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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 xaxis, 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). 
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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Gennadiy

Post subject: Re: math (t.2, qt.20): geometry, x,yplane Posted: 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,yplane (a 2Dfigure). 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 3Dfigure). 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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