Question

Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.

Answer 1

Construction: Draw OP ⊥ CD 

Chord AB = 5cm 

Chord CD = 11 cm 

Chord PQ = 3 cm 

Let OP = x cm 

And  OC = OA = r cm 

WKT perpendicular from center to chord bisects it

`∴CP= PD = 11/2 cm`

and `AQ = BQ=5/2 cm`

In , ΔOCP  by Pythagoras theorem

`OC^2=OP^2+CP^2`

`⇒r^2=x^2+(11/2)^2`         ...(1)

In , OQAΔ by Pythagoras theorem

`OA^2=OQ^2+AQ^2`

`⇒r^2=(x+3)^2+(5/2)^2`          ...(2)

Compare equation (1) and (2)

`(x+3)^3(5/2)^2=x^2+(11/2)^2`

`⇒x^2+9+6x+25/4=x^2+(121/4)`

`⇒x^2+6x-x^2=121/4-25/4-9`

`⇒6x=15`

`⇒x=15/6=5/2`

Put the value of x in equation (1)

`r^2 = (5/2)^2 + (11/2)^2`

⇒ `r^2 = 25/4 + 121/4 = 146/4`

⇒ `r = (sqrt146)/2` cm

∴ Radius of circle = `(sqrt146)/2` cm