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Equilateral triangle BDF is inscribed in equilateral triangl

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Equilateral triangle BDF is inscribed in equilateral triangl  [#permalink]

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New post 19 Jun 2010, 07:17
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Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?

(1) Angle DFE is 90º
(2) The length of AF is 10*sqrt(3)
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Re: DS: Equilateral triangle inscribed in triangle  [#permalink]

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New post 19 Jun 2010, 08:09
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Nusa84 wrote:
Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?
1. Angle DFE is 90º
2. The length of AF is 10*sqrt(3)

Can you help me with this one? Thanks!!!
Image


The problem is easier to solve than to explain, but anyway:

In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side.

(1) Angle DFE is 90º --> angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180-\angle {BDF}-\angle{EDF}=180-60-30\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 30-60-90 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).

So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\)

Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\);

\(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\).

Sufficient.

(2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient.

Answer: A.
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Re: DS: Equilateral triangle inscribed in triangle  [#permalink]

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New post 19 Jun 2010, 08:27
Bunuel wrote:
Nusa84 wrote:
Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?
1. Angle DFE is 90º
2. The length of AF is 10*sqrt(3)

Can you help me with this one? Thanks!!!
Attachment:
Captura.jpg


The problem is easier to solve than to explain, but anyway:

In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side.

(1) Angle DFE is 90º --> angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180-\angle {BCD}-\angle{EDF}=180-60-30\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 30-60-90 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).

So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\)

Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\);

\(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\).

Sufficient.

(2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient.

Answer: A.


As you said, easier to solve, many thanks!!!!! :)
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Re: DS: Equilateral triangle inscribed in triangle  [#permalink]

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New post 27 Jun 2010, 14:19
Good solution in steps given.
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Re: Equilateral triangle BDF is inscribed in equilateral triangl  [#permalink]

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New post 06 Mar 2014, 01:37
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Re: DS: Equilateral triangle inscribed in triangle  [#permalink]

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New post 06 Mar 2014, 18:12
Bunuel wrote:
Nusa84 wrote:
Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?
1. Angle DFE is 90º
2. The length of AF is 10*sqrt(3)

Can you help me with this one? Thanks!!!
Attachment:
Captura.jpg


The problem is easier to solve than to explain, but anyway:

In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side.

(1) Angle DFE is 90º --> angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180-\angle {BCD}-\angle{EDF}=180-60-30\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 30-60-90 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).

So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\)

Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\);

\(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\).

Sufficient.

(2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient.

Answer: A.



\(\angle {BDC}=180-\angle {BCD}-\angle{EDF}=180-60-30\) -

Could you explain the logic for taking EDF?
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Re: Equilateral triangle BDF is inscribed in equilateral triangl  [#permalink]

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New post 06 Mar 2014, 22:55
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Nusa84 wrote:
Attachment:
The attachment Captura.jpg is no longer available
Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?

(1) Angle DFE is 90º
(2) The length of AF is 10*sqrt(3)


Note that in a given equilateral triangle, you can inscribe many other equilateral triangles. The areas of these inscribed triangles will be different hence the leftover areas will be different.
Attachment:
Ques3.jpg
Ques3.jpg [ 13.11 KiB | Viewed 10615 times ]

So depending on what type of triangle is inscribed, the shaded region will be a fraction of the area of triangle ACE.


(1) Angle DFE is 90º

All internal angles of Equilateral triangles are 60. Angle DFE is 90 so triangle DFE is a right triangle with angles 30-60-90 and sides in the ratio 1:√3:2.

Attachment:
Ques4.jpg
Ques4.jpg [ 15.69 KiB | Viewed 10614 times ]

If we assume that side of inscribed triangle is s, we get the side of the large triangle in terms of s too. We also get the area of the right triangle (shaded region) in terms of s. Hence we can easily get the area of the shaded region as a fraction of area of triangle ACE (since they will both be in terms of s^2).
This statement alone will be sufficient.

(2) The length of AF is 10*sqrt(3)
Look at the three triangles shown above again. AF could be 10√3 in any of the three cases. The area of shaded region as a fraction of the area of ACE will be different in each case. Hence this statement alone is not sufficient to answer the question.

Answer (A)
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Re: Equilateral triangle BDF is inscribed in equilateral triangl  [#permalink]

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New post 06 Mar 2014, 23:05
Since you asked for further discussion, here goes.

I more intuited the answer than solved. I know that each interior angle of both equilateral triangles is 60. If I am told that DFE = 90, I can figure out that the other 3 triangles are 30-60-90 and congruent (Each 60-degree angle is across from one side of the smaller equilateral triangle; thus all corresponding sides for the right triangles are the same "x(sqrt3)"). Because I know the relative dimensions of triangles BCD, ABF, DFE, and BDF, I can figure out the relative areas of the triangles. For example, I know I can make BD sqrt3, BC 2, and CD 1. I can figure out the areas of all 4 smaller triangles, add these areas to get ACE's area, and thus determine the fraction of ACE that BCD represents. I didn't actually bother to assign values or do the calculations because it was clear Statement 1 is sufficient.

Eliminate B, C, and E.

Statement 2 only tells me AF. It doesn't provide me with any information about any other sides or angles in the figure. Thus, I would not be able to solve for area. The information is useless. Insufficient.

Eliminate D. The answer is A.
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Re: DS: Equilateral triangle inscribed in triangle  [#permalink]

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New post 07 Mar 2014, 00:34
X017in wrote:
Bunuel wrote:
Nusa84 wrote:
Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?
1. Angle DFE is 90º
2. The length of AF is 10*sqrt(3)

Can you help me with this one? Thanks!!!
Attachment:
Captura.jpg


The problem is easier to solve than to explain, but anyway:

In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side.

(1) Angle DFE is 90º --> angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180-\angle {BCD}-\angle{EDF}=180-60-30\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 30-60-90 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)).

So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\)

Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\);

\(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\).

Sufficient.

(2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient.

Answer: A.



\(\angle {BDC}=180-\angle {BCD}-\angle{EDF}=180-60-30\) -

Could you explain the logic for taking EDF?


It should be \(\angle {BDC}=180-\angle {BDF}-\angle{EDF}=180-60-30\) (those 3 angles make up a straight line, which is 180 degrees).

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Equilateral triangle BDF is inscribed in equilateral triangl  [#permalink]

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