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

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GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4889
GMAT 1: 770 Q49 V46
O is the center of the circle. If line segment DC has length 5, and si  [#permalink]

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Updated on: 08 Aug 2019, 12:57
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Difficulty:

95% (hard)

Question Stats:

35% (03:12) correct 65% (03:51) wrong based on 23 sessions

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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.

Attachment:

LmRAOvn.png [ 7.25 KiB | Viewed 583 times ]

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Originally posted by BrentGMATPrepNow on 08 Aug 2019, 07:00.
Last edited by BrentGMATPrepNow on 08 Aug 2019, 12:57, edited 1 time in total.
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6320
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: O is the center of the circle. If line segment DC has length 5, and si  [#permalink]

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Updated on: 08 Aug 2019, 10:59
1
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 ;

GMATPrepNow wrote:

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.

Attachment:
LmRAOvn.png

Originally posted by Archit3110 on 08 Aug 2019, 10:07.
Last edited by Archit3110 on 08 Aug 2019, 10:59, edited 1 time in total.
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Joined: 20 Jul 2017
Posts: 1506
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: O is the center of the circle. If line segment DC has length 5, and si  [#permalink]

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08 Aug 2019, 10:56
2
GMATPrepNow wrote:

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.

Attachment:
LmRAOvn.png

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

Posted from my mobile device
Director
Joined: 16 Jan 2019
Posts: 614
Location: India
Concentration: General Management
WE: Sales (Other)
Re: O is the center of the circle. If line segment DC has length 5, and si  [#permalink]

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08 Aug 2019, 11:38
1
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

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

$$\frac{24}{x^2+25}=\frac{x^2}{25}$$

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

$$x^2=15$$
$$x=\sqrt{15}$$

Posted from my mobile device
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4889
GMAT 1: 770 Q49 V46
Re: O is the center of the circle. If line segment DC has length 5, and si  [#permalink]

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08 Aug 2019, 12:54
Top Contributor
Archit3110 wrote:
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 ;

GMATPrepNow wrote:

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.

Attachment:
LmRAOvn.png

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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Re: O is the center of the circle. If line segment DC has length 5, and si   [#permalink] 08 Aug 2019, 12:54