Chapter 8, Problem 21SGR

To determine

To find the perimeter of the frame.

Answer to Problem 21SGR

  $4x^2+4x+8$

Explanation of Solution

Given:

  

Formula used:

Perimeter of the rectangle $\text{= }2(l+b)$

Calculation:

Given that the frame of a painting is a rectangle.

From the given figure,

  $\text{Length of rectangle = }2x^2-3x+1$

  $\text{Width of rectangle = }5x+3$

Substituting the above values in the formula of perimeter of rectangle,

Perimeter of the rectangle $\text{= }2(l+b)$

  $\begin{array}{c}\Rightarrow 2((2x^2-3x+1)+(5x+3))\\ \Rightarrow 2(2x^2-3x+5x+(3+1))\\ \Rightarrow 2(2x^2+2x+4)\\ \Rightarrow 4x^2+4x+8\end{array}$

Conclusion:

Therefore, the perimeter of the rectangle is $4x^2+4x+8.$