thE LENGHT OF A RECTANGLE IS 3M MORE THAN ITS WIDHT .iF ITS AREA IS 54mETRE CARRE ,FIND THE DIMENSIONS OF THE RECTANGLE ?
let x = the width
let x+3 = the length
The area of a rectangle is length*width
x(x+3) = 54
x^2 + 3x = 54
x^2 + 3x - 54 = 0
(x+9)(x-6) = 0
x = 6, -9
A negative answer is irrelevant for this problem - the dimensions cannot be negative.
x = 6
x+3 = 9
The width is 6, and the length is 9.
To find the dimensions of the rectangle, we need to solve the problem step by step.
Let's assume that the width of the rectangle is "w" meters.
According to the problem, the length of the rectangle is 3 meters more than its width, which can be expressed as "w + 3" meters.
To find the area of the rectangle, we multiply its length and width. The formula for the area of a rectangle is:
Area = Length * Width
In this case, the area is given as 54 square meters. So, we can set up the equation:
54 = (w + 3) * w
Now, we need to solve this equation to find the dimensions of the rectangle.
Expanding the equation:
54 = w^2 + 3w
Rearranging the equation to make it equal to zero:
w^2 + 3w - 54 = 0
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Let's factor the equation:
(w + 9) (w - 6) = 0
Setting each factor equal to zero:
w + 9 = 0 or w - 6 = 0
Solving for w:
w = -9 or w = 6
Since the width cannot be negative, we discard the solution w = -9.
Thus, the width of the rectangle is 6 meters.
We can substitute this value back into the equation for the length:
Length = width + 3 = 6 + 3 = 9 meters
Therefore, the dimensions of the rectangle are width = 6 meters and length = 9 meters.