length:15 cm
width: 14 cm
Explanation
Step 1
Let
w represents the width of the rectangle
l represents the length of the rectangle
a)The width of a rectangle is 4 less than the length,hence
write this in math terms, replace
[tex]\begin{gathered} \text{width}=length-4 \\ w=l-4\rightarrow equation\text{ (1)} \end{gathered}[/tex]b)the area of the rectangle is 165 square cm
the area of a rectangle is given by
[tex]\begin{gathered} \text{Area}=\text{length}\cdot\text{width} \\ \text{replace} \\ 165cm^2=l\cdot w\rightarrow \\ 165=l\cdot w\rightarrow equiation(2) \end{gathered}[/tex]Step 2
now, solve the equations
[tex]\begin{gathered} w=l-4 \\ 165=lw \end{gathered}[/tex]a) replace the w from equation (1) into equation(2)
[tex]\begin{gathered} 165=lw \\ 165=l(l-4) \\ 165=l^2-4l \\ \text{subtract 165 on both sides} \\ 165-165=l^2-4l-165 \\ l^2-4l-165=0 \end{gathered}[/tex]b) now, we have a quadratic equation in the form
[tex]ax^2+bx+c=0[/tex]hence, we can use the quadratic formula to solve
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]let
a=1
b=-4
c=-165
replace
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-4)\pm\sqrt[]{-4^2-4(1)(-165)}}{2\cdot1} \\ x=\frac{4\pm\sqrt[]{676}}{2}=x=\frac{4\pm26}{2} \end{gathered}[/tex]therefore
[tex]\begin{gathered} x=\frac{4\pm26}{2} \\ x_1=\frac{4+26}{2}=\frac{30}{2}=15 \\ x_2=\frac{4-26}{2}=\frac{-22}{2}=-11 \end{gathered}[/tex]as we are searchinf for a distance, the only valid answer is 15
so, the answer is length= 15 cm
[tex]l=15\text{ cm}[/tex]c) finally, replace the l value into equation (1) to find the width
[tex]\begin{gathered} w=l-4\rightarrow equation\text{ (1)} \\ w=15-4 \\ w=11 \end{gathered}[/tex]therefore, the width is 11 centimeters
I hope this helps you