349/MR00. zadatak / Matematika - Zbirka rešenih zadataka

MR-349 / 349. zadatak

Površina romba je 100 cm², dok njegov oštar ugao iznosi 30°. Izračunajte dijagonale romba!

T = \frac{d_{1} \cdot d_{2}}{2} = 100

d_{1} \cdot d_{2} = 200

\sin \frac{α}{2} = \frac{\frac{d_{2}}{2}}{2}

\sin 15 ° = \frac{d_{2}}{4}

\sin^{2} \frac{α}{2} = \frac{1 - \cos α}{2}

\sin^{2} \frac{30}{2} = \frac{1 - \cos 30 °}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{\frac{2 - \sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}

\sin 15 ° = \sqrt{\frac{2 - \sqrt{3}}{4}}

\sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{d_{2}}{4}

\frac{2 - \sqrt{3}}{4} = \frac{d_{2}^{2}}{16}

d_{2}^{2} = 4 \cdot (2 - \sqrt{3})

d_{2} = \sqrt{4 \cdot (2 - \sqrt{3})} = 2 \sqrt{2 - \sqrt{3}}

d_{2} = \frac{200}{d_{1}} = \frac{200}{2 \sqrt{2 - \sqrt{3}}}

$d_{1} = 2 \sqrt{2 - \sqrt{3}}; d_{2} = \frac{200}{2 \sqrt{2 - \sqrt{3}}}$

MathReference - Zadaci