SOLUTION: In the diagram, ABCD is a square with side length 2, and BEF is an equilateral triangle with side length 3. Find DF.

Question 1210476: In the diagram, ABCD is a square with side length 2, and BEF is an equilateral triangle with side length 3. Find DF.
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
The given information allows us to determine the exact angles between the sides of the square and the triangle:
* **Square ABCD** side length: $AB = BC = 2$.
* **Equilateral Triangle BEF** side length: $BE = BF = 3$.
* **Connecting Angle:** $\angle ABE = 30^\circ$.
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## 1. Distance between A and F ($\mathbf{AF}$)
The distance $AF$ can be found by analyzing $\triangle ABF$.
### Finding Angle $\angle ABF$
The angle $\angle ABF$ is the sum of the angle $\angle ABE$ and the angle $\angle EBF$, assuming the triangle is positioned sequentially adjacent to $AB$:
$$\angle ABF = \angle ABE + \angle EBF$$
* $\angle ABE = 30^\circ$ (Given)
* $\angle EBF = 60^\circ$ (Angle of an equilateral triangle)
$$\angle ABF = 30^\circ + 60^\circ = 90^\circ$$
Since $\angle ABF = 90^\circ$, $\triangle ABF$ is a **right-angled triangle**.
### Using the Pythagorean Theorem
We use the Pythagorean theorem: $AF^2 = AB^2 + BF^2$.
$$AF^2 = 2^2 + 3^2$$
$$AF^2 = 4 + 9$$
$$AF^2 = 13$$
$$AF = \mathbf{\sqrt{13}}$$
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## 2. Distance between C and F ($\mathbf{CF}$)
The distance $CF$ can be found by analyzing $\triangle CBF$.
### Finding Angle $\angle CBF$
First, we find $\angle EBC$, assuming $E$ is *inside* the $\angle ABC$ corner:
$$\angle EBC = \angle ABC - \angle ABE$$
* $\angle ABC = 90^\circ$ (Angle of a square)
* $\angle ABE = 30^\circ$ (Given)
$$\angle EBC = 90^\circ - 30^\circ = 60^\circ$$
For $\angle CBF$, we assume the points $C, E, F$ are positioned such that $\angle CBF = \angle EBC + \angle EBF$ (a typical non-collinear configuration).
$$\angle CBF = \angle EBC + \angle EBF$$
$$\angle CBF = 60^\circ + 60^\circ = 120^\circ$$
### Using the Law of Cosines
We use the Law of Cosines: $CF^2 = CB^2 + BF^2 - 2(CB)(BF) \cos(\angle CBF)$.
$$CF^2 = 2^2 + 3^2 - 2(2)(3) \cos(120^\circ)$$
$$CF^2 = 4 + 9 - 12 \left(-\frac{1}{2}\right)$$
$$CF^2 = 13 - (-6)$$
$$CF^2 = 19$$
$$CF = \mathbf{\sqrt{19}}$$
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Given that $\angle ABF=90^\circ$ simplifies the calculation significantly, it is highly likely that the intended question was to find $AF$.
The answer is:
* The distance **$AF$** is **$\sqrt{13}$**.
* The distance **$CF$** is **$\sqrt{19}$**.
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