If the sides of a square are increased by 3 cm, the area is decreased by 36 cm^2. What were the dimensions of the original square? Please show me the formula and how to work.
That doesn't sound possible.
If the sides of a square are increased, the area must also be increased.
Ms.Sue is correct. You may have copied down the problem incorrectly.
If increasing the sides by three INCREASES the area by 36, and the orginal side length is a, then
(a+3)^2 = a^2 + 36
a^2 + 6x + 9 = a^2 + 36
Subtract out the a^2 terms on both sides.
6x = 27
x = 4.5
I'm sorry, but your initial statement about the sides of the square being increased and the area being decreased is contradictory. In a square, if the sides are increased, the area will also increase. Similarly, if the sides are decreased, the area will decrease.
However, if you have made a mistake in expressing the problem and would like to find the dimensions of the original square when the sides are increased by 3 cm and the area is decreased by 36 cm^2, I can help you with that.
Let's assume the original side length of the square is "x" cm.
When the sides are increased by 3 cm, the new side length becomes "x + 3" cm.
The area of the original square is given by the formula: A = side length^2. Therefore, the original area is x^2 cm^2.
The area of the new square (after increasing the sides) is given by: A' = (x + 3)^2 cm^2.
According to the problem, the area is decreased by 36 cm^2. So, we can set up the equation: x^2 - (x + 3)^2 = 36.
Simplifying this equation step by step:
x^2 - (x + 3)(x + 3) = 36
x^2 - (x^2 + 6x + 9) = 36
x^2 - x^2 - 6x - 9 = 36
-6x - 9 = 36
-6x = 36 + 9
-6x = 45
x = 45 / -6
x = -7.5
However, this negative value for x does not make sense in the context of the problem, as side lengths cannot be negative. This suggests that there might be a mistake in the problem statement or in the calculations.
Please double-check the problem statement or provide any additional information to help in finding the correct solution.