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

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### HideShow timer Statistics 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 [ 18.64 KiB | Viewed 8036 times ]

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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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Senior SC Moderator V
Joined: 22 May 2016
Posts: 2759
In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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dave13 wrote:
carcass wrote: 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, 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. P.S. mikedays , I answered your query via PM. _________________
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##### General Discussion
Intern  B
Joined: 28 Apr 2017
Posts: 38
In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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
$$5\sqrt{2}^2=2x^2$$
Intern  B
Joined: 27 Dec 2017
Posts: 26
Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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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carcass , Can you please let us know . How this question can be solved.

Thanks.
Intern  B
Joined: 01 Nov 2017
Posts: 11
Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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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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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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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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?
VP  D
Joined: 09 Mar 2016
Posts: 1284
In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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, 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 Attachments circle1.jpg [ 13.96 KiB | Viewed 5268 times ]

VP  D
Joined: 09 Mar 2016
Posts: 1284
Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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generis many thanks for great explanation... just in the end the last part about math it was like "tail wagging the dog "  you know i think the root of my gmat quant problem, is in root 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 $$2x^2$$ Senior SC Moderator V
Joined: 22 May 2016
Posts: 2759
In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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dave13 wrote:
generis many thanks for great explanation... just in the end the last part about math it was like "tail wagging the dog "  you know i think the root of my gmat quant problem, is in root 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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Hi generis,

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

Thanks!!
Veritas Prep GMAT Instructor D
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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2
carcass wrote: 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

Responding to a pm:
The tangent-secant theorem tells us that Tangent^2 = Whole Secant * External Secant
So whenever the length of 2 of these is defined, the third is automatically fixed.

Here,
AB = z,
BC = x
BD = y

$$z^2 = y*x$$ (also given in the question so that you don't need to remember the theorem)

1. CD = x cm
y = 2x
We have no numerical value for any length, x, y or z. There is no way we can get the numerical value of x from this information.
Not sufficient.

2. $$z = 5\sqrt{2}$$
$$5\sqrt{2} = \sqrt{xy}$$
xy = 50
x can be 5, y can be 10
If we move BD slightly up, x will be less than 5 and y will be more than 10 (but the product will stay 50)

Value of x is not unique. Not sufficient.

Using both statements,
x*2x = 50
x = 5
Sufficient.

Answer (C)
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Intern  B
Joined: 24 Apr 2017
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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Would someone kindly look at the diagram from the solution given by OG?

Based on ONLY the 2nd data point (2): z = 5√2

How are we able to determine that CD is the diameter of the circle?
And that the triangle AOB is a right-angle triangle?
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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Hi lassakul

Sometimes the explanations in the OG are just complex and not easy to follow or comprehend.

I would advise you to look at it from the power of point theorem as explained in the above posts. Re: In the figure above, AB, which has length z cm, is tangent to the   [#permalink] 22 Oct 2018, 13:22
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# In the figure above, AB, which has length z cm, is tangent to the

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