Chapter 4, Problem 1RVS

To determine

To calculate: The two algebraic expressions for the shaded area of the provided figure which is shown below,

Answer to Problem 1RVS

Solution:

The required area for the shaded region is $8x+15$ or $\left(x+5\right)\left(x+3\right)-x^2$ which is $E$ and $F$.

Explanation of Solution

Given Information:

The provided option is shown below,

A. $9-4x^2$, B. $x^2-{\left(x-6\right)}^2$, C. $\left(x+3\right)\left(x-3\right)$, D. ${10}^2+2^2$, E. $8x+15$, F. $\left(x+5\right)\left(x+3\right)-x^2$, G. $x^2-6x+9$, H. ${\left(3-2x\right)}^2+4x\left(3-2x\right)$, I. ${\left(x+3\right)}^2$, J. ${\left(5x+3\right)}^2-25x^2$, K. ${\left(5-2x\right)}^2+4x\left(5-2x\right)$, L. $x^2-9$, M. $104$, N. $x^2-15$, O. $12x-36$, P. $30x+9$, Q. $\left(x-5\right)\left(x-3\right)+3\left(x-5\right)+5\left(x-3\right)$, R. ${\left(x-3\right)}^2$, S. $25-4x^2$, T. $x^2+6x+9$.

The provided figure is shown below,

Formula used:

The area of rectangle A with length $l$ and width $w$ is given by,

$A=lw$

Calculation:

Consider the provided figure which is shown below,

Now, recall that the area of rectangle A with length $l$ and width $w$ is given by,

$A=lw$

Now, the area of shaded figure will be the difference of the area of entire rectangle and the area of rectangle not shaded.

The dimensions of the rectangle are $x+5$ by $x+3$, thus the area of entire rectangle will be,

$\begin{array}{c}{A}_1=\left(x+5\right)\left(x+3\right)\\ =x^2+3x+5x+15\\ =x^2+8x+15\end{array}$

Also, the dimensions of the area not shaded are $x$ by $x$, thus the area of unshaded rectangle will be,

$\begin{array}{c}{A}_2=x\cdot x\\ =x^2\end{array}$

Therefore, the area of the shaded region will be,

$\begin{array}{c}A=A_1-A_2\\ =x^2+8x+15-x^2\\ =8x+15\end{array}$

Thus, the required polynomial is $8x+15$.

Also, sum of area of inner part and outer shaded rectangle is $\left(x+5\right)\left(x+3\right)-x^2$.

Therefore, the required area for the shaded region is $8x+15$ or $\left(x+5\right)\left(x+3\right)-x^2$ which is $E$ and $F$.