Question

2) In a circle of radius 6 cm, there are two chords of length 10 cm
and 11 cm. If two chords are parallel to each other then find
distance between two chords.​

Answer 1

Step-by-step explanation:

AB=10cm

CD=11cm

we know that distance of chord is perpendicular from centre

Answer 2

Answer:

Let O is center and r is radius of circle r = 10 cm chord AB = 12 cm and chord CD = 16 cm. Draw OP ⊥ AB which cuts chord CD at Q Since AB || CD Thus, OQ || CD AP = BP = 1212AB = 1212 × 12 = 6 cm and CQ = QD = 1212 × CD = 1212 × 16 = 8 cm. In right angled triangle OPA By Pythagoras theorem OA2 = AP2 + OP2 (10)2 = (6)2 + (OP)2 100 = 36 + (OP)2 OP2 = 100 – 36 = 64 OP = 64−−√64 = 8 cm Similarly on right angled triangle OCQ By Pythagoras theorem OC 2 = CQ2 + OQ2 (10)2 = (8)2 + OQ2 100 = 64 + OQ2 OQ2 = 100 – 64 = 36 OQ = 36−−√36 = 6 cm (a) Hence, distance between AB and CD PQ = OP – OQ = 8 – 6 cm = 2 cm (b) Hence, distance between two chords AB and CD PQ = OP + OQ = 8 + 6 cm PQ = 14 cm Thus, Distance between two chords is 14 cmRead more on Sarthaks.com - https://www.sarthaks.com/761185/in-a-circle-of-radius-10-cm-length-of-two-parallel-chords-are-12-cm-and-16-cm-respectively