Chapter 4.9, Problem 3ST

To determine

To find the uniform width of the deck.

Answer to Problem 3ST

The uniform width of the deck is $3$ meters.

Explanation of Solution

Given information:

Length and breadth of the pool are $10\text{ m}$ and $20\text{ m}$ respectively. The area of the deck is $216{\text{ m}}^2$ .

  

Let, the width of the deck is $x\text{ m}$ .

Area of the pool $=10\times 20=200{\text{ m}}^2$ .

Length of the pool with deck $=10+x+x=(10+2x)\text{ m}$ .

Breadth of the pool with deck $=20+x+x=(20+2x)\text{ m}$ .

Area of the pool with deck $\text{=}\left[\left(10+2x\right)\times \left(20+2x\right)\right]{\text{m}}^2$ .

Now, the area of the deck $=216{\text{ m}}^2$ .

  $\therefore$ Area of the pool with deck − Area of the pool $=216{\text{ m}}^2$

  $\left[\left(10+2x\right)\times \left(20+2x\right)\right]$

  $-200=216$

  $\begin{array}{l}200+20x+40x+4x^2=216+200\\ 200+60x+4x^2=416\\ 4x^2+60x=416-200\\ 4x^2+60x=216\end{array}$

  $\begin{array}{l}4\cdot \left(x^2+15x\right)=216\\ x^2+15x=\frac{216}{4}\\ x^2+15x=54\end{array}$

  $\begin{array}{l}{x}^2+15x-54=0\\ x^2+18x-3x-54=0\\ x\cdot \left(x+18\right)-3\cdot \left(x+18\right)=0\end{array}$

  $\begin{array}{l}\left(x+18\right)\cdot \left(x-3\right)=0\\ \\ \therefore x+18=0\therefore x-3=0\\ x=-18x=3\end{array}$

Since $x$ represents a width, so it cannot be negative.

  $\therefore x=-18\text{ cannot be the width of the deck}\text{. }$

So, the only reasonable solution is $x=3$ .

Hence,

The uniform width of the deck is $3$ meters.