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

Question

https://gmatclub.com/forum/download/file.php?id=49655 O is the center of the circle. If line segment DC has length 5, and side AB has length \sqrt{24} , what is the length of x? A) \sqrt{15} B) 2\sqrt{5} C) 2\sqrt{6} D) \sqrt{30} E) 6 Aside: I posted a nearly-identical question a few minutes ago, and realized there was a flaw. So, I deleted the question and posted this corrected version. Sorry to those who attempted my first post. Well, it turns out I wasn't finished making flawed questions!!

Please see my explanation (as to why this is a flawed question) below. LmRAOvn.png

CareerGeek · Accepted Answer

Angle in a semicircle is 90
—> ∠B = 90

Triangles ADB & BDC are similar
—> DB/AB = CD/BC
—> x/√24 = 5/√(5^2 + x^2)
Squaring on both sides,
—> x^2/24 = 25/(5^2 + x^2)
—> 25x^2 + x^4 = 600
—> x^4 + 25x^2 - 600 = 0
—> x^4 + 40x^2 - 15x^2 - 600 = 0
—> (x^2 + 40)(x^2 - 15) = 0
—> x = √15

IMO Option A

Pls Hit kudos if you like the solution

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Archit3110 · Answer

GMATPrepNow 
giving a try
I solved the question using given answer choices for x
let side OD = a
so radius of circle = 5+a 
for ∆ ABD we can say 
(√24)^2= ( x)^2+ ( 5+2a)^2 ------(1)
substituting values of x we can clearly see that options c,d,e wont be valid as in that case the relation (1) wont be valid
amongst a & b
I checked with x= \sqrt{15}  &  2\sqrt{5}  ; there in value of a is coming out to be -ve...
I am not sure whether which one would be correct.. as value of a is coming an integer for option A ( -4,-1) and fraction for option B ( -7/2,-3/2)
IMO A ;

firas92 · Answer

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

BrentGMATPrepNow · Answer

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

bumpbot · Answer

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

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

Prep Toolkit

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My Rewards