# Thread: How can i solve this problem! rectangles

1. ## How can i solve this problem! rectangles

Hi, How's everything guys?!
Guys i have a problem with a basic math question

Question: Length of a rectangle is equal to [2x+1] and Width of the rectangle is equal to [x+5] Calculate the perimeter and area of ​​the rectangle in terms of x.

Length : 2x+1
Width : x+5

Perimiter : 2(2x+1+x+5) => 2(3x +6) => 6x + 12 => x= 6/12 => x=2
Area: (2x+1).(x+5) => (3x+ 10x + x + 5) => (14x + 5) => x = 5/14

2. ## Re: How can i solve this problem! rectangles

Length : 2x+1
Width : x+5

Perimiter : 2(2x+1+x+5) => 2(3x +6) => 6x + 12 => x= 6/12 => x=2
Area: (2x+1).(x+5) => (3x+ 10x + x + 5) => (14x + 5) => x = 5/14
For the perimeter calculation you were doing great for both of these right up til the last statement. You have:

Perimeter = 6x + 12.

That's the answer right there, as it defines the perimter in terms of x, so no need to go on! Your next move was to say that x = 12/6, but that's wrong. Applying basic algebra you might have said this:

Perimter = 6x+12
6x = Perimeter - 12
x = Perimter/6 - 2

But that simply gives x in terms of the perimter, but that's not what they're asking.

As for the area calculation you didn't multiply the two terms correctly. Area = (2x+1)(x+5) = 2x(x+5) + 1(x+5) = 2x^2 +10x + x + 5 = 2x^2 +11x + 5.