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Re: In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is [#permalink]
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Bunuel wrote:

In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is the length of FG?

(1) Ttriangle FGH has 9 times the area of triangle ABC
(2) HF = 9


Kudos for a correct solution.

Attachment:
gsdsq_img1.png


MAGOOSH OFFICIAL SOLUTION:

The information given implies that ∠C = 110° and ∠G = 20°, because the sum of the three angles in each triangle must be 180°, by Euclid’s well-known theorem. Thus, the two triangles have all the same angles, but they are different sizes — they are similar. We know the length of AB, so all we need is the scale factor to determine length of the corresponding side FG.

Statement #1: this statement tells us the ratio of areas is 9 —- this is the square of the scale factor, so k = 3, and from this we can calculate the length of FG. Statement #1, alone and by itself, is sufficient.

Statement #2: this statement gives us a third side, so we can set up a proportion:

AC/AB = FH/FG.

Since we now know three of the terms of that proportion, we can solve for the fourth, FG. Statement #2, alone and by itself, is sufficient.

Answer = D
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Re: In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is [#permalink]
Bunuel wrote:
Bunuel wrote:

In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is the length of FG?

(1) Ttriangle FGH has 9 times the area of triangle ABC
(2) HF = 9


Kudos for a correct solution.

Attachment:
gsdsq_img1.png


MAGOOSH OFFICIAL SOLUTION:

The information given implies that ∠C = 110° and ∠G = 20°, because the sum of the three angles in each triangle must be 180°, by Euclid’s well-known theorem. Thus, the two triangles have all the same angles, but they are different sizes — they are similar. We know the length of AB, so all we need is the scale factor to determine length of the corresponding side FG.

Statement #1: this statement tells us the ratio of areas is 9 —- this is the square of the scale factor, so k = 3, and from this we can calculate the length of FG. Statement #1, alone and by itself, is sufficient.

Statement #2: this statement gives us a third side, so we can set up a proportion:

AC/AB = FH/FG.

Since we now know three of the terms of that proportion, we can solve for the fourth, FG. Statement #2, alone and by itself, is sufficient.

Answer = D


How do we know that the factor is 3 based off the area being 9 times greater? I'm having trouble making this connection without picking random values for the height of the triangle. On that note, is it possible to solve for the height? Using GMAT tested properties.

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Re: In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is [#permalink]
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ak1802 wrote:
Bunuel wrote:
Bunuel wrote:

In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is the length of FG?

(1) Ttriangle FGH has 9 times the area of triangle ABC
(2) HF = 9


Kudos for a correct solution.

Attachment:
gsdsq_img1.png


MAGOOSH OFFICIAL SOLUTION:

The information given implies that ∠C = 110° and ∠G = 20°, because the sum of the three angles in each triangle must be 180°, by Euclid’s well-known theorem. Thus, the two triangles have all the same angles, but they are different sizes — they are similar. We know the length of AB, so all we need is the scale factor to determine length of the corresponding side FG.

Statement #1: this statement tells us the ratio of areas is 9 —- this is the square of the scale factor, so k = 3, and from this we can calculate the length of FG. Statement #1, alone and by itself, is sufficient.

Statement #2: this statement gives us a third side, so we can set up a proportion:

AC/AB = FH/FG.

Since we now know three of the terms of that proportion, we can solve for the fourth, FG. Statement #2, alone and by itself, is sufficient.

Answer = D


How do we know that the factor is 3 based off the area being 9 times greater? I'm having trouble making this connection without picking random values for the height of the triangle. On that note, is it possible to solve for the height? Using GMAT tested properties.

Posted from my mobile device


If two similar triangles have sides in the ratio \(\frac{x}{y}\), then their areas are in the ratio \(\frac{x^2}{y^2}\).

OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{SIDE^2}{side^2}\).
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Re: In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is [#permalink]
Can anyone clarify how do they arrive at the fact that the given triangles are similar? I only see 1 angle that is same and we should have information about another angle or about the ratio of two other sides.
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Re: In the diagram above, ∠A = ∠F = 50°, ∠B = 20°, and ∠H = 110°. What is [#permalink]
ravigupta2912 wrote:
Can anyone clarify how do they arrive at the fact that the given triangles are similar? I only see 1 angle that is same and we should have information about another angle or about the ratio of two other sides.


We can deduce it from the information given.

∠A = 50° ∠B = 20° ---> ∠C = 110
∠F = 50 ∠H = 110° ---> ∠G = 20
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Re: In the diagram above, A = F = 50°, B = 20°, and H = 110°. What is [#permalink]
Should we assume the length of the sides because of the labels? It aren't stated in the question or the question doesn't anything about the labelling.
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Re: In the diagram above, A = F = 50°, B = 20°, and H = 110°. What is [#permalink]
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Re: In the diagram above, A = F = 50°, B = 20°, and H = 110°. What is [#permalink]
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