Two chords

In a circle with radius r = 26 cm two parallel chords are drawn. One chord has a length t1 = 48 cm and the second has a length t2 = 20 cm, with the center lying between them. Calculate the distance of two chords.

Correct result:

v₁ =  34 cm
v₂ =  14 cm

Solution:

$$\begin{gathered} r=26\text{ cm } \\ {t}_1=48\text{ cm } \\ {t}_2=20\text{ cm } \\ {x}_1=\sqrt{r^2-(t_1\mathrm{/}2{)}^2}=\sqrt{26^2-(48\mathrm{/}2{)}^2}=10\text{ cm } \\ {x}_2=\sqrt{r^2-(t_2\mathrm{/}2{)}^2}=\sqrt{26^2-(20\mathrm{/}2{)}^2}=24\text{ cm } \\ {v}_1=x_2+x_1=24+10=34\text{ cm} \end{gathered}$$

$v_2=\mathrm{\mid }{x}_2-x_1\mathrm{\mid }=\mathrm{\mid }24-10\mathrm{\mid }=14\text{ cm}$

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