# Thread: Question with trapezoid inside circle

1. ## Question with trapezoid inside circle

Can someone tell me thoroughly how to do #5. Unless I am mistaken, I did everything right except up to and am stuck at finding the measures of BC and AD.

Any help would be greatly appreciated!
Thanks!

2. ## Question

I just wanted to verify.
The given lengths are DM = 3, OM = 3, and ON = 2, correct?

3. Yes, that is correct.

4. ## Possible Solution

Okay, this is what I came up with:
We are given DM = 3, OM = 3 and ON = 2.
Since the diameter of the circle bisects the trapezoid, we know that MC = 3 as well. We can compute OC (the radius) by using Pythagorean Theorem:

OM^2 + MC^2 = OC^2
3^2 + 3^2 = OC^2
OC = sqrt(18)
OC = 3*sqrt(2)

Now that we have OC, we know that OB is also 3*sqrt(2) because they are both radii of the circle. We are also given ON = 2, so we can compute NB by the Pythagorean Theorem:

ON^2 + NB^2 = OB^2
2^2 + NB^2 = (3*sqrt(2))^2
4 + NB^2 = 18
NB = sqrt(14)

Now that we have NB, we also know that AN = sqrt(14) as well.

This was the somewhat tricky part. I drew a line parallel to MN from point B up to a point outside the circle to the right of point C. I called it X. This forms a rectangle MNBX. Since MN is parallel to XB, we know that XB also equals 5 since opposite sides of rectangles are congruent, and using the fact that MO + ON = 3 + 2 = 5 = MN.

Now, our job is to find CB. Once we find this length our job is done because this will also be equal to the length of DA.

So, we have XB equals 5. But we also know the length of MX = sqrt(14) since it is parallel to NB. Also, MC = 3. So, we can find the length of CX by doing:

CX = sqrt(14) - 3.

Now, we can find CB by using the Pythagorean Theorem:

CX^2 + XB^2 = CB^2
[sqrt(14) - 3]^2 + 5^2 = CB^2
14 - 6*sqrt(14) + 9 + 25 = CB^2
48 - 6*sqrt(14) = CB^2
CB = sqrt[6*(8-sqrt(14))]

So, all the lengths are as follows:
DM = MC = 3
NB = AN = sqrt(14)
CB = DA = sqrt[6*(8-sqrt(14))]

So, perimeter of ABCD = 6 + 2*sqrt(14) + 2*sqrt[6*(8-sqrt(14))]

I hope this was somewhat useful.