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.