadatke.

Računamo stranicu $c$$c$ pomoću površine i kosinusovog poučka:
$P = \frac{1}{2}ab \sin \gamma \implies \sin \gamma = \frac{2P}{ab} = \frac{2 \cdot 12/13}{2} = \frac{12}{13}$$P = \frac{1}{2}ab \sin \gamma \implies \sin \gamma = \frac{2P}{ab} = \frac{2 \cdot 12/13}{2} = \frac{12}{13}$.
Kosinus može biti pozitivan ili negativan: $\cos \gamma = \pm \sqrt{1 - (12/13)^2} = \pm \frac{5}{13}$$\cos \gamma = \pm \sqrt{1 - (12/13)^2} = \pm \frac{5}{13}$.
1. Slučaj $\cos \gamma = 5/13$$\cos \gamma = 5/13$: $c_1 = \sqrt{1^2+2^2 - 2(1)(2)(5/13)} = \sqrt{5 - 20/13} = \sqrt{45/13}$$c_1 = \sqrt{1^2+2^2 - 2(1)(2)(5/13)} = \sqrt{5 - 20/13} = \sqrt{45/13}$.
2. Slučaj $\cos \gamma = -5/13$$\cos \gamma = -5/13$: $c_2 = \sqrt{1^2+2^2 - 2(1)(2)(-5/13)} = \sqrt{5 + 20/13} = \sqrt{85/13}$$c_2 = \sqrt{1^2+2^2 - 2(1)(2)(-5/13)} = \sqrt{5 + 20/13} = \sqrt{85/13}$.

Računamo površinu pobočja pravilne četverostrane piramide:
$a = 12.6$$a = 12.6$ cm, kut između visine pobočke i osnovice $\alpha = 48^{\circ} 31'$$\alpha = 48^{\circ} 31'$.
Visina pobočke $v_1$$v_1$ iz pravokutnog trokuta (pola baze, visina piramide, visina pobočke):
$\cos \alpha = \frac{a/2}{v_1} \implies v_1 = \frac{6.3}{\cos 48^{\circ} 31'}$$\cos \alpha = \frac{a/2}{v_1} \implies v_1 = \frac{6.3}{\cos 48^{\circ} 31'}$.
Površina pobočja (4 trokuta): $P = 4 \cdot \frac{a \cdot v_1}{2} = 2a v_1$$P = 4 \cdot \frac{a \cdot v_1}{2} = 2a v_1$.
$P = 2 \cdot 12.6 \cdot \frac{6.3}{\cos 48^{\circ} 31'} \approx 239.67$$P = 2 \cdot 12.6 \cdot \frac{6.3}{\cos 48^{\circ} 31'} \approx 239.67$.