Two noncongruent triangles have equal area. One triangle has sides 5, 5, and 4. The other has sides 5, 5, and x. Find the value of x.
I managed to get a quadratic equation, but I have no idea how to solve it. Any tips here?
This equation can be solved by completing the square or using the quadratic formula
Completing the square
$\displaystyle x^2-10x+8\sqrt{21}=0$
$\displaystyle (x^2-10x+25)+8\sqrt{21}-25=0$
$\displaystyle (x-5)^2+8\sqrt{21}-25=0$
$\displaystyle (x-5)^2=25-8\sqrt{21}$
$\displaystyle x-5=\sqrt{25-8\sqrt{21}}$
$\displaystyle x=5\pm\sqrt{25-8\sqrt{21}}$
Quadratic Formula
for $\displaystyle ax^2+bx+c=0$
then $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
so for $\displaystyle x^2-10x+8\sqrt{21}=0$
then $\displaystyle x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)8\sqrt{21}}}{2\times1}$
If a triangle has sides of length a, b, c, and C is the angle opposite side c, then the area of the triangle is $\displaystyle \tfrac12ab\sin C$. If a = b = 5 then the area is $\displaystyle 12.5\sin C$. The only way that two different triangles could have that same area is if $\displaystyle \sin C$ is the same for both triangles. But that would mean that the two angles are supplementary: $\displaystyle C_2 = \pi - C_1$ (or $\displaystyle C_2 = 180 - C_1$ if you prefer degrees to radians).
Now use the cosine rule. You are told that one of the two triangles (say the one with angle $\displaystyle C_1$) has the third side equal to 4. The cosine rule says that $\displaystyle \cos C_1 = \frac{5^2+5^2-4^2}{2\times5\times5} = \frac{34}{50}$. For the other triangle, with third side x and angle $\displaystyle C_2$, the formula says $\displaystyle \cos C_2 = \frac{5^2+5^2-x^2}{2\times5\times5} = \frac{50-x^2}{50}$. But $\displaystyle C_1$ and $\displaystyle C_2$ are supplementary, and so $\displaystyle \cos C_2 = -\cos C_1$. Therefore $\displaystyle \frac{50-x^2}{50} = -\frac{34}{50}$. That's your quadratic equation. Now all you have to do is to solve it.