Question 1113459
the wires are cut into lengths of x and 10-x.


they are then shaped into squares.


the wires form the perimeter of the squares.


the length of each side of the wire with length x, is equal to x/4.


the length of each side of the wire with length (10-x) is equal to (10-x) / 4


the area of each of the squares is equal to s^2, where s is the length of each side of the square.


the area of the square formed by the wire of length x would be s^2 = (x/4)^2.


the area of the square formed by the wire of length (10-x), would be s^2 = ((10-x)/4)^2.


the total area of both square is therefore (x/4)^2 + ((10-x)/4)^2


since (a/b)^2 is equal to a^2/b^2, the total area therefore becomes equal to x^2/4^2 + (10-x)^2/4^2


that simplifies to x^2/16 + (10-x)^2/16


that simplifies to (x^2 + (10-x)^2) / 16


(10 - x)^2 is equal to (10 - x) * (10 - x) which is equal to 10*10 -10x - 10x + x^2)


that simplifies to 100 - 20x + x^2.


therefore (x^2 + (10-x)^2) / 16 becomes (x^2 + 100 - 20x + x^2)/ 16.


that simplifies to (2x^2 - 20x + 100) / 16


that simplifies to (x^2 - 10x + 50) / 8


that looks like selection C.


you can confirm by assigning an arbitrary value to x that is within the limits of the problem.


the total length of the wire is 10.
if you assume that x = 8, then you have lengths of 8 and 2.
the wires are then formed into 2 squares.
the first square has a side length of 8/4 = 2
the second square has a side length of 2/4 = .5
the area of the first square is 2^2 and the area of the second square is .5^2.
2^2 + .5^2 = 4 + .25 which is equal to a total area of 4.25.


using the final formula of (x^2 - 10x + 50) / 8, and replacing x with 8, we get (8^2 - 10*8 + 50) / 8 becomes (64 - 80 + 50) / 8 which becomes 34/8 which becomes 4.25.


selection C looks good.