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If the length of fred's piece of paper is represented by 2x-6 ad the width is represented by 3x-5, then what is the perimeter and area of fred's paper?

Answer 1

Savannah Aguilar

Area #=6x^2-28x+30#

Perimeter #=10x-22#

So to start off, the perimeter is

#P = 2l+2w#

Then you input the width for #w# and the length for #l#. You get

#P = 2(2x-6) + 2(3x - 5)#

#P = 4x - 12 + 6x - 10#

#P = 10x - 22#

for the perimeter. For the area, you multiply.

#A=L*W #

#A = (2x-6)(3x-5)#

# = 6x^2-10x-18x+30#

#= 6x^2-28x+30#

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Answer 2

Savannah Aguilar

The perimeter of Fred's piece of paper can be found by adding up all the sides, which are represented by ( 2x - 6 ) for the length and ( 3x - 5 ) for the width:

[ \text{Perimeter} = 2(2x - 6) + 2(3x - 5) ]

To find the area of Fred's paper, multiply the length and width:

[ \text{Area} = (2x - 6)(3x - 5) ]

You can simplify these expressions further by distributing and combining like terms to get the final values for the perimeter and area.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.