Question

The length of a rectangle is 7 more than twice the width, and the area of the rectangle is 99 yd^2 . Find the dimensions of the rectangle

Answer 1

Answer:

Width of a rectangle = $\frac{11}{2}$ yards

Length of a rectangle = $18$ yards

Step-by-step explanation:

Let width be $x$ yards.

Length of a rectangle = $7+2x$

Area of a rectangle = length × width

Put area of the rectangle as $99\,\,yd^2$

Therefore,

$99=x(7+2x)\\\\99=7x+2x^2\\\\2x^2+7x-99=0\\\\2x^2+18x-11x-99=0\\2x(x+9)-11(x+9)=0\\\\(2x-11)(x+9)=0\\\\x=\frac{11}{2},x=-9$

Being a dimension, $x$ cannot be negative.

So,

$x=\frac{11}{2}$

Width = $x=\frac{11}{2}$ yards

Length of a rectangle = $7+2x=7+2(\frac{11}{2})=7+11=18$ yards