Question

the length of a rectangle is 11 yd more than twice the width and the area of the rectangle is 63 yd squared. find the dimensions of the rectangle.

Answer 1

the length of a rectangle is 11 yd more than twice the width and the area of the rectangle is 63 yd squared. find the dimensions of the rectangle.​

Let

L ------> the lenght

W ----> the width

we know that

the area of rectangle is

A=L*W

A=63 yd2

63=L*W -------> equation 1

and

L=2W+11 ------> equation 2

substitute equation 2 in equation 1

63=(2W+11)*w

2W^2+11w-63=0

solve the quadratic equation using the formula

a=2

b=11

c=-63

substitute

`w=\frac{-11\pm\sqrt[]{11^2-4(2)(-63)}}{2(2)}`
`\begin{gathered} w=\frac{-11\pm\sqrt[]{625}}{4} \\ \\ w=(-11\pm25)/(4) \\ \end{gathered}`

the solutions for W are

w=3.5 and w=-9 (is not a solution, because is negative)

so

Find the value of L

L=2W+11 -------> L=2(3.5)+11

L=18

therefore

the dimensions are

Length is 18 yards

Width is 3.5 yards