+85 Square A and Square B are both 2009 by 2009 squares. Square A has both its length and width increased by an amount x, while Square B has its length and width decreased by the same amount x. What is the minimum value of x such that the difference in area between the two new squares is at least as great as the area of a 2009 by 2009 square?
Thanks so much!
+85 The new area of Square A is (2009 +x)^2, while the new area of Square B is (2009 - x)^2. The difference is (2009+x)2−(2009−x)2=(2009+x+2009−x)(2009+x−2009+x)=(2⋅2009)(2x)
For this to be at least as great as the area of a 2009 by 2009 square, 2(2009)2(x)≥20092⇒x≥20094.
+921 Let's analyze the areas of the new squares and set up an inequality to find the minimum value of x.
Original Areas:
Area of Square A: 2009 * 2009 (square units)
Area of Square B: 2009 * 2009 (square units)
New Areas:
New area of Square A: (2009 + x) * (2009 + x) (square units)
New area of Square B: (2009 - x) * (2009 - x) (square units)
Difference in Area:
We want the difference in area between the new squares to be at least as great as the area of a 2009 by 2009 square. So, we can set up an inequality:
(New area of Square A) - (New area of Square B) ≥ Area of a 2009 by 2009 square
Expanding the Inequality:
[(2009 + x) * (2009 + x)] - [(2009 - x) * (2009 - x)] ≥ 2009 * 2009
Simplifying the Inequality:
Expand both squares and subtract like terms:
4x^2 ≥ 2009 * 2009
Solving for x:
Divide both sides by 4:
x^2 ≥ (2009 * 2009) / 4
Take the square root of both sides (remembering that x can be positive or negative since the area cannot be negative):
x ≥ ± (2009 / 2)
Finding the Minimum Value:
We want the minimum non-negative value of x. Since the area cannot be negative, we discard the negative solution. Therefore:
x ≥ 2009 / 2
+85 The new area of Square A is (2009 +x)^2, while the new area of Square B is (2009 - x)^2. The difference is (2009+x)2−(2009−x)2=(2009+x+2009−x)(2009+x−2009+x)=(2⋅2009)(2x)
For this to be at least as great as the area of a 2009 by 2009 square, 2(2009)2(x)≥20092⇒x≥20094.
