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 Post subject: GRE Coordinate Geometry (Select One)Posted: Wed Aug 17, 2011 1:11 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
The two lines y = x and x = -4 intersect on the coordinate plane. If z represents the area of the figure formed by the intersecting lines and the x-axis, what is the side length of a cube whose surface area is equal to 6z?
A. 16
B. 8√2
C. 8
D. 2√2
E. (√2)/3

(D) 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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They mentioned the surface is 6z not z.
In your calculation you backsolve for surface = z = 8.
Have I missed something?

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 Post subject: Re: math (t.2, qt.20): geometry, x,y-planePosted: Thu Aug 18, 2011 6:27 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
In this problem we deal with two different figures.

1. The right triangle on the x,y-plane (a 2D-figure). The question statement tells us its area equals z. We can calculate this area and we do so to find the value of z.

2. Some specific cube (a 3D-figure). The question statement tells us its surface area equals 6z. We already calculated that z equals 8. So the area of the cube is 6 × 8 = 48.
We denoted the side of the cube by s (Our goal is to find the value of s). So its surface area is 6 × s². (The surface area of a cube is composed of its six sides. Each side is an "s × s" square).
6 × s² = 6 × 8
s² = 8

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