Question

The length of a rectangular field is 9 m more than the twice of its width. If the area of the field is $810 \, m^2$, find the length and width of the field.

Hint:

In word problems, always form an equation step-by-step. For rectangular fields, width is often taken as $x$, and length is expressed in terms of $x$. Then apply area = length × width.

Answer 1

Let the width of the rectangular field be $x$ metres.
Then, length of the field $= 2x + 9$ metres.

Step 1: Write the area equation
\[ \text{Area} = \text{Length} \times \text{Width} \] \[ 810 = (2x+9)\cdot x \]

Step 2: Form the quadratic equation
\[ 810 = 2x^2 + 9x \] \[ 2x^2 + 9x - 810 = 0 \]

Step 3: Simplify
Divide throughout by 2: \[ x^2 + \tfrac{9}{2}x - 405 = 0 \] Multiply through by 2 to avoid fraction: \[ 2x^2 + 9x - 810 = 0 \]

Step 4: Solve using quadratic formula
\[ x = \frac{-9 \pm \sqrt{9^2 - 4(2)(-810)}}{2(2)} \] \[ x = \frac{-9 \pm \sqrt{81 + 6480}}{4} \] \[ x = \frac{-9 \pm \sqrt{6561}}{4} \] \[ x = \frac{-9 \pm 81}{4} \] So, \[ x = \frac{72}{4}=18 \text{or} x = \frac{-90}{4}=-22.5 \] Width cannot be negative, so $x=18$.

Step 5: Find length
\[ \text{Length} = 2x + 9 = 2(18)+9 = 45 \] \[ \boxed{\text{Width = 18 m, Length = 45 m}} \]