Geometry Problem

IGNORE THE OTHER ANSWER!!!

Guys, Sorry, I realized i made a mistake in one of the steps, the answer is not corrrect .

In my solution, i substitued x-10, instead of 10-x for AF. Instead \({AB}^{2}=49+{(10-x)}^{2}-2*7*(10-x)*\cos({60}^{\circ})\). So Simplifying, \({AB}^2={x}^2-13x+79\).

Ssing pythagorean theorem we form a equation, our previous responses, \(37+{x}^2-3x+9={x}^{2}-13x+79\). Solving, we get \(x=\frac{33}{10}\). Ignore the other answer!!! this answer is correct. The answer is 33/10.

Looking at this problem, we don't see we have that many options, so we use our backup plan for any geometry problem: trigonometry. 

Set AE as x. The side length of triangle DEF is 10, so BF = 7, and AF = 10-x.

We want to write some equation, with x, so we look at the triangle in the middle, BCA, and hope to use the pythagorean theorem.

Now in each of these triangles, BDC, AEC, BFA, we have 2 sides and an angle, and we want to find the side opposite to the angle.

Luckily, for us we have the Law of Cosines!

\({BC}^{2}=9+49-2*3*7*\cos({60}^{\circ})\). Simplifying, we get, \({BC}^{2}=37\). 

Similarly, \({AC}^2=9+{x}^{2}-2*3*x*\cos({60}^{\circ})\). Simplifying, we, get \({AC}^{2}=9+{x}^{2}-3x\).

Also, \({AB}^2=49+{(x-10)}^{2}-2*7*(x-10)*cos({60}^{\circ})\). Simplifying, we get, \({AB}^{2}={x}^{2}-27x+219\).

Using pythagorean theorem, we have \(37+{x}^2-3x+9={x}^{2}-27x+219\). Solving and cancelling, we get \(x=\frac{173}{24}\). Our answer is 173/24.

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