Let's denote the width of the rectangle as \( w \). According to the problem, the length \( l \) of the rectangle is twice the width:
\[ l = 2w \]
If the width is reduced by 1 cm and the length is reduced by 2 cm, the new dimensions will be:
- New width: \( w - 1 \)
- New length: \( 2w - 2 \)
The area of the modified rectangle is given as 15 cm². Therefore, we can set up the equation:
\[ (w - 1)(2w - 2) = 15 \]
Now, let's expand this equation:
\[ w(2w - 2) - 1(2w - 2) = 15 \] \[ 2w^2 - 2w - 2w + 2 = 15 \] \[ 2w^2 - 4w + 2 = 15 \]
Now, we can simplify and rearrange the equation:
\[ 2w^2 - 4w + 2 - 15 = 0 \] \[ 2w^2 - 4w - 13 = 0 \]
To simplify the equation, we can divide everything by 2:
\[ w^2 - 2w - \frac{13}{2} = 0 \]
Now we will use the quadratic formula to find \( w \). The quadratic formula states that for an equation in the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = -\frac{13}{2} \):
First, we calculate \( b^2 - 4ac \):
\[ b^2 = (-2)^2 = 4 \] \[ 4ac = 4 \cdot 1 \cdot -\frac{13}{2} = -26 \] \[ b^2 - 4ac = 4 + 26 = 30 \]
Now, we can compute the width \( w \):
\[ w = \frac{-(-2) \pm \sqrt{30}}{2 \cdot 1} = \frac{2 \pm \sqrt{30}}{2} = 1 \pm \frac{\sqrt{30}}{2} \]
Now let's find the positive solution, since dimensions must be positive:
\[ w = 1 + \frac{\sqrt{30}}{2} \]
Now, substitute back to find the length:
\[ l = 2w = 2 \left(1 + \frac{\sqrt{30}}{2}\right) = 2 + \sqrt{30} \]
Thus, the original dimensions of the rectangle are:
- Width: \( 1 + \frac{\sqrt{30}}{2} \) cm
- Length: \( 2 + \sqrt{30} \) cm
Now you have the dimensions of the original rectangle.
1 answer