Thermal Expansion in an aluminum plate

[SoccaCrazy24]

Question 1:
A hole in an aluminum plate has a diameter or 1.178 cm at 23.00 degrees Celsius.
(A) What is the diameter of the hole at 199.0 degrees Celsius?
(B) What temperature is the diameter of the hole equal to 1.176 cm?

For (A) I used the equation Change in Area = 2 * alpha * A * Change in Temp.
alpha = 24e-6
A= pi * r^2 = 1.09e-4
Change in Temp = 176
So Change in Area = 2 * 24e-6 * 1.09e-4 * 176 = 9.207e-7
so change in Diameter = sqroot (A/pi) = 5.414e-4m
This doesn't seem right... am i using the right equation? and then i would use the same equation for (B)

Question 2:
At 12.25 degrees Celsius a brass sleeve has an inside diameter of 2.196 cm and a steel shaft has a diameter of 2.199 cm. It is desired to shrink-fit the sleeve over the steel shaft.
(A) To what temperature must the sleeve be heated in order for it to slip over the shaft?
(B) Alternatively, to what temperature must the shaft be cooled before it is able to slip through the sleeve?

If I use the same equation from question 1 shall i receive the same answer?

 

[SoccaCrazy24]

not to be repetitve or a forum whore? but does anyone have a clue?