Chapter 6, Problem 9MPS

To determine

To Solve:

The given problem.

Answer to Problem 9MPS

The dimensions of the rectangle are 6 cm and 3 cm.

Explanation of Solution

Given:

Given statement is :

Find the dimensions of a rectangle whose perimeter is 18 cm and whose area is $18{\text{ cm}}^2$ .

Calculation:

Let

l = length of the rectangle

w = width of the rectangle

Perimeter of the rectangle $=2(l+w)=18$

Area of the rectangle = $lw=18$

Solve both the equations:

  $\begin{array}{l}2(l+w)=18\\ \Rightarrow l+w=\frac{18}{2}\mathrm{..............}[\text{divide each side by 2]}\\ \Rightarrow l+w=9\\ \Rightarrow l=9-w\mathrm{................}[\text{subtract w from each side]}\\ \text{Substitute l=9-w in the equation:}\\ lw=18\\ \Rightarrow (9-w)w=18\\ \Rightarrow 9w-w^2=\mathrm{18...........}[\text{distribute w]}\\ \Rightarrow w^2-9w+18=\mathrm{0......}[{\text{subtract (9w-w}}^2)from\text{ each side]}\\ \Rightarrow w^2-6w-3w+18=\mathrm{0..}\left[\begin{array}{l}factorize\text{ the middle term in such a way that the sum of the }\\ \text{factors is the middle term and the product of the factors is the product of the}\\ \text{first and the last term}\text{.}\end{array}\right]\\ \Rightarrow w(w-6)-3(w-6)=0\\ \Rightarrow (w-3)(w-6)=0\\ \Rightarrow w-3=0\text{ or }w-6=\mathrm{0..........}[ab=0\iff a=0orb=0]\\ \Rightarrow w=3\text{ or }w=6\\ If\text{ w=3:}\\ l=9-w=9-3=6\\ If\text{ w=6:}\\ l=9-w=9-6=3\end{array}$

So, the dimensions of the rectangle are 6 cm and 3 cm.