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# Geometry Problem

0
21
2
+101

Triangle DEF is equilateral. Find AE.

Feb 12, 2024

#2
+394
+2

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.

Feb 13, 2024

#1
+394
+1

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.



Feb 12, 2024
#2
+394
+2
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.