O is the center of the circle. If line segment DC has length 5, and si

A perpendicular drawn from the right angle to the hypotenuse in a right traingle divides the triangle into two similar triangles each of which are also similar to the original triangle

By this property, since AC is the diameter which makes ABC a right angled triangle,

ABC, ADB and BDC are similar triangles

From BDC, \(BC=\sqrt{x^2+25}\)

From ABC and BDC

\(\frac{AB}{BC}=\frac{BD}{DC}\)
\(\frac{√24}{\sqrt{x^2+25}}=\frac{x}{5}\)

\(\frac{24}{x^2+25}=\frac{x^2}{25}\)

\(x^4+25x^2-600=0\)
\((x^2+40)(x^2-15)=0\)

\(x^2=15\)
\(x=\sqrt{15}\)

Answer is (A)

Posted from my mobile device

This is a great approach. In fact, if I had used that approach, I would have seen that the diagram I created cannot exist in our Universe.

When creating the question, I used the similar triangles approach used by Dillesh4096 and firas92 asmd and my calculations yielded an answer of \(\sqrt{15}\) (answer choice A)
However, this means my diagram makes no sense whatsoever.
Notice that, if \(x=\sqrt{15}\), then we can use the Pythagorean Theorem to see that side BC has length \(\sqrt{40}\)
And all of this means that side AC has length 8
In other words, the diameter of the circle is 8, which means the radius is 4.
This is where things get ugly. If the radius is 4, how can side DC = 5?

Ughh!!!!!!
That said, the above solutions are great/valid.
The only issue is that diagram is nonsensical.
My apologies.

Kudos for everyone!!!

Cheers,
Brent

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