Posted by **Bo** on Friday, May 25, 2012 at 3:17pm.

in circle p below the lengths of the parallel chords are 20,16, and 12. Find measure of arc AB..... the chord with a length of 20 is the diameter. the chords with lengths 16 and 12 are below the diameter torwards the bottom of the circle. arc AB is the arc made up of the endpoints of the chords with lengths 16 and 12.

- math -
**MathMate**, Friday, May 25, 2012 at 5:42pm
Take a piece of paper and note the following on your diagram. It will be difficult to follow without an annotated diagram.

Label the (horizontal) diameter as CD, with C on the left.

Label the chord 16 units long as PA (P on the left).

Label the chord 12 units long as QB (Q on the left).

Label the centre of the circle as O.

Construct a (vertical) diameter perpendicular to CD. Label the ends as E (on top) and F (at the bottom).

Label the intersection PA and EF as X, and the distance OX=x.

Label the intersection QB and EF as Y, and the distance OY=y.

Label the radius of the circle as r = OD.

Confirm that:

OD=OE=OC=OF=r=10

PX=XA=8

QY=YB=6

We don't know yet the distances x and y.

By the theorem of intersection of chords, we have

PX*XA = FX*XE

8²=(10-x)(10+x)

Solve for x to get 6.

θ1=∠AOD=sin^{-1}(6/10)

Similarly,

QY*YB=(10-y)(10+y)

6²=(10-y)(10+y)

Solve for y to get 8.

θ2=∠BOD=sin^{-1}(6/10)

⇒

φ=∠BOA=θ2-θ1

Arc length of AB

=rφ/2π

=10φ/2π

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