Question

You are placing the 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 a diameter of 30 inches. A 3 inches wide non-slip tape is placed along the perimeter of the bottom of the doormat.

You will save about...inches

Answer 1

The amount of tape saved by placing the non-slip tape along the inside semicircle of the doormat is 9.42 inches

Explanation:

Here we have;

Diameter of the semicircle shaped doormat = 30 in.

Width, w of the non-slip tape to be applied to the inside perimeter of the doormat = 3 in.

Therefore, length of the outer and inner arc formed by width of the non-slip tape is given as follows;

Outer perimeter = π×r = π × D/2 = π × 30/2 = 15·π in.

Inner perimeter = π × (D-2×w)/2 = π × (30 - 2×3)/2 = π × 24/2 = 12·π in.

Average length of arc = (15·π + 12·π)/2 = 13.5·π in.

Total length of non-slip tape to be applied to the inside perimeter of the doormat = Non-slip tape applied to the arc + Non-slip tape applied to the diameter line of the doormat

Total length of non-slip tape to be applied to the inside perimeter of the doormat = 13.5·π in. + 30 in. = 72.4 in.

Whereby the tape is applied on the edge of the doormat, we have;

Total length of tape = (π·30/2 + π×(30 + 6)/2)/2 + 30 =81.84 in.

Tape saved = 81.84 - 72.4 = 9.42 in