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

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

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04 Jul 2018, 17:49
5
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.

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.

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

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)

Hope that helps.

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##### General Discussion
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Joined: 28 Apr 2017
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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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26 Jul 2017, 14:51
3
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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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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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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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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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.

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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17 May 2018, 08:24
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Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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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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.

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 ]

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Posts: 1284
Re: In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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 "
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 $$2x^2$$
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In the figure above, AB, which has length z cm, is tangent to the  [#permalink]

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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 "
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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24 Aug 2018, 03:51
Hi generis,

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

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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26 Sep 2018, 00:04
1
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.

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

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20 Oct 2018, 01:46
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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22 Oct 2018, 13:22
1
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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