In the figure (attached) here, lines l and m are parallel. I : GMAT Problem Solving (PS)
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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]

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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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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 $$m = 9*\sqrt{2}$$

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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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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

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

PA/PQ=2/3=l/m
m=(3/2) 6root2
=9root2
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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)

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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]

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22 Dec 2013, 09:17
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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]

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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)

Hope its clear
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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]

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

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01 May 2016, 09:18
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Re: In the figure (attached) here, lines l and m are parallel. I [#permalink]

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01 May 2016, 22:25
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bhandariavi wrote:
Attachment:
Untitled.png
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

First of all, the figure should make you think of similar triangles. Evaluate whether the triangles are similar - they have a common angle. Also, the two sides, l and m are parallel so corresponding angles of the transversal will be equal. So triangle PAB is similar to triangle PQR (say the unnamed lower right vertex of the triangle is R). So by similar triangles, PA /PQ = 2/3 implies l/m = 2/3 too.

Now, AB subtends 45 degree angle at the circle so it will subtend a 90 degree angle at the centre of the circle which means that arc AB is 1/4 of the total circumference of the circle. Since length of arc AB is $$3\pi$$, circumference of the circle is $$4*3\pi = 12\pi$$. So radius is 6.

Now consider the triangle formed by l and the central angle subtended by AB. This will be an isosceles right triangle so 45-45-90 triangle giving ratio of sides as $$1:1:\sqrt{2}$$. The length of the two legs will be 6 each so l $$= 6*\sqrt{2}$$.

Since $$l/m = 2/3 = 6*\sqrt{2}/m$$

$$m = 9*\sqrt{2}$$

Guessing:
If nothing comes to mind or time is very short, improve your chances of correct answer by eliminating some options easily. Note that length of l will be something less than arc AB so less than $$3\pi$$.
So m will be $$3/2 * 3\pi = 4.5*3 = 13.5$$ (approximately)
So m is a bit less than 13.5
Options (A), (C) and (D) are too large so they are out. Option (E) is the closest.
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Re: In the figure (attached) here, lines l and m are parallel. I   [#permalink] 01 May 2016, 22:25
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