Finding the length of the median on a triangle with side lengths 8, 5, and 6

Question

I got a problem that goes like this: Triangle ABC has side lengths AB = 6, AC = 5, and BC = 8, Draw the median AD where BD = DC = 4, what is the length of AD?

I found this post that says you can get this length by using this formula: $Mc = 1/2 * √(2*a^2+2*b^2-c^2)$

But I want to know if it is possible to solve simply by using the Pythagoras theorem (this is an 8th grade math question so..)

Also, does anyone know how to write line AB in mathjax or something that I can type in? Thanks so much for the help!

Answer 1

constructing the nhight fro Point $A$ to $BC$ with the Point on $BC$ is equal $E$ and let $$ED=x$$ then we get $$x^2+h_a^2=m_a^2$$ $$\left(\frac{a}{2}-x\right)^2+h_a^2=m_a^2$$ $$\left(\frac{a}{2}+x\right)^2+h_a^2=m_a^2$$ where $$a=8$$ then unknowns are $x,h_a,m_a$ can you finish?

Answer 2

You can use the Pythagoras Theorem generalization, the law of cosines. This gives you a relationship between the 3 sides and one angle of ANY triangle.

Given the triangle ABC, you can you the law of cosines to get the angle at the B corner. Then again use the law of cosines with the ABD triangle to get the length of AD.

Answer 3

Credits to Jack D'Aurizio for explaining to me what that formula meant.

In a parallelogram the sum of the squared lengths of the diagonals equals the sum of the squared lengths of the sides (polarization identity). So basically if I cut and paste the parallelogram into a rectangle, the diagonals are the hypotenuse of the rectangle, which is the sum of the new four sides (which equal to the original four sides) squared.

So, I'll extend AD to E, where $\Delta BEC \cong \Delta ABC$, there I've made a parallelogram ABEC where BC and AE are the diagonals. This means $2*(5^2 + 6^2) = 8^2 + 2x^2$

which solves to $x = \pm\sqrt19$,

$x = \sqrt19$

Answer 4

You just need Heron's formula and Pythgoras theorem.

Let $AE$ be the height. Use Heron's formula to find area of triangle. Let it be $S$. Then $\frac 12*8*AE=S$. You will get $AE$. Then use Pythgoras theorem in $\Delta AEC$ to find $EC$. Then you can get $DE$. Then use Pythgoras theorem again in $\Delta AED$ to find $AD$.