Question

The length of a rectangle is twice it's width .if the width is reduced by 1cm and the length is also reduced by 2cm ,the area will be 15cm .find the dimensions of the original rectangle

Answer 1

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.