Chapter 8, Problem 3PT

To determine

To find the total area to be carpeted.

Answer to Problem 3PT

1y13y

  $2x^2+6x-10$

Explanation of Solution

Given:

Dimensions of room 1:

  

Dimensions of room 2:

  

Formula used:

  $\text{Area of rectangle}=l\times b$

Calculation:

Dimensions of room 1:

Length = x + 3

Breadth = x

From the given figure, the area of room 1 is calculated as,

  $\begin{array}{c}\text{Area of room 1}=l\times b\\ =(x+3)\times x\\ =x^2+3x\end{array}$

Therefore, the area of room 1 is $x^2+3x.$

Dimensions of room 1:

Length = x + 5

Breadth = x − 2 From the given figure, the area of room 2 is calculated as,

  $\begin{array}{c}\text{Area of room 2}=l\times b\\ =(x+5)\times (x-2)\\ =(x\times x)+(x\times -2)+(5\times x)+(5\times -2)\\ =x^2-2x+5x-10\\ =x^2+3x-10\end{array}$

Therefore, the area of room 2 is $x^2+3x-10.$

The sum of the areas of room 1 and room 2 gives the total area.

  $\begin{array}{l}\Rightarrow (x^2+3x)+(x^2+3x-10)\\ \Rightarrow (2x^2+6x-10)\end{array}$

Conclusion:

Therefore, the total area to be carpeted for both the rooms is $2x^2+6x-10.$