Question

You are placing non-slip tape along the perimeter of the bottom of a semicircle-shaped doormat. How much tape will you save applying the tape to the perimeter of the inside semicircle of the doormat? Round your answer to the nearest hundredth, if necessary. A semicircle-shaped doormat with diameter of 30 inches. A 3 inches wide non-slip tape is placed along the perimeter of the bottom of the doormat.

Answer 1

The amount of tape to be saved is 15.43 inches.

Explanation:

Semicircle is half of a given circle, consisting of an arc and a diameter. Its perimeter is determined by adding the length of its arc to the diameter.

perimeter of a semicircle = length of its arc + its diameter

length of the arc = half of the circumference of its circle

circumference of a circle = 2
`\pi`r

half of the circumference of a circle =
`\pi`r

length of the arc of the semicircle = half of the circumference of a circle =
`\pi`r

The initial diameter of the semicircle = 30 inches

⇒ its radius, r = 15 inches

Amount of tape to be saved = Initial perimeter of the semicircle - perimeter of the inside of the semicircle

Initial perimeter of the semicircle =
`\pi`r + length of the diameter

=
`(22)/(7)` × 15 + 30

= 47.1429 + 30

= 77.1429 inches

Initial perimeter of the semicircle is 77.1429 inches.

The inside semicircle is formed, since a tape of width 3 inches has been placed along the perimeter of the initial doormat. Then,

diameter of the inside semicircle = 30 - 6

= 24 inches

⇒ its radius = 12 inches

perimeter of the inside semicircle =
`\pi`r + length of the diameter

=
`(22)/(7)` × 12 + 24

= 37.7143 + 24

= 61.7143 inches

perimeter of the inside semicircle is 61.7143 inches.

Amount of tape to be saved = 77.1429 - 61.7143

= 15.4286

The amount of tape to be saved is 15.43 inches.