If perimeter is 62 cm and area is 198 squared cm what are the dimensions?
2(x+y) = 62
xy = 198
x+y=31
y = 31-x
x(31-x) = 198
x^2-31x+198 = 0
(x-9)(x-22) = 0
...
To find the dimensions of a shape given its perimeter and area, you can use a system of equations. Let's assume that the shape we are dealing with is a rectangle.
Let's say the length of the rectangle is L and the width is W.
We are given two pieces of information:
1. The perimeter of a rectangle is given by the formula: P = 2(L + W) = 62 cm
2. The area of a rectangle is given by the formula: A = L * W = 198 cm²
We can use these two equations to solve for the dimensions of the rectangle.
From the first equation: 2(L + W) = 62, we can simplify it to L + W = 31 and rearrange it to W = 31 - L.
Substituting this value of W into the second equation, we get (31 - L) * L = 198.
Now we have a quadratic equation that can be solved to find the values of L (length) and W (width).
Expanding and rearranging the equation, we have:
31L - L² = 198.
Rearranging it further, we have:
L² - 31L + 198 = 0.
Now we can solve this quadratic equation using factoring, completing the square, or using the quadratic formula.
The factors of 198 that add up to -31 are -18 and -11. Therefore, the equation can be factored as:
(L -18)(L - 11) = 0.
Setting each factor equal to zero, we get L - 18 = 0 or L - 11 = 0.
So, the possible values for L are L = 18 or L = 11.
If L = 18, then W = 31 - L = 31 - 18 = 13.
If L = 11, then W = 31 - L = 31 - 11 = 20.
Therefore, the dimensions of the rectangle can be either 18 cm by 13 cm or 11 cm by 20 cm.