Chapter 4.10, Problem 12P

To determine

To Find: the dimension of original rectangle.

Answer to Problem 12P

No solution.

Explanation of Solution

Given:

Length is $8\text{cm}$ longer than its width.

If width is increased by $3\text{cm}$ and width is decreased by $3\text{cm}$ then area is increased by $4{\text{cm}}^2$ .

Calculation:

Step 1.

Area of rectangle.

Step 2.

Since, it is given as.

  $\begin{array}{l}L-W=8\\ L=8+W\end{array}$

Equation as per given condition.

  $\begin{array}{l}\text{New}\text{area}-\text{original}\text{area=4}\\ \text{L}\times \text{W-}\left(L-3\right)\left(W+3\right)=4\\ \left(8+W\right)W-\left(8+W-3\right)\left(W+3\right)=4\\ \left(8+W\right)W-\left(W+5\right)\left(W+3\right)=4\\ W^2+8W-\left(W^2+3W+5W+15\right)=4\\ W^2+8W-\left(W^2+8W+15\right)=4\\ -15=4\end{array}$

It is not possible because it leads to inconsistency therefore it has no solution.

So the answer is no solution.