Find the length of the altitude drawn to the hypotenuse. The triangle is not drawn to scale.a.130b.√31c.31d.√130If you could explain how you did it that would be great.

We can solve this by understanding triangle rules and seeing the two right triangles that together form the whole right triangle. Lastly we'll use some basic Trig.
Let's start by labeling the triangles:
Let's make the bottom left angle (next to 5) "b"
Then we'll call the small angle at the top, which is the apex of the small triangle, "a". The height of the triangles we'll call "x".
So for the angles at this point, if a + b (as we've called them) are the other 2 angles of a right triangle, then a + b = 90.
Why? because all (3) angles of any triangle always sum (add up) to the 180.
if a + b = 90, then a = 90-b
Now let's look at the triangle on the right: it has 90 as it's left angle, and b as apex (top) angle, the angle on the right must add to b to make 90, so it's 90-b
plug "a" in for "90-b" --> now we have an a, b, and 90 for the small right and for the small left triangle.
For the left triangle: take the tangent (tan) of angle a, which = the opposite side (O) ÷ adjacent side (A) --> tan(a) = O/A = 5/x
For the other triangle: tangent of that angle a = O/A --> tan(a) =x/26
Now if tan(a) = 5/x AND tan(a) = x/26, we can plug one into the other:
tan(a) = x/26
5/x = x/26
cross-multiply
x * x = 5 * 26
[tex] {x}^{2} = 130 \\ \sqrt{ {x}^{2} } = \sqrt{130} \\ x = \sqrt{130} [/tex]