Question

In a circle two parallel chords of length 4 and 6 centemeter are 5cm apart what is the radius of the circle?

Answer 1

We know that chords can be bisected perpendicularly.

Let's construct a triangle using the shorter chord and a segment (the 5cm long gap) of a common bisector of the two chords (radius). You should get an isosceles of altitude 5cm, two sides of √29cm and a base of 4cm.

Arctan(2/5) to find half the angle opposite to the base (remember, the altitude here bisects at 90°).

Let the distance between the centre and the 6cm chord be x. Now, using the supplementary angle of arctan(2/5) and one of the √29cm, we can find the radius. Use the cosine rule (c^2=a^2+b^2–2abcosC), letting a be x, b be √29cm and C be arctan(2/5). c, not C, is the radius.

I'll be ignoring units from now; bear with me.

The radius squared is also x^2+9, if we construct and split an isosceles with the 6cm chord into two congruent right angled triangles. After simultaneous equations, you should get 20-2x√29cos(arctan(2/5)) to equal 0. x turns out to be 2.

So the radius is √13(cm, if you're pedantic about units).

Edit: I completely forgot about another case. This other case is when the centre is between the two chords. This case should be easier, as by Pythagoras' theorem, (5-x)^2+4=r^2 and x^2+9=r^2. x is the distance between the 6cm chord and the centre. 5-x is the distance between the centre and the 4cm one. The radius happens to be the same here.

Edit 2: I realised how roundabout my first case was, and yes, it is fine to only use Pythagoras' theorem there. Just be careful about x being negative (I think it has something to do with flipping a chord along an axis from Case 2). You should still get √13.