Ineed an answer and explanation for question 4 !

Probably the best way to do it, is just to plug in each value until you get one that works.  Otherwise, use algebra:

Let the line in the middle of the triangle be "a".  Using pythagorean theorem.

3^2 + a^2 = 6^2
a^2 = 36-9 = 27
a = sqrt(27)

Now let the base of the triangle be "b".

We have two equations using pythagorean theorem:
First equation:

a^2 + y^2 = b^2
(sqrt(27))^2 + y^2 = b^2
27 + y^2 = b^2

Second equation:
6^2 + b^2 = (y+3)^2

Since we have b^2 in both equations, let's substitute the b^2 from the first equation into the second equation:

6^2 + 27 + y^2 = (y+3)^2
36 + 27 + y^2 = y^2 +6y + 9
63 + y^2 = y^2 +6y +9
54 + y^2 = y^2 + 6y
54 = 6y
y = 9

Therefore the answer is H.

A^2 + b^2 = c^2

3^2 + b^2 = 6^2
9 + b^2 = 36                    a = 3, b = 5.19, c = 6
b^2 = 27                          a = 6. b = ?, c = 3 + y
sqrt( b^2 = 27)                If side a of triangle 1 is three, and side a of triangle 2 is 
b = 5.19                           6, than triangle 2 is double that of triangle 1. So:
                                         c = [(2 *6) = (3 + y)]
                                         c = [12 = 3 + y]
                                         c = [9 = y]
                                         a = 6, b = 10.38, c = 12,  y = 9

Just a second zzz