Math Help - Triangle areas?

1. Triangle areas?

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?

2. Originally Posted by Rinnie
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?
what is your equation? how did you arrive at it?

3. my equation is x^2 - 10x + 8*sqrt21.

4. This equation can be solved by completing the square or using the quadratic formula

Completing the square

$x^2-10x+8\sqrt{21}=0$

$(x^2-10x+25)+8\sqrt{21}-25=0$

$(x-5)^2+8\sqrt{21}-25=0$

$(x-5)^2=25-8\sqrt{21}$

$x-5=\sqrt{25-8\sqrt{21}}$

$x=5\pm\sqrt{25-8\sqrt{21}}$

for $ax^2+bx+c=0$

then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

so for $x^2-10x+8\sqrt{21}=0$

then $x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)8\sqrt{21}}}{2\times1}$

5. don't forget to throw away any negative values you arrive at!

Area can't be negative.

6. That's what I tried to do, but I'm confused about how to deal with the sqrt21 under the sqrt sign.

7. Originally Posted by Rinnie
That's what I tried to do, but I'm confused about how to deal with the sqrt21 under the sqrt sign.
You may need to make an approximation for $\sqrt{21} = 4.583$ and continue you calculations.

8. Using the quadratic forumula, I end up with 100 - 146.something in under the square root sign which makes it a negative though.

9. This would suggest your equation is not correct.

Originally Posted by pickslides
what is your equation? how did you arrive at it?

10. Originally Posted by Rinnie
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?
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 $\tfrac12ab\sin C$. If a = b = 5 then the area is $12.5\sin C$. The only way that two different triangles could have that same area is if $\sin C$ is the same for both triangles. But that would mean that the two angles are supplementary: $C_2 = \pi - C_1$ (or $C_2 = 180 - C_1$ 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 $C_1$) has the third side equal to 4. The cosine rule says that $\cos C_1 = \frac{5^2+5^2-4^2}{2\times5\times5} = \frac{34}{50}$. For the other triangle, with third side x and angle $C_2$, the formula says $\cos C_2 = \frac{5^2+5^2-x^2}{2\times5\times5} = \frac{50-x^2}{50}$. But $C_1$ and $C_2$ are supplementary, and so $\cos C_2 = -\cos C_1$. Therefore $\frac{50-x^2}{50} = -\frac{34}{50}$. That's your quadratic equation. Now all you have to do is to solve it.