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The width of a rectangle is 4 less than the length. If the area of the rectangle is 165cm?,solve for the length and width of the rectangle.

The width of a rectangle is 4 less than the length. If the area of the rectangle is-example-1
User ChRoNoN
by
8.4k points

1 Answer

5 votes

length:15 cm

width: 14 cm

Step-by-step 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


\begin{gathered} \text{width}=length-4 \\ w=l-4\rightarrow equation\text{ (1)} \end{gathered}

b)the area of the rectangle is 165 square cm

the area of a rectangle is given by


\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}

Step 2

now, solve the equations


\begin{gathered} w=l-4 \\ 165=lw \end{gathered}

a) replace the w from equation (1) into equation(2)


\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}

b) now, we have a quadratic equation in the form


ax^2+bx+c=0

hence, we can use the quadratic formula to solve


x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

let

a=1

b=-4

c=-165

replace


\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=(4\pm26)/(2) \end{gathered}

therefore


\begin{gathered} x=(4\pm26)/(2) \\ x_1=(4+26)/(2)=(30)/(2)=15 \\ x_2=(4-26)/(2)=(-22)/(2)=-11 \end{gathered}

as we are searchinf for a distance, the only valid answer is 15

so, the answer is length= 15 cm


l=15\text{ cm}

c) finally, replace the l value into equation (1) to find the width


\begin{gathered} w=l-4\rightarrow equation\text{ (1)} \\ w=15-4 \\ w=11 \end{gathered}

therefore, the width is 11 centimeters

I hope this helps you

User Kamal Palei
by
8.0k points
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