SOLUTION: What will be the length and the breadth for an area of 1202.45 square metres?
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Question 1011073: What will be the length and the breadth for an area of 1202.45 square metres?
Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x=length
y=width
xy=1202.45
There are an infinite number of ways this can occur.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the area is equal to length * width.
your equation is length * width = 1202.45
take any value for the length that is less than 1202.45 and more than 0 and you can then find the corresponding width.
likewise take any value for the width that is less than 1202.45 and more than 0 and you can then find the corresponding length.
for example:
if width is 5, then length is 1202.45 / 5 = 240.49
length * width = 5 * 240.49 = 1202.45
usually these types of problems have some other restriction such as width = 2 * length.
if width = length * 2, you can replace width with 2 * length in the equation of length * width = area to get:
length * width = 1202.45 becomes length * (2 * length) = 1202.45.
this becomes 2 * length squared = 1202.45.
divide both sides of this equation by 2 to get length squared = 1202.45 / 2.
takt the square root of this equation to get length = sqrt(1202.45/2)
if the length is sqrt(1202.45/2), then the width = 2 * sqrt(1202.45/2).
the area is equal to length * width which is equal to sqrt(1202.45/2) * 2 * sqrt(1202.45/2) which is equal to 2 * 1202.45/2 which is equal to 1202.45.
since there was no specification as to what the restrictions on length and width would be, you could choose any valid value of length and solve for width or you could choose any valid value of width and solve for length.
length * width = area
length = area / width.
width = area / length.
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