Question

The area of a rectangle is 21 yd^2, and the length of the rectangle is 1 yd less than twice the width. Find the dimensions of the rectangle.

Answer 1

Length = 6yd

width = 3.5yd

Explanation:

Let, width = b

So, length = 2b-1

area = 21yd

We know,

area of rectangle = length X width

So,

$21yd^{2} = (2b-1) X b$

$2b^{2} - b - 21 = 0$

On solving the above equation, we get

Width, b = 3.5yd

length, l = 6yd