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In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and

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In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and [#permalink]

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In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and the shaded region has an area of 48. What is the length of QW?

A. \(2\sqrt{2}\)

B. 3

C. \(\sqrt{10}\)

D. 3.2

E. 3.5

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[Reveal] Spoiler:
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[Reveal] Spoiler: OA

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Re: In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and [#permalink]

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New post 16 Mar 2015, 00:57
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Hi
this question basically tests the concept of similar triangles
plz have a look on the attached pic
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Re: In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and [#permalink]

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Bunuel wrote:
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In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and the shaded region has an area of 48. What is the length of QW?

A. \(2\sqrt{2}\)

B. 3

C. \(\sqrt{10}\)

D. 3.2

E. 3.5

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

We know the larger and smaller triangles are similar, because they share the angle at P, and one other angle in each is 70º. What’s tricky is that they have different orientations, so that side PT actually corresponds to side PR. We know PR:PT = 3, so that’s the scale fact. Every length in triangle PRS is three times more than the corresponding side in triangle PTQ. If lengths are multiplied by 3, area is multiplied by 3 squared, or 9. Let’s say that the area of triangle PTQ is A. Then the area of triangle PRS is 9A. The shaded area is the difference of the two triangle areas, or 8A. If 8A = 48, this means A = 6, and that’s the area of triangle PTQ.

Well, PT = 4 is a base of PTQ, and QW is a corresponding altitude: call its length h.

A = 0.5bh

6 = 0.5(4)h

6 = 2h

3 = h

The length of QW is 3.

Answer = (B)
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Re: In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and [#permalink]

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New post 23 Mar 2015, 04:01
It is probably worth mentioning that formula of 1/2*a*b*sin(a^b) is a good way to remember the relation between sides and the areas of triangles: as in, 1/2*PQ*PT*sin(P)/(1/2*PS*PR*sin(P) = PQ*PT/(PS*PR) = PQ*PT/(3*PQ*3*PT) = 1/3*1/3 = 1/9 in our case.

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Re: In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and [#permalink]

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New post 19 Feb 2016, 20:26
oh man..took me some time to solve..would definitely not solve on the actual test..spent about 10 minutes to solve it...
we know that angles PTQ and QRS are 70 degrees. we also know that angle P is shared by the triangle PQT and PRS.
since these 2 triangles have 2 similar angles, the triangles must be similar.
the "base", or the segment from the 70 degree angle to the P angle is 3 times greater than PT, thus, the height of the PRS triangle must be 3 times greater than QW.

ok, so the area of the shaded region is:
A(PRS)-A(PQT) = 48
area of PQT=4QW/2 = 2QW
area of PRS=12*3QW/2 = 18QW.

now, 18QW-2QW=16QW=48.
QW=3.

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Re: In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and [#permalink]

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Re: In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and   [#permalink] 08 Sep 2017, 09:16
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In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and

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