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How to find the area and the perimeter of a rectangle when the with is 3ab^4 and the width is 5a^2b^4

2 Answers

5 votes

Answer:

AREA of rect. - l x b

= 3ab^4 x 5a^2b^4

PERIMETER - 2(l+b)

=2(3ab^4 + 5a^2-b^4

User Wolfer
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5 votes

Explanation:

Area of a rectangle = length x width

Perimeter of a rectangle = 2 x (length + width)

Width = 3ab^4

Length = 5a^2b^4

Area = length x width

Area = (5a^2b^4) x (3ab^4)

Area = 15a^3b^8

Therefore, the area of the rectangle is 15a^3b^8.

Perimeter = 2 x (length + width)

Perimeter = 2 x (5a^2b^4 + 3ab^4)

Perimeter = 2 x a^2b^4 (5 + 3)

Perimeter = 16a^2b^4

Therefore, the perimeter of the rectangle is 16a^2b^4.

User Jillian
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