Question

1. find the perimeter and area of the following rectangle. (2a² - 4a + 5)(6 + 2a) 2a² - 4a + 5 6 + 2a a = 4a³+4a² - 14a + 30

Explanation:

Step1: Recall perimeter formula

The perimeter $P$ of a rectangle is $P = 2(l + w)$, where $l=2a^{2}-4a + 5$ and $w=6 + 2a$.
\[

$$\begin{align*} P&=2((2a^{2}-4a + 5)+(6 + 2a))\\ &=2(2a^{2}-4a+2a + 5 + 6)\\ &=2(2a^{2}-2a + 11)\\ &=4a^{2}-4a + 22 \end{align*}$$

\]

Step2: Recall area formula

The area $A$ of a rectangle is $A=l\times w$. We already have the expansion of $(2a^{2}-4a + 5)(6 + 2a)$ as $A = 4a^{3}+4a^{2}-14a + 30$ from the given work.

Answer:

Perimeter: $4a^{2}-4a + 22$
Area: $4a^{3}+4a^{2}-14a + 30$