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# In the figure (attached) here, lines l and m are parallel. I

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In the figure (attached) here, lines l and m are parallel. I [#permalink]  22 Mar 2011, 20:59
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In the figure (attached) here, lines l and m are parallel. If the arc AB = 3π, and the ratio of PA to PQ is 2/3, what is the length of m?

A. 15 * Sqroot2
B. 8 * Sqroot2
C. 15
D. 9pi
E. 9 * Sqroot2
[Reveal] Spoiler: OA

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Re: Geometry Tough [#permalink]  22 Mar 2011, 21:48
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bhandariavi wrote:
In the figure (attached) here, lines l and m are parallel. If the arc AB = 3π, and the ratio of PA to PQ is 2/3, what is the length of m?
Choices
A
15 * Sqroot2
B
8 * Sqroot2
C
15
D
9pi
E
9 * Sqroot2

The angle subtended by arc AB at the centre would be 90 degrees. We are given that arc AB is $$2\pi$$. Let the radius of the circle be r, then we are given that $$(90/360)*2\pi*r=3\pi$$

or $$r=6$$.

Now, length l is the hypotenuse of an isosceles right angled triangle that has two sides as r, so $$l = 6*\sqrt{2}$$

Now, ratio of PA to PQ is 2/3, so ratio of l:m would be 2:3 as well, by Basic proportionality theorem.

So, $$m = 3/2*6*\sqrt{2}$$
or $$m = 9*\sqrt{2}$$

Answer E
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Re: Geometry Tough [#permalink]  22 Mar 2011, 21:56
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bhandariavi wrote:
In the figure (attached) here, lines l and m are parallel. If the arc AB = 3π, and the ratio of PA to PQ is 2/3, what is the length of m?
Choices
A
15 * Sqroot2
B
8 * Sqroot2
C
15
D
9pi
E
9 * Sqroot2

Let's name the other vertex of line m as R.

$$\angle{QPR}=\angle{APB}$$
$$\angle{PQR}=\angle{PAB}$$ $${l || m}$$
$$\angle{PRQ}=\angle{PBA}$$ $${l || m}$$

According to AAA similarity criterion, $$\triangle{QPX}\approx\triangle{APB}$$
Means QPX and APB are similar triangle.

The ratio will be same for the sides of two similar triangles;

$$\frac{l}{m}=\frac{PA}{PQ}=\frac{2}{3}$$

Now, let's get to finding the length of l.

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Meaning thereby; arc AB will subtend an angle of 90 degree with the center.

Length of an arc;
$$\frac{\theta}{360}*2*\pi*r=3*\pi$$
$$\frac{90}{360}*2*\pi*r=3*\pi$$
$$r=6$$

Another interesting thing to note is that; if arc AB subtended an angle of 90 degree with the center, line segment AB becomes the hypotenuse of a right angled triangle with sides as radius.

Thus;
$$(AB)^2=6^2+6^2$$
$$l=\sqrt{72}$$
$$l=6\sqrt{2}$$

Substitute in the first equation;
$$\frac{l}{m}=\frac{2}{3}$$
$$\frac{6\sqrt{2}}{m}=\frac{2}{3}$$
$$m=9\sqrt{2}$$

Ans: "E"
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Re: Geometry Tough [#permalink]  22 Mar 2011, 22:00
Angle at the center for arc AB = 90deg
=> (90/360)*2 pi R = arc AB
=> (90/360)*2 pi R = 3 pi
=> radius = 6

triangle OAB is right isosceles, right angled at O
OA=OB=radius=6
AB = 6root2 (pythagoras)

PA/PQ=2/3=l/m
m=(3/2) 6root2
=9root2
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Re: Geometry Tough [#permalink]  22 Mar 2011, 22:23
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m/l = PQ/PA (Similar triangles)

=> m/l = 3/2

AB = 3pi

AB subends a right triangle at center

so 90/360 = 3pi/2pir (r is radius of circle)

=> 1/4 = 3/2r

=> r = 6

Noe there is an isoceles triangle formed by AB at center

and l = AB (hypotenus) = root(2) * 6

So m = 3/2 * root(2) * 6 = 9root(2)

Answer - E
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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]  22 Dec 2013, 09:17
Hello from the GMAT Club BumpBot!

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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]  05 Feb 2014, 11:25
1. First of all redraw figure
2. Central angle of inscribed angle of 45 is 90. Therefore since radius makes this an isosceles triangle we have an isosceles right triangle.
3. We need to find the measurements, so we need the radius, we are given that the arc represents 1/4 of the circumference (Since central angle is 90) therefore, 2 pi * r / 4 = 3 pi --> r = 6
4. The hypothenuse of our isosceles right triangle will be 6 sqrt (2), as per the relation of measure in right isosceles triangles 45-45-90.
5. Since the ratio of the small triangle to the large triangle is 2:3 and since both sides are parallel, then 3/2 * 6sqrt (2) = 9sqrt (2)

Hence answer is E

Hope its clear
Cheers
J
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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]  20 Apr 2015, 16:06
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Re: In the figure (attached) here, lines l and m are parallel. I   [#permalink] 20 Apr 2015, 16:06
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# In the figure (attached) here, lines l and m are parallel. I

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