Question

the sides of rectangle are 2a2+b and 6a-5b Find its perimeter?​

Answer 1

Answer:

length of rectangle = 2a2+b

beardth of rectangle = 6a-5b

p = 2(l+b)

p = 2(2a2+b + 6a-5b)

p = 2(4a+b+1ab

p = 5ab +1ab

p = 6ab

perimeter = 6ab

Answer 2

Given,

• Length of the rectangle = 2a² + b
• Breadth of the rectangle = 6a-5b

To find,

• The perimeter of the rectangle

Solution,

The sides of the rectangle are 2a²+b and 6a-5b, then the perimeter is 4a² + 12a - 8b.

As we know,

Perimeter of the rectangle = 2(length +breadth)

                                            = 2( 2a² + b + 6a-5b)

                                            = 2(2a² + 6a -4b)

                                            = 4a² + 12a - 8b

∴ The perimeter of the rectangle is  4a² + 12a - 8b.

Hence, the sides of the rectangle are 2a²+b and 6a-5b, then the perimeter is 4a² + 12a - 8b.