How do you find the area of a rectangle with sides #6a^2b^4# and #3ab^2#?

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Answer 1 · 2016-10-18T21:23:25

#A = 18 a^3b^6#

The area of a rectangle is found by multiplying the length by the breadth.

#A = lxxb#

If the sides are #6a^2b^4 and 3ab^2#, then the area is the product if these sides.

#A= 6a^2b^4 xx 3ab^2#

Just multiply the numbers and add the indices of like bases

#A = 18 a^3b^6#

Do not be put off by the question! The lengths of the sides of the rectangle are given using variables and powers.

Do the same as you would with numbers - just multiply the terms.

Answer 2 · 2016-10-18T21:26:54

#A=18a^3b^6#

In this exercise we will using the product of powers with same base that's :

#color(red)(b)^4*color(red)b^2=color(red)b^color(blue)(4+2)#

Knowing the area of rectangle that's :
#A_(rec)=L * W#

where #L# is the length of a rectangle and #W# is its width
Here,
#L=6a^2b^4 and W=3ab^2#

#A=(6a^2b^4)*(3ab^2)#
#A=(6*3)(a^2*a)(b^4*b^2)#

#A=18a^color(blue)(2+1)b^color(blue)(4+2)#

#A=18a^3b^6#