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In the figure above, AB, which has length z cm, is tangent to the

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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post Updated on: 26 Jul 2017, 09:00
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In the figure above, AB, which has length z cm, is tangent to the circle at point A, and BD, which has length y cm, intersects the circle at point C. If BC = x cm and \(z = \sqrt{xy}\), what is the value of x ?

(1) CD = x cm

(2) \(z = 5 \sqrt{2}\)


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Originally posted by carcass on 26 Jul 2017, 08:58.
Last edited by Bunuel on 26 Jul 2017, 09:00, edited 1 time in total.
Formatted the question.
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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 26 Jul 2017, 14:51
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Sol- we are given and also by second theorem that z^2=xy
1- this gives y=2x--> z^2=2x^2
2-\(z=5\sqrt{2}\)...not given x or y

Combining we know
\(5\sqrt{2}^2=2x^2\)
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 19 Jan 2018, 00:14
From 1 statement we know y = 2x
And then z = x√3 but to calculate z we need it's value
From 2 statement we know will get z
So it's together sufficient

study mode
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 19 Jan 2018, 01:33
carcass , Can you please let us know . How this question can be solved.

Thanks.
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 19 Jan 2018, 04:08
Given: z = Root(X*Y) -> to answer question we need to have information about z and y

1) y = 2x -> z=Root(x*2x) = Root (2x^2) = x*root(2) -> however INSUFF. as we have no information about x
2) z=5*root(2) clearly Insuff as we don't have any details about x or y

However combining (1) + (2) you get x*root(2) = 5*root(2) => x=5 , therefor SUFFICIENT
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 19 Jan 2018, 06:53
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prashant0099 wrote:
carcass , Can you please let us know . How this question can be solved.

Thanks.


Hi

You can go through the solutions posted in this thread.
You can also read about circles theory here:

https://gmatclub.com/forum/math-circles-87957.html
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 17 May 2018, 08:24
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Bunuel, chetan2u could you please share your approach here? Thanks.
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 26 Jun 2018, 14:22
In the OG2018 explanation of this answer it is said that only by looking at statement 2) one can conclude that CD is a diameter of the circle. How do we get to that conclusion?
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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 04 Jul 2018, 04:55
amanvermagmat wrote:
prashant0099 wrote:
carcass , Can you please let us know . How this question can be solved.

Thanks.


Hi

You can go through the solutions posted in this thread.
You can also read about circles theory here:

https://gmatclub.com/forum/math-circles-87957.html



hey generis today is international day of indepencence :) so i tried to solve the above DS question independently, :grin: i ve read whole thread also visited this link...

but i smply couldnt understand how to approach the problem...

i know formula of area of circle PiR^2 and also circumference 2PiR

i know what is tangent, what is diametr and radius, but still couldnt wrap my mind aroud this problem what rule to apply here , can you help please :-)

have a fantastic day :)
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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 04 Jul 2018, 17:49
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dave13 wrote:
carcass wrote:
Image
In the figure above, AB, which has length z cm, is tangent to the circle at point A, and BD, which has length y cm, intersects the circle at point C. If BC = x cm and \(z = \sqrt{xy}\), what is the value of x ?

(1) CD = x cm

(2) \(z = 5 \sqrt{2}\)

Attachment:
circle.jpg

amanvermagmat wrote:
prashant0099 wrote:
carcass , Can you please let us know . How this question can be solved.
Thanks.

Hi
You can go through the solutions posted in this thread.
You can also read about circles theory here:

https://gmatclub.com/forum/math-circles-87957.html

hey generis today is international day of indepencence :) so i tried to solve the above DS question independently, :grin: i ve read whole thread also visited this link...

but i smply couldnt understand how to approach the problem...

i know formula of area of circle PiR^2 and also circumference 2PiR

i know what is tangent, what is diametr and radius, but still couldnt wrap my mind aroud this problem what rule to apply here , can you help please :-)

have a fantastic day :)

dave13 , cosmopolitanism meets metaphor meets smiley face ... Classic. :-)

This question is not about area or circumference.
It involves the "power of a point" theorem, one version of which is called the "secant-tangent theorem."
I'll put reference material below.

BUT: you do not need to use the secant-tangent theorem.

I will use the theorem in Method II because a harder question would not give us \(z=\sqrt{xy}\)
The power of a point theorem has three iterations, is very useful, and is not hard.
Initially it might appear so. Read material linked below. :-)

We do not need the theorem because this part is given: \(z=\sqrt{xy}\)
The variables are lengths, thus positive. Square both sides: \(z^2=xy\)
See below. That equation is the essence of the secant-tangent theorem.

I. WITHOUT the theorem - Unique value for \(x\)?

(1) \(CD = x\) cm
\(y - CD = x\)
\(y - x = x\)
\(y=2x\)

Given: \(z=\sqrt{xy}\)
\(z=\sqrt{x*2x}=\sqrt{2x^2}=x\sqrt{2}\)
\(z=x\sqrt{2}\)
Two variables, one equation. Insufficient.
OR: we know nothing about \(z\), hence nothing about \(x\)

Or: test numbers that make this statement true: \(z=x\sqrt{2}\)
If you can get two different values for \(x\), #1 is insufficient
Let \(x = 25\) and \(z = 25\sqrt{2}\)
Let \(x = 2\) and \(z = 2\sqrt{2}\)
Two values of x = not unique
INSUFFICIENT alone. Cross off A and D

(2) \(z = 5\sqrt{2}\)
\(z=\sqrt{xy}\)
\(z = 5\sqrt{2}\)
\(5\sqrt{2}=\sqrt{xy}\)
Two variables, one equation. Insufficient.
OR: we know nothing about \(y\), hence nothing about \(x\)

OR test numbers.
\(z=\sqrt{xy}\)
\((5\sqrt{2})=\sqrt{xy}\)
Let \(x = 5\) and \(y = 10\)
Let \(x = 10\) and \(y = 5\)
#2 does not yield a unique solution for \(x\). Insufficient. Cross off B.

