Question

the length of a rectangle is 11 m less than three times the width, and the area of the rectangle is 70 m². find the dimensions of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ meters. Then the length $l$ of the rectangle is $3w - 11$ meters.

Step2: Set up the area - equation

The area $A$ of a rectangle is given by $A=l\times w$. We know that $A = 70$ and $l=3w - 11$, so we have the equation $(3w - 11)w=70$.
Expanding this gives $3w^{2}-11w - 70 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 3$, $b=-11$, $c = - 70$), we can use the quadratic formula $w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-11)^{2}-4\times3\times(-70)=121 + 840=961$.
Then $w=\frac{11\pm\sqrt{961}}{2\times3}=\frac{11\pm31}{6}$.
We get two solutions for $w$:
$w_1=\frac{11 + 31}{6}=\frac{42}{6}=7$ and $w_2=\frac{11-31}{6}=\frac{-20}{6}=-\frac{10}{3}$. Since the width cannot be negative, we take $w = 7$ meters.

Step4: Find the length

Substitute $w = 7$ into the length formula $l=3w - 11$. Then $l=3\times7-11=21 - 11 = 10$ meters.

Answer:

Width: 7 m, Length: 10 m