Two rectangles have the same width. The length of one is 1 foot longer than the width. The length of the other is 2 feet longer than the width. The larger rectangle has 5 more square feet than the smaller. What is the width of the rectangles
Area=lw
w=width
w=width of 1st, w=width of 2nd (since their widths are equal)
l=length
l=w+1 length of 1st
l=w+2 length of 2nd
Area of 1st = w(w + 1)
Area of 2nd = w(w + 2) - 5
Therefore,
w(w + 1) = w(w + 2) - 5
Solve for w
if the length is 7 meters longer than its width. what is the width of the object if the perimeter is equal to 98
To find the width of the rectangles, let's assume the width of both rectangles is represented by "x" (in feet).
According to the given information, the length of one rectangle is 1 foot longer than the width. Hence, the length of this rectangle would be x + 1.
Similarly, the length of the other rectangle is 2 feet longer than the width. Therefore, the length of this rectangle is x + 2.
Now, we can calculate the areas of both rectangles. The area of a rectangle is determined by multiplying its length and width.
The area of the smaller rectangle = x * (x + 1)
The area of the larger rectangle = x * (x + 2)
It is given that the larger rectangle has 5 more square feet than the smaller. So, we can set up the equation:
(x * (x + 2)) - (x * (x + 1)) = 5
Now let's solve this equation step by step:
(x² + 2x) - (x² + x) = 5
x² + 2x - x² - x = 5
x - x = 5
x = 5
Therefore, the width of both rectangles is 5 feet.