Question

Given area of a rectangle: 7x^2+57x+8 Find Perimeter:

Answer 1

Answer:

Perimeter of the rectangle is 16x+18 units.

Step-by-step explanation:

The area of rectangle is

$7x^2+57x+8$

First we have to find the factor form of above expression.

Splitting the middle term, we get

$7x^2+56x+x+8$

$7x(x+8)+1(x+8)$

$(x+8)(7x+1)$

Area of a rectangle is the product of its dimensions. It means dimensions of the rectangle are (x+8) and (7x+1).

Perimeter of a rectangle is

$Perimeter=2(l+b)$

$Perimeter=2[(x+8)+(7x+1)]$

$Perimeter=2[8x+9]$

$Perimeter=16x+18$

Therefore, the perimeter of the rectangle is 16x+18 units.