Circumscribed circle ABC

Triangle ABC, with sides a = 15 cm, b = 17.4 cm, and c = 21.6 cm, is circumscribed by a circle. Calculate the area of the segments determined by the sides of the triangle.

Final Answer:

S₁ =  30.9391 cm²
S₂ =  52.8996 cm²
S₃ =  158.1036 cm²

Step-by-step explanation:

$$\begin{gathered} a=15\text{cm} \\ b=17.4\text{cm} \\ c=21.6\text{cm} \\ {a}^2=b^2+c^2-2bc\cdot \cos \alpha \\ A=\arccos (\frac{b^2+c^2-a^2}{2\cdot b\cdot c})=\arccos (\frac{17.4^2+21.6^2-15^2}{2\cdot 17.4\cdot 21.6})\doteq 0.761\text{rad} \\ \alpha =A\to \text{°}=A\cdot \frac{180}{\pi }\text{°}=0.761\cdot \frac{180}{\pi }\text{°}=43.60282\text{°} \\ B=\arccos (\frac{a^2+c^2-b^2}{2\cdot a\cdot c})=\arccos (\frac{15^2+21.6^2-17.4^2}{2\cdot 15\cdot 21.6})\doteq 0.9273\text{rad} \\ \beta =B\to \text{°}=B\cdot \frac{180}{\pi }\text{°}=0.9273\cdot \frac{180}{\pi }\text{°}=53.1301\text{°} \\ \gamma =180-\alpha -\beta =180-43.6028-53.1301\doteq 83.2671{}^{\circ } \\ r=\frac{a}{2\cdot \sin \alpha }=\frac{a}{2\cdot \sin 43.602818972704°}=\frac{15}{2\cdot \sin 43.602818972704°}=\frac{15}{2\cdot 0.689655}=10.875\text{cm} \\ {x}_1=\sqrt{r^2-(a/2{)}^2}=\sqrt{10.875^2-(15/2{)}^2}=\frac{63}{8}=7.875\text{cm} \\ {x}_2=\sqrt{r^2-(b/2{)}^2}=\sqrt{10.875^2-(17.4/2{)}^2}=\frac{261}{40}=6.525\text{cm} \\ {x}_3=\sqrt{r^2-(c/2{)}^2}=\sqrt{10.875^2-(21.6/2{)}^2}=\frac{51}{40}=1.275\text{cm} \\ {h}_1=r-x_1=10.875-7.875=3\text{cm} \\ {h}_2=r-x_2=10.875-6.525=\frac{87}{20}=4.35\text{cm} \\ {h}_3=r-x_3=10.875-1.275=\frac{48}{5}=9.6\text{cm} \\ {a}^2=r^2+r^2-2r^2\cdot \cos {\phi }_1 \\ {\phi }_1=\arccos (\frac{2\cdot r^2-a^2}{2\cdot r^2})=\arccos (\frac{2\cdot 10.875^2-15^2}{2\cdot 10.875^2})\doteq 1.522\text{rad} \\ {\phi }_2=\arccos (\frac{2\cdot r^2-b^2}{2\cdot r^2})=\arccos (\frac{2\cdot 10.875^2-17.4^2}{2\cdot 10.875^2})\doteq 1.8546\text{rad} \\ {\phi }_3=\arccos (\frac{2\cdot r^2-c^2}{2\cdot r^2})=\arccos (\frac{2\cdot 10.875^2-21.6^2}{2\cdot 10.875^2})\doteq 2.9066\text{rad} \\ {S}_0=\pi \cdot r^2=3.1416\cdot 10.875^2\doteq 371.5424{\text{cm}}^2 \\ {S}_1=S_0\cdot \frac{{\phi }_1}{2\pi }-\frac{1}{2}\cdot a\cdot x_1=371.5424\cdot \frac{1.522}{2\cdot 3.1416}-\frac{1}{2}\cdot 15\cdot 7.875=30.9391{\text{cm}}^2 \end{gathered}$$

$S_2=S_0\cdot \frac{{\phi }_2}{2\pi }-\frac{1}{2}\cdot b\cdot x_2=371.5424\cdot \frac{1.8546}{2\cdot 3.1416}-\frac{1}{2}\cdot 17.4\cdot 6.525=52.8996{\text{cm}}^2$

$$\begin{gathered} S_3=S_0\cdot \frac{{\phi }_3}{2\pi }-\frac{1}{2}\cdot c\cdot x_3=371.5424\cdot \frac{2.9066}{2\cdot 3.1416}-\frac{1}{2}\cdot 21.6\cdot 1.275\doteq 158.1036{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ {t}_1={\phi }_1+{\phi }_2+{\phi }_3=1.522+1.8546+2.9066\doteq 6.2832\text{rad} \\ s=\frac{a+b+c}{2}=\frac{15+17.4+21.6}{2}=27\text{cm} \\ S=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{27\cdot (27-15)\cdot (27-17.4)\cdot (27-21.6)}=\frac{648}{5}=129.6{\text{cm}}^2 \\ {S}_4=S_1+S_2+S_3+S=30.9391+52.8996+158.1036+129.6\doteq 371.5424{\text{cm}}^2 \\ {S}_4=S_0 \end{gathered}$$

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