Question

In a circle of radius `'r'`, chords of lengths `a` and `b` cms, subtend angles `q` and `3q` respectively at the centre, then `r = asqrt(a/(2a-b))` cm. True or False?

Answer 1

The given relation is false

Explanation:

We know that perpendicular dropped from the centre of the circle to a chord bisects the chord and the angle subtended by the chord at the centre of the circle.

So here we have

`/_AOM=1/2/_AOB and AM=1/2AB`
If `/_AOB=x , OA =r and AB=d` then `sin(x/2)=(AM)/(OA)=(d/2)/r=d/(2r)`
By the problem

When `d=a and x=q`

We get

`sin(q/2)=a/(2r)......(1)`

Again when `d=b and x=3q`

We get

`sin((3q)/2)=b/(2r)......(2)`

Dividing (2) by (1) we get

`sin((3q)/2)/sin(q/2)=b/a`

`=>(3sin(q/2)-4sin^3(q/2))/sin(q/2)=b/a`

`=>3-4sin^2(q/2))=b/a`

`=>3-4(a/(2r))^2=b/a`

`=>3-a^2/r^2=b/a`

`=>a^2/r^2=3-b/a`

`=>a^2/r^2=(3a-b)/a`

`=>r^2=a^3/(3a-b)`

`=>r=asqrt(a/(3a-b))`

Hence the given relation is false