Is there a relationship between area or perimeter of a rectangle
1 answer:
Hello,
For a given perimeter (P) there are an infinity of Area (A)
Let's say x the length, and y the wide of the rectangle
P=2(x+y)
A=xy
k=x-y >=0
As (x+y)²-4xy=(x-y)²: A²-4P=k² or P=(A²-k²)/4
In primus, you will find a graph (abacus) giving P for a A and k given.
Negative Area or P are excluded.(just remind the first quadrant, A>=0 and P>=0)
For a given perimeter (P) there are an infinity of Area (A)
Let's say x the length, and y the wide of the rectangle
P=2(x+y)
A=xy
k=x-y >=0
As (x+y)²-4xy=(x-y)²: A²-4P=k² or P=(A²-k²)/4
In primus, you will find a graph (abacus) giving P for a A and k given.
Negative Area or P are excluded.(just remind the first quadrant, A>=0 and P>=0)
You might be interested in
Answer:
The equation of a parabola is
Step-by-step explanation:
(h,k) is the vertex and (f,k) is the focus.
Thus, f = 1, k = −4.
The distance from the focus to the vertex is equal to the distance from the vertex to the directrix: f - h = h - 2.
Solving the system, we get h = 3/2, k = -4, f = 1.
The standard form is:
The general form is:
The vertex form is:
The axis of symmetry is the line perpendicular to the directrix that passes through the vertex and the focus: y = -4.
The focal length is the distance between the focus and the vertex: 1/2.
The focal parameter is the distance between the focus and the directrix: 1.
The latus rectum is parallel to the directrix and passes through the focus: x = 1.
The length of the latus rectum is four times the distance between the vertex and the focus: 2.
The eccentricity of a parabola is always 1.
The x-intercepts can be found by setting y = 0 in the equation and solving for x.
x-intercept:
The y-intercepts can be found by setting x = 0 in the equation and solving for y.
y-intercepts: