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)
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Answer:
1348 light years.
Step-by-step explanation:
Please find the attachment.
Let x represent the distance between KA-7 and KA-11.
We have been given that the first system, KA-7, is 1200 light years away while the second system, KA-11, is 1700 light years away. Lucy sees an angle of 52 degrees between KA-7 and KA-11.
We can see from our attachment that Lucy, KA-7 and KA-11 forms a triangle and we will use law of cosines to solve for x.
Upon substituting our given values in above formula we will get,
Let us take square root of both sides of our equation.
Therefore, KA-7 and KA-11 are approximately 1348 light years apart.
