Question

The length of a rectangle is x square +x and its width is 10-x . find its perimeter​

Answer 1

Step-by-step explanation:

Given:-

In a rectangle

Length=${x }^{2}+x$

width=10-x

To find:-

The perimeter of the rectangle

Solution:-

As we know in a rectangle

$\large{\boxed{Perimeter=2(l+b)}}$

According to formula

$perimeter = 2{( {x}^{2} + x) + (10 - x)} \\ = 2( {x}^{2} + x + 10 - x) \\ = 2( {x}^{2} + 10 + {\cancel{x}} -{\cancel{ x}}) \\ = 2( { x}^{2} + 10) \\ = {2x}^{2} + 20$

Answer 2

Answer

perimeter of rectangle = 2×(l+b)

2× X^2+ X +10-X

2X^2+20 ans