Question

a rectangle has an area of 40 square units. the length is 6 units greater than the width.what are the dimensions of the rectangle?8 by 510 by 411 by 913 by 7

Answer 1

Given:

Area of the rectangle = 40 square units

The length is 6 units greater than the width.

Required:

To find the dimensions of the rectangle.

Step-by-step explanation:

Let the width of the rectangle = x unit

Since the length is 6 units greater than the width.

So length = x+6 unit

Now the area of the rectangle = length x width

`\begin{gathered} 40=(x+6)* x \\ 40=x^2+6x \\ x^2+6x-40=0 \end{gathered}`

This equation is a quadratic equation, we will solve it by using the middle term splitting method.

`\begin{gathered} x^2+6x-40=0 \\ x^2-4x+10x-40=0 \\ x(x-4)+10(x-4)=0 \\ (x-4)(x+10)=0 \\ x=4,-10 \end{gathered}`

Since dimensions can not be negative so we will take the positive value.

width = 4 unit

length = 4+6 = 10 unit

Final answer:

The dimensions of the rectangle are 4 and 10.

Thus option (B) is the correct answer.