C is the answer.
We multiply the top and bottom of this by the conjugate (switch the sign of the square root):
(2-sqrt8)(4-sqrt12)/(4+sqrt12)(4-sqrt12)
Multiying the top we have
8-2sqrt12-4sqrt8+sqrt(8×12)
Multiplying the bottom we have:
16-4sqrt12+4sqrt12-12
On top, we simplify the radicals and get
8-4sqrt3-8sqrt2+4sqrt6
On bottom, we have
16-12=4
Now we divide everything on top by 4 and get
2-sqrt3-2sqrt2+sqrt6
Given:
QSR is a right triangle.
QT = 10
TR = 4
To find:
The value of q.
Solution:
Hypotenuse of QSR = QT + TR
= 10 + 4
= 14
Geometric mean of similar right triangle formula:
Do cross multiplication.
Switch the sides.
Taking square root on both sides.
The value of q is 
.
Area of a parallelogram is A= length x width. So 9.5 x 4. 38 in
I think the answer to this one would be B
Dilation since then the lengths are changed. The image would then be only similar; not congruent.
