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express the area of the figure, a, as a function of one of its dimensio…

Question

express the area of the figure, a, as a function of one of its dimensions, x. write the functions equation as a polynomial in standard form. a(x) = □

Explanation:

Step1: Calculate area of rectangle

The rectangle has length $(x - 7)+(x + 3)=2x - 4$ and width $x + 1$. Using the formula for the area of a rectangle $A_{rect}=l\times w$, we have $A_{rect}=(2x - 4)(x + 1)$.
\[

$$\begin{align*} (2x-4)(x + 1)&=2x\times x+2x\times1-4\times x - 4\times1\\ &=2x^{2}+2x-4x - 4\\ &=2x^{2}-2x - 4 \end{align*}$$

\]

Step2: Calculate area of left - hand triangle

The left - hand triangle has base $x - 7$ and height $x$. Using the formula for the area of a triangle $A_{triangle}=\frac{1}{2}bh$, we get $A_{left - triangle}=\frac{1}{2}(x - 7)x=\frac{1}{2}x^{2}-\frac{7}{2}x$.

Step3: Calculate area of right - hand triangle

The right - hand triangle has base $x + 3$ and height $x$. Using the formula for the area of a triangle $A_{triangle}=\frac{1}{2}bh$, we have $A_{right - triangle}=\frac{1}{2}(x + 3)x=\frac{1}{2}x^{2}+\frac{3}{2}x$.

Step4: Calculate total area

The total area $A(x)$ of the figure is the sum of the areas of the rectangle, the left - hand triangle, and the right - hand triangle.
\[

$$\begin{align*} A(x)&=(2x^{2}-2x - 4)+(\frac{1}{2}x^{2}-\frac{7}{2}x)+(\frac{1}{2}x^{2}+\frac{3}{2}x)\\ &=(2x^{2}+\frac{1}{2}x^{2}+\frac{1}{2}x^{2})+(-2x-\frac{7}{2}x+\frac{3}{2}x)-4\\ &=(2 + \frac{1}{2}+\frac{1}{2})x^{2}+(-2-\frac{7}{2}+\frac{3}{2})x-4\\ &=3x^{2}+(-2 - 2)x-4\\ &=3x^{2}-4x - 4 \end{align*}$$

\]

Answer:

$3x^{2}-4x - 4$