Could someone possibly me with this? terri has 19 feet of tape to mark n area in the classroom where the students may read. which of these shapes has a perimeter or circumference that would use most or all of the tape? a) square with side length of 5 feet b) circle with the radius of 3 feet c) right triangle with each leg length of 6 feet d) rectangle with a length of 8 feet and a width of 3 feet

A  4*5  ft = 20 foot - no good

B  circumference  of circle = 2* pi *3 = 6 pi  = 18.85 ft so this will use a lot of the 19 ft.

C.  right triangle  triangle  - perimeter = 6 + 6 + 6*sqrt2 = 20.49  - no good

D.  rectangle  perimeter = 16 + 8 = 24

Answer is choice   B

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

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