S Chand ICSE Mathematics Class 9 Solution Thirteenth Chapter Circle Exercise 13A
(1) Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.
(2) In Fig. 13.30, the radius of the given circle, with centre C, is 6 cm if the chord AB is 3 cm away from the centre, calculate its length.
(3) In Fig. 13.31, CD is a diameter which meets the chord AB in E, such that AE = BE = 4 cm. If CE is 3 cm, find the radius of the circle.
(4) In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords, if they are on
(i) the same side of the centre
(ii) opposite sides of the centre.
(5) The radius of a circle is 2.5 cm. AB, CF are two parallel chords 3.9 cm apart. If AB = 1.4 cm, find CF.
(6) A chord distant 2 cm from the centre of a circle is 18 cm long. Calculate the length of a chord of the same circle which is 6 cm distant from the centre.
(7) In Fig. 13.32, circles are concentric with centre O. Find AC.
(8) The length of the common-chord of two equal intersecting circles is 10 cm and the distance between the two centres is 6 cm. Find the radius of each circle (Fig. 13.33)
(9) In Fig. 13.35, CD is the perpendicular bisector of the chord AB. If AB = 2 cm and CD = 4 cm, calculate the radius of the circle.
(10) In Fig 13.35, CD is the perpendicular bisector of the chord AB. If AB = 2 cm and CD = 4 cm, calculate the radius of the circle.
(11) In Fig. 13.36, OD is perpendicular to the chord AB of a circle, whose centre is O. Prove that CA = 2OD.
(12) In Fig. 13.37, OMNP is a square. A circle drawn with centre O cuts the square in X and Y. Prove that: △OXM ≅ △OYP. Hence prove that NX = NY.