Question

AB and CD are two chords of a circle on the same
side of the centre such that AB = 8 cm, CD = 10 cm
and AB || CD. If the distance between AB and CD
is 2 cm, then find the radius of the circle.

Pls send the ans ASAP

Answer 1

Answer:

Step-by-step explanation:

We know perpendicular to chord from centre bisects it.

Perpendiculars to AB and CD are drawn from centre, meeting at L and M, which are the distances from the centre.

The centre (let's say O) is joined with A and C

Now, we get OAL and OCM as two right triangles.

AL = 1/2AB = 4 cm

CM = 1/2CD = 5 cm

Now, AL^2 + LO^2 = AO ^2 = r^2, where r is the radius.

        CM^2 + MO^2 = CO^2 = r^2

Therefore, AL^2 + LO^2 = CM^2 + MO^2

LO^2 -  MO^2  =  CM^2 - AL^2              

(LO + MO)(LO - MO) = 5^2 - 4^2 = 25-16 = 9

LM(LO+MO) = 9

LO+MO = 9/2 = 4.5

LO+MO = 4.5, LO-MO = 2

So, LO = 3.25, MO = 0.75

So, r = √(5^2 + 1.25^2) = √(25 + 25/16) = (5√17)/4