Chord 24

A chord of length t = r√2 divides a circle with radius r into two circular segments. What is the ratio of the areas of these two segments?

Final Answer:

x =  1:10

Step-by-step explanation:

$$\begin{gathered} t=r\cdot \sqrt{2} \\ \sin \alpha =t/2:r \\ \alpha =\frac{180°}{\pi }\cdot \arcsin (\frac{\sqrt{2}/2}{1})=45{}^{\circ } \\ \Phi =2\cdot \alpha =2\cdot 45=90{}^{\circ } \\ z=\sqrt{r^2-(t/2{)}^2}=\sqrt{1-1/2}=\sqrt{1/2}=\sqrt{2}/2 \\ {S}_1=\frac{\Phi }{360}\cdot \pi =\frac{90}{360}\cdot 3.1416\doteq 0.7854 \\ {S}_2=z\cdot z \\ {S}_2=2/4=\frac{1}{2}=0.5 \\ A=S_1-S_2=0.7854-0.5\doteq 0.2854 \\ B=\pi r^2-A \\ B=\pi -A=3.1416-0.2854\doteq 2.8562 \\ {x}_0=A:B=0.2854:2.8562\doteq 0.0999=33146:331717 \\ x=1:10=0.1 \end{gathered}$$

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