Question

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

Answer 1

The sides are;

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

The perimeter is;

`P=4a-2`

Step-by-step 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* 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* 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`