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# Equilateral triangle BDF is inscribed in equilateral

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Equilateral triangle BDF is inscribed in equilateral [#permalink]  19 Jun 2010, 08:17
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60% (02:42) correct 40% (00:00) wrong based on 2 sessions
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)

A) Statement 1 alone is sufficient, but statement 2 is not sufficient.
B) Statement 2 alone is sufficient, but statement 1 is not sufficient.
C) Both statements together are sufficient, but neither statement alone is sufficient.
D) Each statement alone is sufficient.
E) Statements 1 and 2 together are not sufficient.

Can you help me with this one? Thanks!!!
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Re: DS: Equilateral triangle inscribed in triangle [#permalink]  19 Jun 2010, 09:09
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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!!!
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.

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Re: DS: Equilateral triangle inscribed in triangle [#permalink]  19 Jun 2010, 09:27
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.

As you said, easier to solve, many thanks!!!!!
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Re: DS: Equilateral triangle inscribed in triangle [#permalink]  27 Jun 2010, 15:19
Good solution in steps given.
Re: DS: Equilateral triangle inscribed in triangle   [#permalink] 27 Jun 2010, 15:19
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