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.

Question

asked Mar 10, 2023 111k views

1 Answer

Ronney · Answer 1 · 2023-03-16T08:14:35+0000

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