Hello, I just need a little assistance in Geo, my grade right now is decent(B) so I would like it to stay there, or possibly raise a little higher. If

the length of FH is 2

Kenya.....(1) can be solved with something called "Pick's" Theorem...it can be used when the vertices of a figure lie on integer values for x and y....just like yours do.....!!!

The "points" described below  are the occasions where the grid lines intersect

The area is given by

Boundary points/ 2  +  interior points  - 1

The number of boundary points  =  14

The number of interior points  = 20

So

The area  is

14/2 + 20 - 1  =

7 + 20 - 1  =

26 sq units

Here's a pic of the boundary points and the interior points:

3. The corresponding angles in triangle ABC are congruent to the corresponding angles in triangle FGH. The measures of two corresponding sides of the triangles are given. 

AB=6, AC=8, FG=x+10, and FH=4x. What is the length of FH?

We have, by similar triangles, that

AB/AC  = FG/FH

6 / 8  =  [ x + 10] / [4x]        cross-multiply

6 * 4x  =  8 [x + 10 ]     simplify

24x   =   8x + 80     subtract 8x from both sides

16x  = 80      divide  both sides by 16

x  = 5

So.....FH  = 4(5)   = 20

Actually....you probably divided the figure into a trapezoid and a triangle or as a rectangle and two triangles??....I just thought  it might be kinda' neat to show you another way.....plus.....it gives me a chance to "show off"....LOL!!!!