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 $l = 3w - 11$ meters.

Step2: Set up area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 70$ m², we substitute $l$ and $A$ into the formula: $(3w - 11)w=70$.

Step3: Expand the equation

Expand $(3w - 11)w$ to get $3w^{2}-11w = 70$. Rearrange it to the standard quadratic - form $3w^{2}-11w - 70 = 0$.

Step4: 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 have 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 $w = 7$ m.

Step5: Find the length

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

Answer:

Length: 10 m
Width: 7 m