Question

AB is a diameter of a circle with centre o.AB=34cm and CD is a chord of length 30 cm then the distance of CD from AB​

Answer 1

Answer:

Step-by-step explanation:

if AB is 34 cm then its radius is 17 cm ,because the radius is half the diameter

if CD is the chord then draw a line perpendicular from center o to CD and name the point of intersection as p. Now join O TO D.

now you have got a right angle triangle whose sides are OP,PD AND OD.similarly OD becomes the radius of the circle and the hypotenuse of the triangle so, OD=17cm and OP=15cm(because a line which is drawn perpendicularly from the centre to the chord divides the chord equally)

then by pythogoras theorem

17²=15²+OP²

289=225+OP²

289-225=OP²

64=OP²

OP²=8²

OP=8cm

and OP is also the distance between AB and CD.

Therefore,the distance between AB and CD is 8cm.

tq...............................

Answer 2

Step-by-step explanation:

if AB is 34 cm then its radius is 17 cm ,because the radius is half the diameter

if CD is the chord then draw a line perpendicular from center o to CD and name the point of intersection as p. Now join O TO D.

now you have got a right angle triangle whose sides are OP,PD AND OD.similarly OD becomes the radius of the circle and the hypotenuse of the triangle so, OD=17cm and OP=15cm(because a line which is drawn perpendicularly from the centre to the chord divides the chord equally)

then by pythogoras theorem

17²=15²+OP²

289=225+OP²

289-225=OP²

64=OP²

OP²=8²

OP=8cm

and OP is also the distance between AB and CD.

Therefore,the distance between AB and CD is 8cm.