Two parallel chords of a circle of radius $2$ are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the center, angles of

$\text{Given: circle radius } R = 2$ $\text{two parallel chords, distance apart } = \sqrt{3} + 1$ $\text{angles subtended at the centre: } \dfrac{\pi}{3} \text{ and } \dfrac{2\pi}{3}$ $\text{To find: lengths of the two chords}$ $\text{and their perpendicular distances from the centre}$ $1.\ \text{drop a perpendicular from the centre to a chord}$ $2.\ \text{it bisects the chord; right triangle with hypotenuse } R, \text{ angle } \dfrac{\theta}{2}$ $3.\ \text{write } d = R\cos\dfrac{\theta}{2}, \quad L = 2R\sin\dfrac{\theta}{2}$ $\text{perpendicular from the centre bisects the chord}$ $\cos\dfrac{\theta}{2} = \dfrac{d}{R} \Rightarrow d = R\cos\dfrac{\theta}{2}$ $\sin\dfrac{\theta}{2} = \dfrac{L/2}{R} \Rightarrow L = 2R\sin\dfrac{\theta}{2}$ $\theta_1 = \dfrac{\pi}{3}: \quad d_1 = 2\cos\dfrac{\pi}{6} = \sqrt{3}$ $L_1 = 2(2)\sin\dfrac{\pi}{6} = 2$ $\theta_2 = \dfrac{2\pi}{3}: \quad d_2 = 2\cos\dfrac{\pi}{3} = 1$ $L_2 = 2(2)\sin\dfrac{\pi}{3} = 2\sqrt{3}$ $d_1 + d_2 = \sqrt{3} + 1 = \text{given distance apart}$ $\Rightarrow \text{the chords lie on opposite sides of the centre}$ $\text{Lengths: } L_1 = 2 \ (\theta = \pi/3), \quad L_2 = 2\sqrt{3} \ (\theta = 2\pi/3)$ $\boxed{\text{Chords: } 2 \text{ and } 2\sqrt{3}}$