Question 976442: A Rectangular lot measures (5w+32)cm by (5w-32)cm is added to both dimensions how is area affected : EXPLAIN
1. What is Given
2.Solutions
A.Original Area
B.New Area
3.Explain
Need answers now !
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The dimensions are (5w+32)cm by (5w-32)cm . I don't know what is added.
The area as it stands is 25w^2-1024. If w=10, the area is 2500-1024=1476, an 82X18 rectangle
If you add 10 cm, the lot is (5w+42)(5w-22)=25w^2+100w-964. If w=10, area is 3500-964=2536, a 92X28 rectangle. It is quadratic, however, and not simply linear, like the perimeter.
Areas increase as the square, but it is more complicated in a rectangle because of internal products. Double the side length of a square and you increase the area 4 times. Triple it and you increase the area 9 times. With a rectangle, it is not that simple, but the This is basically how you do these problems. Again, without knowing what is added, I can't tell.
Example:
2*5 lot =10 square units.
add x to each.
(2+x)(5+x)=10 + 7x +x^2
if x =1, the lot's new area is 3*6=18
x^2+7x+10=18.
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