You are placing non-slip tape along the perimeter of the bottom of a semicircle-shaped doormat. How much tape will you save appl
1 answer:
Answer:
The amount of tape to be saved is 15.43 inches.
Step-by-step 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 = 2r
half of the circumference of a circle = r
length of the arc of the semicircle = half of the circumference of a circle = 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 = r + length of the diameter
= × 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 = r + length of the diameter
= × 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.
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The rest of the question is the attached figure
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solution:
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As show in the attached figure
∠M = ∠R = 54.4°
∠N = ∠T = 71.2°
∠O = 180° - (∠M + ∠N) = 180° - (54.4°+71.2°) = 54.4°
∠S = 180° - (∠R + ∠T) = 180° - (54.4°+71.2°) = 54.4°
∠O = ∠S = 36°
∴ Δ MNO is similar to Δ RTS
So, the correct statement:
The triangles each have two given angle measures and one unknown angle measure.
