Question

AB and CD are two parallel chords of a circle of length 24cm and 10cm respectively and lie on the same side of its centre O. If the distance between AB and CD is 7cm, find the radius of the circle....

Plz help me out its really important ​

Answer 1

Given AB = 24 cm and CD = 10 cm are two chords of the circle.

E and F are the mid points of AB and CD respectively.

Again, given distance between two chords = 7 cm

Let distance between O and F = x

Then distance between O and E = 7 - x

Now, in triangle OEB,

from Pythagoras theorem,

      OB2 = OE2 + ED2

=> OB2 = (7 - x)2 + 122

=> OB2 = (7 - x)2 + 144 ..........1

Again

in triangle OFD,

from Pythagoras theorem,

      OD2 = OF2 + FD2

=> OD2 = x2 + 52

=> OD2 = x2 + 25 .............2

 Since OB = OD {radius}

    (7 - x)2 + 144 = x2 + 25

=> 49 + x2 - 14x + 144 = x2 + 25

=> 49 - 14x + 144 =  25

=> 14x = 49 + 144 - 25

=> 14x = 193 - 25

=> 14x = 168

=> x = 168/14

=> x = 12

fron equation 2, we get 

      OD2 = 122 + 25

=> OD2 = 144 + 25

=> OD2 = 169

=> OD = √169

=> OD = 13

So, the radius of the circle is 13 cm