Two parallel chords of a circle, which are on the same side of centre,subtend angles of `72^@ and 144^@` respetively at the centre Prove that the perpendicular distance between the chords is half the radius of the circle.

Let PQ and RS be two parallel chords of a given circle of radius 6 cm, lying on the same side of the center. If the chords subtends angles of 72^(@) and 144^(@) at the center and the distance between the chords is d, then d^(2) is equal to

AB and CD are two parallel chords of a circle of radius 6 cm lying on the same side of the centre. If the chords subtend angles of 72^circ and 144^circ at the centre, find the distance between the chords.

Two parallel chords are drawn on the same side of the centre of a circle of radius R .It is fouind that they subtend angles of theta and 2 theta at the centre of the circle.The prependicular distance between the chords is

2 parallel chords 24 cm, and 18 cm, are on the same side of the centre of a circle. If the distance between the chords is 3 cm, calculate the radius of the circle.

Two parallel chords are drawn on the same side of the centre of a circle of radius R. It is found that they subtend and angle of theta and 2theta at the centre of the circle. The perpendicular distance between the chords is

Two parallel chords are drawn on the same side of the centre of a circle of radius R. It is found that they subtend and angle of theta and 2theta at the centre of the circle. The perpendicular distance between the chords is

Two parallel chords are drawn on the same side of the centre of a circle of radius R. It is found that they subtend and angle of theta and 2theta at the centre of the circle. The perpendicular distance between the chords is

Let p and q be the length of two chords of a circle which subtend angles 36^@ and 60^@ respectively at the centre of the circle . Then , the angle (in radian) subtended by the chord of length p + q at the centre of the circle is (use pi=3.1 )