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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<h2>Given :</h2>
- height (h) = 2 yd
- radius (r) = 3 yd
_____________________________
Answer:
a =
x =
Step-by-step explanation:
Let's solve for a.
8x−5x+4−x=ax+8
Step 1: Flip the equation.
ax+8=2x+4
Step 2: Add -8 to both sides.
ax+8+−8=2x+4+−8
ax=2x−4
Step 3: Divide both sides by x.
Let's solve for x.
8x−5x+4−x=ax+8
Step 1: Add -ax to both sides.
2x+4+−ax=ax+8+−ax
−ax+2x+4=8
Step 2: Add -4 to both sides.
−ax+2x+4+−4=8+−4
−ax+2x=4
Step 3: Factor out variable x.
x(−a+2)=4
Step 4: Divide both sides by -a+2.
Hope this helps!
brainliest?