The two lines y = x and x = -4 intersect on the coordinate plane. If z

Line y = x, is a line passing through Ist and IInd Quadrant via origin. And Line x = -4 is a vertical line.
The area intersecting these two lines is a triangle with base = 4 and height = 4. Hence the area, z = 1/2 * 4 * 4 = 8

Given, Surface area of Cube = 6 * side^2 = 6z ==> side^2 = 8
side = 2\sqrt{2}

Option D

800score Official Solution:

The first step to solving this problem is to actually graph the two lines. The lines intersect at the point (-4, -4) and form a right triangle whose base length and height are both equal to 4. As you know, the area of a triangle is equal to one half the product of its base length and height: A = (1/2)bh = (1/2)(4 × 4) = 8; so z = 8.

The next step requires us to find the length of a side of a cube that has a face area equal to 8. As you know the 6 faces of a cube are squares. So, we can reduce the problem to finding the length of the side of a square that has an area of 8. Since the area of a square is equal to s², where s is the length of one of its side, we can write and solve the equation s² = 8. Clearly s = √8 = 2√2 , or answer choice (D).

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