+2446 We meet again, Rick!
The shaded region, I would presume anyway, also happens to be the area of a rectangle. Let's take a guess and say that the left figure has more area. To confirm (or disprove) this, one must find the areas of the rectangles.
| \(\text{left }\boxed{\quad}=a(b+2)\) | |
| \(\text{right}\boxed{\quad}=b(a+2)\) | Let's assume that the left rectangle has more area and see what happens. |
| \(a(b+2) < b(a+2)\) | Now, solve. This will allow you to figure out the relationship between a and b. Distribute. |
| \(ab+2a < ab+2b\) | The ab's cancel out. |
| \(2a<2b\) | Divide by 2 on both sides. |
| \(a < b\) | |
What is this result telling you or me? It tells you that, in order for the leftward rectangle to have more area, a must be less than b. Of course, this cannot be; the original problem states that \(a>b\) . We have reached a contradiction. Therefore, the rightward rectangle has more area!