C or E?
Combine #1 and #2:
#1, expanded:
\(y = 2x\)
\(z=\sqrt{xy}\)
\(z=\sqrt{x*2x}=\sqrt{2x^2}\)

Combined #1 and #2
\(5\sqrt{2}=\sqrt{2x^2}\)
One variable, one equation. Sufficient. Or solve
\(5\sqrt{2}=x\sqrt{2}\)
\(x=5\) That works

Answer C

II. Secant-tangent theorem

In the diagram, point B lies outside the circle.
AB (= z) is tangent to the circle
BD is a secant, a line that intersects a circle at two points
BD (= y) has two segments, BC (= x) and CD

The secant-tangent theorem states
\((AB)(AB)=(BC)(BD)\)
\((AB)^2=(BC)(BD)\)

In this case, using the theorem with given lengths, we have
\(z^2=xy\)

We need information about \(y\) and \(z\) to get a unique value for \(x\)

(1) \(CD = x\)
Inspect the diagram: \(y=2x\), or

\(y - CD = x\)
\(y - x = x\)
\(y=2x\)
From secant-tangent theorem:
\(z^2=xy\)
\(z^2=(x*2x)=2x^2\)
Two variables, one equation.
Insufficient.

Test numbers:
Let \(x = 1\) and \(z = \sqrt{2}\)
Let \(x = 2\) and \(z = \sqrt{8}\)

Or, we know nothing about value of \(z\), so we still know nothing about \(x\).
#1 alone is insufficient. Cross off A and D

(2) \(z = 5 \sqrt{2}\)
From the secant-tangent theorem, we know \(z^2=xy\)
\((5\sqrt{2})^2=xy\)
\(50=xy\)
Two variables, one equation. Insufficient.

OR, we know nothing about \(y\) and hence nothing about \(x\)

OR test numbers.
\(50=xy\)
Values: \(x=2\) and \(y=25\)
\(x=10\) and \(y=5\)
\(x\) does not have one unique value - insufficient

(2) alone is insufficient. Cross off B.

(C) or (E)?
Combine #1 and #2
\(z^2=2x^2\) and \(z=5\sqrt{2}\), so
\((5\sqrt{2})^2=2x^2\)
One variable, one equation: sufficient
If you need to, do the math:
\((5\sqrt{2})^2=2x^2\)
\(50=2x^2\) --> \(x=\sqrt{25}\) --> \(x=5\) (length cannot be (-5)

Answer C

Hope that helps. :-)

Good reference material:
A simple overview and ways to remember each power theorem is HERE
Secant-tangent with a simple proof is HERE.
Rules for Chords, Secants, and Tangents in a Circle are HERE.
Power of a Point Description and Problems are HERE
Power of a Point Theorem - material is short and dense

Then see the the page linked above, at the Power of a Point section.


P.S. mikedays , I answered your query via PM. :-)
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 06 Jul 2018, 09:58
generis many thanks for great explanation... :) just in the end the last part about math it was like "tail wagging the dog " :lol: :-)
you know i think the root of my gmat quant problem, is in root :lol:

If you need to, do the math. (i would love to do the math) :)

\((5\sqrt{2})^2=2x^2\) why after you sqauare both sides

\(50=2x^2\) --> the 2x^2 remains intact? :?

OR why didnt you write initially

\((5\sqrt{2})^2=(2x^2)^2\) you squared RHS, why didnt you apply the same pattern to \(2x^2\) :?
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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 06 Jul 2018, 13:25
1
dave13 wrote:
generis many thanks for great explanation... :) just in the end the last part about math it was like "tail wagging the dog " :lol: :-)
you know i think the root of my gmat quant problem, is in root :lol:

If you need to, do the math. (i would love to do the math) :)

\((5\sqrt{2})^2=2x^2\) why after you sqauare both sides

\(50=2x^2\) --> the 2x^2 remains intact? :?

OR why didnt you write initially

\((5\sqrt{2})^2=(2x^2)^2\) you squared RHS, why didnt you apply the same pattern to \(2x^2\) :?

dave13 - too much staring at numbers? :)

I would bet you missed the difference between \(z^2\) and \(z\)

The equation is: \(z^2=2x^2\)

\(z\) by itself is: \(z=5\sqrt{2}\) -- Not squared yet. Plug \(z\) INTO the \(z^2\) equation

\(z^2=2x^2\)
\((z)*(z) =2x^2\) Leave RHS alone. That's what \(z\)-squared equals :)
\(z=5\sqrt{2}\)
\((z)*(z) =2x^2\)
\((5\sqrt{2})*(5\sqrt{2})=2x^2\)
\(5*5*\sqrt{2}*\sqrt{2}=2x^2\)
\(25*2=2x^2\)
\(50=2x^2\)
\(\frac{50}{2}=x^2\)
\(25=x^2\)
. Take the square root of both sides:

\(x=5\) (lengths cannot be negative)

I hope that helps. :-)
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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New post 24 Aug 2018, 03:51
Hi generis,

Thanks for your explanation. Could you copy here your explanation to mikedays' query?

Thanks!!
Re: In the figure above, AB, which has length z cm, is tangent to the &nbs [#permalink] 24 Aug 2018, 03:51
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