Question

The length of a rectangle is 8cm greater than twice its width. Find the dimensions if the area

Answer 1

Answer:

Length $14\ cm$

Width $3\ cm$

Step-by-step explanation:

Let

x-----> the length of rectangle

y ----> the width of rectangle

we know that

The area of rectangle is equal to

$A=xy$

$A=42\ cm^{2}$

so

$42=xy$ ----> equation A

$x=2y+8$ ---> equation B

substitute equation B in equation A and solve for y

$42=(2y+8)y\\ \\2y^{2}+8y-42=0$

Solve the quadratic equation by graphing

The solution is $y=3\ cm$

see the attached figure

Find the value of x

$x=2(3)+8=14\ cm$

therefore

Length $14\ cm$

Width $3\ cm$