Kruhové úseče

Trojuholníku ABC so stranou a = 15 cm, b = 17,4 cm, c = 21,6 cm je opísaná kružnica. Vypočítajte obsah kruhových úsečí určených stranami trojuholníka.

Správna odpoveď:

S₁ =  30,9391 cm²
S₂ =  52,8996 cm²
S₃ =  158,1036 cm²

Postup správneho riešenia:

$$\begin{gathered} a=15\text{cm} \\ b=17,4\text{cm} \\ c=21,6\text{cm} \\ {a}^2=b^2+c^2-2bc\cdot \cos \alpha \\ A=\arccos (\frac{b^2+c^2-a^2}{2\cdot b\cdot c})=\arccos (\frac{17,4^2+21,6^2-15^2}{2\cdot 17,4\cdot 21,6})\doteq 0,761\text{rad} \\ \alpha =A\to \text{°}=A\cdot \frac{180}{\pi }\text{°}=0,761\cdot \frac{180}{\pi }\text{°}=43,60282\text{°} \\ B=\arccos (\frac{a^2+c^2-b^2}{2\cdot a\cdot c})=\arccos (\frac{15^2+21,6^2-17,4^2}{2\cdot 15\cdot 21,6})\doteq 0,9273\text{rad} \\ \beta =B\to \text{°}=B\cdot \frac{180}{\pi }\text{°}=0,9273\cdot \frac{180}{\pi }\text{°}=53,1301\text{°} \\ \gamma =180-\alpha -\beta =180-43,6028-53,1301\doteq 83,2671{}^{\circ } \\ r=\frac{a}{2\cdot \sin \alpha }=\frac{a}{2\cdot \sin 43,602818972704°}=\frac{15}{2\cdot \sin 43,602818972704°}=\frac{15}{2\cdot 0,689655}=10,875\text{cm} \\ {x}_1=\sqrt{r^2-(a/2{)}^2}=\sqrt{10,875^2-(15/2{)}^2}=\frac{63}{8}=7,875\text{cm} \\ {x}_2=\sqrt{r^2-(b/2{)}^2}=\sqrt{10,875^2-(17,4/2{)}^2}=\frac{261}{40}=6,525\text{cm} \\ {x}_3=\sqrt{r^2-(c/2{)}^2}=\sqrt{10,875^2-(21,6/2{)}^2}=\frac{51}{40}=1,275\text{cm} \\ {h}_1=r-x_1=10,875-7,875=3\text{cm} \\ {h}_2=r-x_2=10,875-6,525=\frac{87}{20}=4,35\text{cm} \\ {h}_3=r-x_3=10,875-1,275=\frac{48}{5}=9,6\text{cm} \\ {a}^2=r^2+r^2-2r^2\cdot \cos {\phi }_1 \\ {\phi }_1=\arccos (\frac{2\cdot r^2-a^2}{2\cdot r^2})=\arccos (\frac{2\cdot 10,875^2-15^2}{2\cdot 10,875^2})\doteq 1,522\text{rad} \\ {\phi }_2=\arccos (\frac{2\cdot r^2-b^2}{2\cdot r^2})=\arccos (\frac{2\cdot 10,875^2-17,4^2}{2\cdot 10,875^2})\doteq 1,8546\text{rad} \\ {\phi }_3=\arccos (\frac{2\cdot r^2-c^2}{2\cdot r^2})=\arccos (\frac{2\cdot 10,875^2-21,6^2}{2\cdot 10,875^2})\doteq 2,9066\text{rad} \\ {S}_0=\pi \cdot r^2=3,1416\cdot 10,875^2\doteq 371,5424{\text{cm}}^2 \\ {S}_1=S_0\cdot \frac{{\phi }_1}{2\pi }-\frac{1}{2}\cdot a\cdot x_1=371,5424\cdot \frac{1,522}{2\cdot 3,1416}-\frac{1}{2}\cdot 15\cdot 7,875=30,9391{\text{cm}}^2 \end{gathered}$$

$S_2=S_0\cdot \frac{{\phi }_2}{2\pi }-\frac{1}{2}\cdot b\cdot x_2=371,5424\cdot \frac{1,8546}{2\cdot 3,1416}-\frac{1}{2}\cdot 17,4\cdot 6,525=52,8996{\text{cm}}^2$

$$\begin{gathered} S_3=S_0\cdot \frac{{\phi }_3}{2\pi }-\frac{1}{2}\cdot c\cdot x_3=371,5424\cdot \frac{2,9066}{2\cdot 3,1416}-\frac{1}{2}\cdot 21,6\cdot 1,275\doteq 158,1036{\text{cm}}^2 \\ \text{ Skuˊsˇka spraˊvnosti: } \\ {t}_1={\phi }_1+{\phi }_2+{\phi }_3=1,522+1,8546+2,9066\doteq 6,2832\text{rad} \\ s=\frac{a+b+c}{2}=\frac{15+17,4+21,6}{2}=27\text{cm} \\ S=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{27\cdot (27-15)\cdot (27-17,4)\cdot (27-21,6)}=\frac{648}{5}=129,6{\text{cm}}^2 \\ {S}_4=S_1+S_2+S_3+S=30,9391+52,8996+158,1036+129,6\doteq 371,5424{\text{cm}}^2 \\ {S}_4=S_0 \end{gathered}$$

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