Question

A rectangle has an area of a^2 - a - 132. Findthe sides and then find the perimeter

Answer 1

Answer:

The sides are;

$\begin{gathered} l=a-12 \\ b=a+11 \end{gathered}$

The perimeter is;

$P=4a-2$

Explanation:

The area of the rectangle is given as;

$A=a^2-a-132$

recall that area of a rectangle is length multiplied by breadth;

$A=l\times b$

let us factorise the given equation;

$\begin{gathered} A=a^2-a-132 \\ A=a^2-12a+11a-132 \\ A=a(a-12)+11(a-12) \\ A=(a-12)(a+11) \end{gathered}$

Comparing the factorised equation to the formula for area;

$\begin{gathered} A=l\times b=(a-12)(a+11) \\ l=a-12 \\ b=a+11 \end{gathered}$

The perimeter is given as;

$P=2(l+b)$

substituting the values of l and b;

$\begin{gathered} P=2(a-12+a+11) \\ P=2(2a-1) \\ P=4a-2 \end{gathered}$

Therefore; the sides are;

$\begin{gathered} l=a-12 \\ b=a+11 \end{gathered}$

The perimeter is;

$P=4a-2$