With part a, i’re tried the question but i’m not sure if it’s correct

tsa of semi-cylinder is 21.21m²

Step-by-step explanation:

tsa of cylinder = 2πr² + 2πrh

radius = diameter ÷2

= 1.5m

tsa = (2×π×1.5²) + (2×π×1.5×3)

= 4.5π + 9π

= 13.5π

≈ 42.412m²(3d.p) (cylinder)

semi = 42.41 ÷ 2

≈ 21.21m² (2d.p)

(Correct me if i am wrong)

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