In the square, x represents a positive whole number. find the value of x such that the area is equal to half the perimeter of the square.
A = x^2 P = 4x
A = 1/2 P
x^2 = 2x
x^2 -2x =0
(x)(x-2) = 0
x= 0 and x = 2 Ans x=2
To solve this problem, let's denote the side length of the square as "s".
The area of a square is given by the formula A = s^2, and the perimeter is given by the formula P = 4s.
We need to find the value of x such that the area of the square is equal to half of its perimeter:
A = 1/2 * P
Substituting the formulas for area and perimeter:
s^2 = 1/2 * 4s
Simplifying the equation:
s^2 = 2s
Now, let's solve for "s" by subtracting 2s from both sides:
s^2 - 2s = 0
Factoring out an "s":
s(s - 2) = 0
This equation tells us that either s = 0 or (s - 2) = 0. However, since the problem specifies that "s" represents a positive whole number, we can ignore the solution s = 0.
So, we are left with the equation (s - 2) = 0:
s - 2 = 0
Solving for "s":
s = 2
Therefore, the value of "x" that makes the area of the square equal to half of its perimeter is x = 2.
To find the value of x, we first need to know the formulas for finding the area and perimeter of a square.
The area of a square is given by the formula: Area = side length * side length, or A = s^2.
The perimeter of a square is given by the formula: Perimeter = 4 * side length, or P = 4s.
According to the problem, the area of the square is equal to half the perimeter. Mathematically, this can be represented as:
A = (1/2)P
Substituting the formulas for area and perimeter of a square, we can rewrite the equation as:
s^2 = (1/2) * (4s)
Now, let's solve this equation step by step to find the value of x:
s^2 = 2s (simplify the right side: 1/2 * 4s = 2s)
s^2 - 2s = 0 (subtract 2s from both sides)
s(s - 2) = 0 (factor out the common term)
Now, we have two possible solutions for s:
s = 0 (from s = 0)
s - 2 = 0 (from s - 2 = 0)
Solving the second equation:
s = 2
Since the problem states that x represents a positive whole number, we can conclude that x = 2 is the value that satisfies the condition of the area being equal to half the perimeter of the square.